Chemical Bonding
March 24, 2017 | Author: Ankur Agarwal | Category: N/A
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chemical bonding...
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41
CHEMICAL BONDING
CHEMICAL BONDING Ionic Bond or Electrovalent Bond An ion is an atom or group of atoms which has acquired charge due to the loss or gain of one or more electrons. When an atom gains an electron to form a negative ion (anion), it will increase in size. On the other hand, when an atom loses an electron to give positive ion (cation), it will contract. The electron lost or gained is always from the outermost shell. When two atoms, one of which can lose one or more electrons to attain a noble gas configuration and the other can receive these electrons and thereby acquire a noble gas configuration, they are said to be bonded by an ionic bond. Since the loss and gain of electrons by atoms results in the formation of ions, ionic bond is formed when two ions interact with each other and are thus held together by electrostatic attraction. The formation of potassium chloride (KCl), is illustrated below.
K (1s 2 2s 2 2p 6 3s 2 3p 6 4s1 ) loses K (1s 2 2s 2 2p 6 3s 2 3p 6 ) 1 electron
gains Cl (1s 2 2s2 2p6 3s 2 3p5 ) Cl (1s 2 2s 2 2p 6 3s 2 3p6 ) 1 electron
(i) (ii)
(Ar configuration)
From the above illustrations, it is clear that the formation of an ionic compound is obviously related to the ease of formation of the cations and anions from the neutral atoms which depends on two main factors: Ionization energy: Lower the value of ionization energy of an atom, greater will be the ease of formation of the cation from it. Electron affinity: Higher the electron affinity of an atom, greater the ease of formation of the anion from it.
Lattice Energy When one mole of an ionic solid is formed from its constituent gaseous ions, the energy released is called the lattice energy. Energetics of Formation of Ionic Substances: The energy included in the formation of an ionic compound from its constituent elements may be considered as shown by the Born-Haber Cycle for the formation of one mole of sodium chloride from sodium and chlorine.
Na (s ) Sublimatio n Na ( g ) Na (g ) e S
I
1 on of e Cl 2 (g ) Dissociati Cl ( g ) Addition Cl (g ) 1 / 2 D * E A 2
Na (g ) Cl (g ) Crystal formation NaCl (S) U
Where
S = heat of sublimation of sodium metal I = ionization energy of sodium D = heat of dissociation of molecular chlorine Ea = electron affinity of chlorine, and U = lattice energy of sodium chloride The amount of heat liberated in the overall reaction is the heat of formation of sodium chloride. From 1 D – Ea – U 2 The most important of these energy terms are I, Ea and U, since these are considerably greater than the remaining terms S and D.
the above H = S + I +
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42
CHEMICAL BONDING
More the negative value of the heat of formation, greater would be the stability of the ionic compound produced. Thus on the basis of the above equation, formation of an ionic compound is favoured by a) low ionization energy (I) of the metal. b) high electron affinity (EA) of the other element. c) higher lattice energy (U) of the resulting compound. Formation of Ions with Higher Charges: Formation of a cation with unit positive charge is easy if the first ionization energy is low as in the case of alkali metals. Alkaline earth metals ionizes in two successive steps. + – Mg Mg + e 2+ – Mg+ Mg + e But energy needed to ionize alkaline earth metals are higher than alkali metals. However, bipositive ions like Mg2+, Ca2+, Sr2+ and Ba2+ are quite common. Formation of a tripositive ion like Al3+ requires much more energy (= 5138 kJ) which is not available ordinarily. Successive ionization energies of aluminium are: E1 E1 = 577kJ Al Al e
E2 Al Al2 e
E2 = 1816kJ
E3 = 2745kJ It is on this account that most of aluminium compounds are covalent. In solution, however, aluminiumis known to give hydrated ions [Al.6H2O]3+. This is possible because of the high heat of hydration of Al3+. The energy liberated during hydration of ions is sufficient for ionization. Similarly, anions with unit negative charge (e.g. Cl–1, Br–, I–) are very common. This is because the electron affinity of these atoms is positive and quite high. Formation of anions carrying two units of negative charge (e.g. S2–, O2–) is not so easy as their electron affinities are negative i.e., energy is needed to add second electron. Formation of anions carrying three units of negative charge (e.g. N3–, P3–) is almost rare. E3 Al2 Al3 e
Characteristics of Electrovalent Compounds Melting and Boiling Point: Due to the powerful electrostatic force between the ions in a crystal of an electrovalent compound considerable energy is needed to overcome these forces and break down the crystal lattice. Hence such compounds possess high melting and boiling points. Electrical Conductivity: When an electrovalent compound is molten or dissolved in a solvent of high dielectric constant e.g., water, the binding forces in the crystal lattice disappear and the component ions become mobile. Under the influence of applied electrical field, the ions get charged and thus act as charge carrier of the current. Hence their melts or solutions conduct electricity. Solubility: Ionic compounds are soluble in polar solvents like water because of molecules of the polar solvent interact strongly with the ions of the crystal and the solvation energy is sufficient to overcome the attraction between the ions in the crystal lattice. Dissolution is also favoured by the high dielectric constant of the solvents such as water, since this weakens the interionic attractions in the resulting solutions. Non-polar solvents like benzene and carbon tetrachloride do not solvate the ions as their dielectric constants are low. Ionic compounds are, therefore insoluble in non-polar solvents. Ionic compounds like sulphates and phosphates of barium and strontium are insoluble in water (because lattice energy is greater than hydration energy). This can be attributed to the high lattice energies of these compounds due to polyvalent nature of both the cation and the anion. In these cases, hydration of ions fails to liberate sufficient energy to offset the lattice energy.
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43
CHEMICAL BONDING
Covalent Bond (By Mutual Sharing of Electrons) The covalent bond is formed when two atoms achieve stability by the sharing of an electron pair, each contributing one electron to the electron pair. The arrangement of electrons in a covalent molecule is often shown by a Lewis structure in which only valency shells (outer shells) are depicted. For sake of clarity, the electrons on different atoms are denoted by dots and crosses. Polarity of Bonds: A covalent bond is set up by sharing of electrons between two atoms. It is further classified as polar or non-polar depending upon the fact whether the electron pair is shared unequally between the atoms or shared equally. For example, the covalent bonds in H2 and Cl2 are called non-polar as the electron pair is equally shared between the two atoms. H
H
Cl Cl
Hydrogen molecule
Chlorine molecule
H
(Both formed by equal sharing of electrons between the atoms, i.e., by non-polar bonds)
F d
In the case of hydrogen fluoride the bond is polar as the electron pair is unequally shared. Fluorine has a greater attraction for electrons or has higher electronegativity than hydrogen and the shared pair of electrons is nearer to the fluorine atom than hydrogen atom. The hydrogen end of the molecule, therefore, appears positive with respect to fluorine. Bond polarities affect both physical and chemical properties of compounds containing polar bond. The polarity of a bond determines the kind of reaction that can take place at that bond and even affects the reactivity at nearby bonds. The polarity of bonds can lead to polarity of molecules and affect melting point, boiling point and solubility. Dipole Moment: It is vector quantity and is defined as the product of the magnitude of charge on any of the atom and the distance between the atoms. It is represented by . Magnitude of dipole moment | | ( chargeqin esu) ( Dis tanrce in ) The unit = 10–18 (esu) cm (D) is used in practice. In SI units charge q is measured in coulombs (C) and the distance, r in metre, m 1C = 2.998 × 109 esu and 1 m = 102 cm 1 Cm = 2.998 × 109 × 102 = 2.998 × 1011 (esu) cm Therefore in SI system, the unit of dipole moment is coulomb metre
2.998 1011 2.998 10 29 D 1 Cm = 18 10
1 3.336 10 30 Cm 29 2,998 10 Dipole moment is a vector quantity and is often indicated by an arrow parallel to the line joining the point of charge and pointing towards the negative end e.g., H F . or
1D
Experimental dipole moment
% Ionic character of a covalent bond = Theoretical dipole moment assuming 100% ionic character 100
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44
CHEMICAL BONDING
Illustration 1: The dipole moment of KCl is 3.336 × 10–29 Cm. The interatomic distance K+ and Cl– ion in KCl is 260 pm. Calculate the dipole moments of KCl, if there were opposite charges of the fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl. Solution:
From the given data q = 1.602 × 10–19C r = 260 pm = 260 × 10–12 m = 2.6 × 10–10 m Magnitude of dipole moment for 100% ionic character || = qr = (1.602 × 10–19) (2.6 × 10–10) = 4.165 × 10–29 Cm Actual dipole moment = 3.336 × 10–29 Cm
3.336 10 29 100 = 80.1% % of ionic bond = 4.165 10 29 The bond is 80.1% ionic.
i) ii)
In general a polar bond is established between two atoms of different radii and different electronegativities while positive centres (nuclei) of different magnitudes combine to share an electron pair. Greater the values of the dipole moment, greater is the polarity of the bond. The following points may be borne in mind regarding dipole moments: In case a molecule contains two or more polar bonds, its dipole moment is obtained by the vectorial addition of the dipole moments of the constituent bonds. A symmetrical molecule is non-polar even though it contains polar bonds. For example, carbon dioxide, methane and carbon tetrachloride, being symmetrical molecules, have zero dipole moments. Dipole moment of methyl chloride is a vectorial addition of dipole moments of three C – H bonds and one C – Cl bond.
H
1.75d
F
H
Cl
Cl
C
C
C
H
H H
0D Hydrogen fluoride
Methane
Cl
Cl Cl
H
H H
0D
1.86D
Carbon tetrachloride
Methyl chloride
Dipole moments of some molecules
Dipole moment gives valuable information about the structure of molecules. For example, carbon dioxide is assigned a linear structure since its dipole moment is zero. We have seen that in a polar covalent bond between two atoms (say A and B), there is a partial separation of charge. This bond is, therefore, said to have a partial ionic character. Greater the difference of electronegativity between A and B, greater is the degree of ionic character (or polarity measured by dipole moment of AB) of the bond. Pauling gave a fairly accurate rule by which the nature of the bond can be predicted. According to this rule, “If the difference on the electronegativity scale between the two atoms is 1.9, the bond is 50% ionic in character. When the difference is greater than 1.9, the bond is correspondingly more ionic”. For example, when the electro negativity difference is 0.8, 1.2, 2.2 and 2.6, the corresponding partial ionic character is 12%, 25%, 61% and 74% respectively.
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45
CHEMICAL BONDING
Characteristics of Covalent Compounds Melting Point and Boiling Point: In covalent compounds, except those consisting of giant molecules, the molecules are less powerfully attracted to each other, as a result of which their melting points and boiling points are relatively low compared to ionic compounds, e.g., SiCl 4 (b.p. 33K) and NaCl (b.p. 1713K) (Covalent compound)
(ionic compound)
Conductivity: Covalent substances (whether of the “molecular lattice” or “giant molecule” type) do not conduct electricity in the fused state since there are no free electrons or ions to carry the current. However, substances like graphite which consists of separate layers conduct electricity because the electrons have a passage in between the two flat layers. Solubility: The characteristic solubility of covalent compounds in non-polar solvents such as benzene and carbon tetrachloride can be described to the similar covalent nature of the molecules of solute and solvent (i.e., like dissolves like). Covalent compounds in solution react more slowly as compared with the ionic compounds which react instantaneously in solution. The solubility of covalent compounds is, however, very much dependent upon the size of the molecule. Thus covalent substances having giant molecules are insoluble in virtually all solvents due to the big size of the molecule unit.
Fajan’s Rules
i)
When two oppositely charged ions approach each other closely, the positively charged cation attracts the outermost electrons of the anion and repel its positively charged nucleus. This results in the distortion or polarization of the anion followed by some sharing of electrons between the two ions, i.e., the bond becomes partly covalent in character. Charge on Either of the ions: As the charge on the cation increases, its tendency to polarize the anion increases. This brings more and more covalent nature in the electrovalent compound. Whereas with the increasing charge of anion, its ability to get polarized, by the cation, also increases. For example, in the case of NaCl, MgCl2 and AlCl3 the polarization increases, thereby covalent character becomes more and more as the charge on the cation increases. Similarly, lead forms two chlorides PbCl2 and PbCl4 having charges +2 and +4 respectively. PbCl4 shows covalent nature. Similarly among NaCl, Na2S, Na3P, the charge of the anions are increasing, therefore the increasing order of covalent character. NaCl < Na2S < Na3P
ii)
Size of the cation: Polarisation of the anion increases as the size of the cation decreases i.e., the electrovalent compounds having smaller cations show more of the covalent nature. For example, in the case of halides of alkaline earth metals, the covalent character decreases as we move down the group. Hence melting point increases in the order of BeCl2 < MgCl2 < CaCl2 < SrCl2 < BaCl2
iii)
Size of anion: The larger the size of the anion, more easily it will be polarized by the cation i.e., as the size of the anion increases for a given cation, the covalent character increases. For example, in the case of halides of calcium, the covalent character increases from F– anion to I– anion i.e. CaF CaCl CaBr CaI increasing covalent character
2 2 2 2
Similarly, in case of trihalides of aluminium, the covalent character increases with increase in size of halide anion i.e. AlF3
AlCl3
AlBr3
AlI3
Covalent character increases as the size of the halide ion increases
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CHEMICAL BONDING
Nature of the cation: Cations with 18 electrons (s2p6d10) in outermost shell polarize an anion more strongly than cations of 8 electrons (s2p6) type. The d electrons of the 18 electron shell screen the nuclear charge of the cation less effectively than the s and p electrons of the 18-electron shell. Hence the 18-electron cations behave as if they had a greater charge. Copper (I) and Silver (I) halides are more covalent in nature compared with the corresponding sodium and potassium halides although charge on the ions is the same and the sizes of the corresponding ions are similar. This illustrates the effect of 18-electron configuration of Cu+ (3s2, p6, d10) and Ag+ (4s2, p6, d10) ions. Illustration 4: The decomposition temperature of Li2CO3 is less than that of Na2CO3. Explain. Solution: As Li+ ion is smaller than Na+ ion, thus small cation (Li+) will favour more covalent character in Li2CO3 and hence it has lower decomposition temperature than that of Na2CO3. iv)
BRAIN TEASER 1:
SnCl2 is solid whereas SnCl4 is liquid, why?
Hydrogen Bonding In 1920, Latimer and Rodebush introduced the idea of “hydrogen bond” to explain the nature of association in liquid state of substance like water, hydrogen fluoride, ammonia, formic acid etc. In a hydrogen compound, when hydrogen is bonded to highly electronegative atom (such as F, O, N) by a covalent bond, the electron pair is attracted towards electronegative atom so strongly that a dipole results i.e., one end carries a positive charge (H-end) and other end carries a negative charge (X-end). H
X
X H
or
Electro-negative atom
If a number of such molecules are brought nearer to each other, the positive end of one molecule and negative end of the other molecule will attract each other and weak electrostatic force will develop. Thus, these molecules will associate together to form a cluster of molecules.
X H X H X H X H X H The attractive force that binds hydrogen atom of one molecule with electronegative atom of the other molecule of the same or different substance is known as hydrogen bond.
Hydrogen bonding is of two types: a) Intermolecular hydrogen bonding: This type of bonding results between the positive and negative ends of different molecules of the same or different substances. Example H
i)
Ammonia
N
H
H
H N H
H N H
H N H
H N H
H
H
H
H
H
H O H
H
ii)
Water
O
O
iii)
Acetic acid
H O H
O
H3C
CH3 O
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H
H O H
H
O
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47
CHEMICAL BONDING
This type of hydrogen bonding increases the boiling point of the compound and also its solubility in water. The increase in boiling point is due to association of several molecules of the compound. b)
Intramolecular hydrogen bonding: This type of bonding results between hydrogen and an electronegative element both present in the same molecule. This type of bonding is generally present in organic compounds. Examples are o-nitro-phenol, o-hydroxy benzoic acid, etc. O N
H
O
-
O
H
-
O
O o-Nitrophenol
O o-Hydroxy benzoic acid
This type of bonding decreases the boiling point of the compound. The solubility of the compound also decreases. Hence compound becomes more volatile. Properties Explained by Hydrogen Bonding a) Strength of certain acids and bases can be explained on the basis of hydrogen bonding. b) Solubility: An organic substance is said to be insoluble in water if it does not form hydrogen bonding with water. The organic compound like alkanes, alkenes, ethers, etc., are insoluble in water as they do not form hydrogen bonding with water, while alcohols and acids are soluble because they readily form hydrogen bonds with water. i) Melting and boiling points of hydrides of N, O and F. If the melting points and boiling points of the hydrides of the elements of IVA, VA, VIA and VIIA groups are plotted against the molecular weights of these hydrides, we shall get the plots as shown in figure (a) and (b). From these plots it may be seen that although in case of SbH3, AsH3, PH3 (VA group elements hydrides), H2Te, H2Se, H2S (VI A group elements hydrides) and HI, HBr, HCl (VIII group elements hydrides) there is a progressive decrease in their mp’s and b.p’s with the decrease in their molecular weights, the mp’s and b.p’s of NH3, H2O and HF hydrides suddenly increase with a further decrease of their molecular weights. The sudden increase in mp’s and bp’s in these hydrides is due to the inter-molecular H-bonding in between H and F in case of HF, in between H and O in case of H2O and in between H and N in case of NH3 respectively. The existence of H-bonding in these molecules gives polymerized molecules (NH3)n. Thus mp’s and bp’s of these molecules are suddenly raised. Having no power to form H-bonds, the simple carbon family hydrides (SnH4, GeH4, SiH4 and CH4) show a decrease in their bp’s and mp’s with the decrease in their molecular weights.
100
100 H2O
H2O 0 H 2Se
NH 3
H 2 Se
H2S
HBr -100
HI
VIA VIIIA VA
SbH 3
HCl
HF
AsH 3 PH3
SnH4
Boiling points (°C) increasing
Melting points (°C) increasing
HF H 2Se
NH3
H2S
SbH3
HI HBr -100
GeH 4
PH3
VA VIIIA
SnH4
IVA
SnH4
IVA CH 4
SnH 4
-200
H 2 Se
HCl
GeH 4 CH 4
VIA
0
-200
Molecular weight increasing (a)
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Molecular weight increasing (b)
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ii)
CHEMICAL BONDING
Ice has less density than water. The explanation of this fact is as follows: In the crystal structure of ice,
the O-atom is surrounded by four H-atoms. Two H-atoms are linked to O-atom by covalent bonds as shown (by normal covalent bond) and the remaining two H-atoms are linked to O-atom by two Hbonds shown by dotted lines. Thus in ice every water molecule is associated with four other water molecules by H-bonding in a tetrahedral fashion. Ice has an open cage like structure with a large empty space due to the existence of H-bonds. As ice melts at 0°C, a number of H-bonds are broken down and the space between water molecules decreases so that water molecules move closer together. The density of water increases, from 0° to 4°C, and at 4°C it is maximum. Above 4°C the increase in kinetic energy of the molecules is sufficient to cause the molecules to begin to disperse and the result is that the density decrease with increasing temperature.
H
H
2.76Å
1.80Å -
H
0.96Å 0.96Å
water molecule H
H
Open cage-like tetrahedral crystal structure of ice. Circles indicate oxygen atoms. Bonds represented by solid line are normal covalent bonds while those represented by dotted lines are hydrogen bonds.
H H
H
H H
BRAIN TEASER 2:
Although HF forms stronger hydrogen bond than H2O, Hv of H2O is greater than that of HF why?
Coordinate Bond It is a special type of covalent bond in which both the shared electrons are contributed by one atom only. It may be defined as “a covalent bond in which both electrons of the shared pair are contributed by one of the two atoms”. Such a bond is also called as dative bond. A coordinate or a dative bond is established between two such atoms, one of which has a complete octet and possesses a pair of valence electrons while the other is short of a pair of electrons. xx
A
+ B xx xx
A
xx Bx xx x
or
A
B
This bond is represented by an arrow () pointing towards acceptor atom. The atom which contributes electron pair is called the donor while the atom which accepts it is called acceptor. The compound consisting of the coordinate bond is termed coordinate compound. Some examples of coordinate bond formation are given below:
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CHEMICAL BONDING
i)
Formation of ammonium ion: Hydrogen ion (H+) has no electrons and thus accepts a lone pair donated by nitrogen. H
H H
N
+ H
+
H
H
H
H
ii)
N
Formation of CO: Carbon has four valency electrons and oxygen has six. They combine to form two double bond and a coordinate bond as to achieve their octet completed. C
xx + xxOxx
C
Oxx
Acceptor Donor
Characteristics of Coordinate Compounds: The properties of coordinate compounds are intermediate between the properties of electrovalent compounds and covalent compounds. The main properties are described below: i) Melting and Boiling Points: Their melting and boiling points are higher than purely covalent compounds and lower than ionic compounds. ii) Solubility: These are sparingly soluble in polar solvents like water but readily soluble in non-polar (organic) solvents. iii) Conductivity: Like covalent compounds, these are also bad conductors of electricity. The solutions or fused mass do not allow the passage of electricity.
Valence Shell Electron Pair Repulsion (VSEPR) Theory
1. 2.
3. 4.
In 1957 Gillespie and Nyhom gave this theory to predict and explain molecular shapes and bond angles more exactly. The theory was developed extensively by Gillespie as the Valence Shell Electron Pair Repulsion (VSEPR) theory. This may be summarized as: The shape of the molecule is determined by repulsions between all of the electron pairs present in the valence shell. A lone pair of electrons takes up more space round the central atom than a bond pair, since the lone pair is attracted to one nucleus whilst the bond pair is shared by two nuclei. It follows that repulsion between two lone pairs is greater than repulsion between a lone pair and a bond pair, which in turn is greater than the repulsion between two bond pairs. Thus the presence of lone pairs on the central atom causes slight distortion of the bond angles from the ideal shape. If the angle between a lone pair, the central atom and a bond pair is increased, it follows that the actual bond angles between the atoms must be decreased. The order of repulsion between lone pairs and bond pairs of electrons follows the order as: Lone pair - lone pair repulsion > lone pair – bond pair repulsion > bond pair – bond pair repulsion. The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms. Double bonds cause more repulsion than single bonds, and triple bonds cause more repulsion than a double bond.
Effect of Lone Pairs: Molecules with four electron pairs in their outer shell are based on a tetrahedron. In CH4 there are four bonding pairs of electrons in the outer shell of the C atom, and the structure is a regular tetrahedron with bond angle H – C – H of 109°28’. In NH3 and N atom has four electron pairs in the outer shell, made up of three bond pairs and one lone pair. Because of the lone pair, the bond angle H – N – H is reduced from the theoretical tetrahedral angle of 109°28’ to 107°28’. In H2O the O atom has four electron pairs in the outer shell. The shape of the H2O molecule is based on a tetrahedron with two corners occupied by bond pairs and the other two corners occupied by lone pairs. The presence of two lone pairs reduces the bond angle further to 104°27’. NARAYANA II T ACAD EMY- C P 3 ,
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CHEMICAL BONDING
In a similar way, SF6 has six bond pairs in the outer shell and is a regular octahedron with bond angles of exactly 90°. In BrF5, the Br also has six outer pairs of electrons, made up of five bond pairs and one lone pair. The lone pair reduces the bond angles to 84°30’. Whilst it might be expected that two lone pairs would distort the bond angles in an octahedral as in XeF4 but it is not so. Actual bond angle is 90°, reason being that the lone pairs are trans to each other in the octahedron, and hence the atoms have a regular square planar arrangement. Molecules with five pairs of electrons are all based on a trigonal bipyramid. Lone pairs distort the structures as before. The lone pairs always occupy the equatorial positions (in an triangle), rather than the axial positions (up and down).Thus in I 3 ion, the central I atom has five electron pairs in the outer shell, made of two bond pairs and three lone pairs. The lone pairs occupy all three equatorial positions and the three atoms occupy the top, middle, and bottom positions in the trigonal bipyramid, thus giving a linear arrangement with a bond angle of exactly 180°. Effect of Electronegativity: NF3 and NH3 both have structures based on a tetrahedron with one corner occupied by a lone pair. The high electronegativity of F push the bonding electrons further away from N than in NH3. Hence the lone pair in NF3 causes a greater distortion from tetrahedral and gives a F – N – F bond angle of 102°30’, compared with 107°48’ in NH3. The same effect is found in H2O (bond angle 104°27’) and F2O (bond angle 102°). The effects of bonding and lone pairs on bond angles
Orbitals on BeCl2 BF3 CH4 NH3 NF3 H 2O F2O PCl5 SF4 ClF3 XeF2 SF6 BrF5 XeF4
Shape 2 3 4 4 4 4 4 5 5 5 5 6 6 6
Number of bond pairs Linear 2 Plane triangle 3 Tetrahedral 4 Pyramidal 3 Pyramidal 3 Bent (V-shape) 2 Bent (V-shape) 2 Trigonal bipyramid 5 Trigonal bipyramid 4 T-shape 3 Linear 2 Octahedral 6 Square pyramidal 5 Square planar 4
Number of lone pairs 0 0 0 1 1 2 2 0 1 2 3 0 1 2
central atom Bond angle 180° 120° 109°28 107°48 102°30 104°27 102° 120° & 90° 101°36 & 86°33 87°40 180° 90° 84°30 90°
Some examples using the VSEPR Theory Phosphorus pentachloride PCl 5 : Gaseous PCl5 is covalent. (The electronic structure P is 1s22s22p63s23p3). All five outer electrons are used to form bonds to the five Cl atoms. In the PCl5 molecule the valence shell of the P atom contains five electron pairs: hence the structure is a trigonal bipyramid. There are no lone pairs, so the structure is not distorted. However, a trigonal bipyramid is not a completely regular structure, since some bond angels are 90° and others 120°. Symmetrical structures are usually more stable than asymmetrical ones.
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CHEMICAL BONDING
Note: Thus PCl5 is highly reactive, and in the solid state it splits into [PCl4]+ and [PCl6]– ions, which have tetrahedral and octahedral structures respectively. Cl
Cl
Cl P Cl
Cl Structure of PCl5 molecule
Chlorine trifluoride ClF3: The chlorine atom is at the centre of the molecule and determines its shape.
The electronic configuration of Cl is 1s22s22p63s23p5. Three electrons form bonds to F, and four electrons do not take part in bonding. Thus in ClF3, the Cl atom has five electron pairs in the outer shell, hence the structure is a trigonal bipyramid. There are three bond pairs and two lone pairs. It was noted previously that a trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore follows that all the corners are not equivalent. Lone pair occupy two of the corners, and F atoms occupy the other three corners. Three different arrangements are theoretically possible, as shown in figure below. The most stable structure will be the one of lowest energy, that is the one with the minimum repulsion between the five orbitals. The great repulsion occurs between two lone pairs. Lone pair bond pair repulsions are next strongest, and bond pair-bond pair repulsions are weakest. Groups at 90° repel each other strongly, whilst groups 120° apart repel each other much less.
F
F Cl F
I
F
F
Cl
Cl
F
F
F F
II
III
Chlorine trifluoride molecule
Structure I is the most symmetrical, but has six 90° repulsions between lone pairs and atoms. Structure II has one 90° repulsion between two lone pairs, plus three 90° repulsions between lone pairs and atoms. These factors indicate that structure III is the most probable. The observed bond angles are 80°40, which is close to the theoretical 90°. This confirms that the correct structure is III, and the slight distortion from 90° is caused by the presence of the two lone pairs. As a general rule, if lone pairs occur in a trigonal bipyramid they will be located in the equatorial position (round the middle) rather than the axial positions (top and bottom), since this arrangement minimizes repulsive forces.
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CHEMICAL BONDING
Sulphur hexafluoride SF6: The electronic structure of S is 1s22s22p63s23p6. All six of the outer electrons
are used to form bonds with the F atoms. Thus in SF6, the S has six electron pairs in the outer shell: hence the structure is octahedral. There are no lone pairs, so the structure is completely regular with bond angles of 90°. F F
F S
F
F F
Valence Bond Theory
1. 2. 3. 4.
This theory was proposed by Linus Pauling, who was awarded the Noble Prize for Chemistry 1954. Atoms with unpaired electrons tend to combine with other atoms which also have unpaired electrons. In this way the unpaired electrons are paired up, and the atoms involved, all attain a stable electronic arrangement. This is usually a full shell of electrons(i.e. a noble gas configuration). Two electrons shared between two atoms constitute a bond. The number of bonds formed by an atom is usually the same as the number of unpaired electrons in the ground state, i.e. the lowest energy state. However, in some cases the atom may form more bonds than this. This occurs by excitation of the atom (i.e. providing it with energy) when electrons which were paired in the ground state are unpaired and promoted into suitable empty orbitals. This increases the number of unpaired electrons, and hence it increases number of bond which can be formed. A covalent bond results from the pairing of electrons (one from each atom). The spins of the two electrons must be opposite (antiparallel) because of the Pauli exclusion principle that no two electrons in one atom can have all four quantum numbers the same. In HF, H has a singly occupied s-orbital that overlaps with a singly filled 2p orbital on F. In H2O, the O atom has two singly filled 2p orbitals, each of which overlaps with a single occupied sorbital from two H atoms. In NH3, there are three singly occupied p orbitals on N which overlap with s orbitals from three H atoms. In CH4, the C atom in its ground state has the electronic configuration 1s2, 2s2, 2p1x , 2p1y and only has two unpaired electrons, and so can form only two bonds. If the C atom is excited, then the 2s electrons may be unpaired, giving 1s2, 2s1, 2p1x , 2p1y , 2p1x . There are now four unpaired electrons which overlap with singly occupied s orbitals on four H atoms. 2p 2s
2px
2py 2p z
Electronic structure of carbon atom - groun state 1s Carbon atom - excited state Carbon atom having gained four electrons from H atoms in CH4 molecule.
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2s
2p
sp 3 hybridisation
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CHEMICAL BONDING
CH4 molecule uses its three p-orbitals px, py and pz, which are mutually at right angles to each other, and the s orbital is spherically symmetrical. Hence they form tetrahedral structure. CH4 H – C – H = 109°28
Sigma and Pi Bonds ( and Bonds) 1.
A covalent bond is formed by the overlapping of atomic orbitals. Covalent bonds formed are of two types depending upon the way the orbitals overlap each other. Sigma bond ( bond): The bond formed by the overlapping of two half filled atomic orbitals along their axis is known as sigma bond. bond is a strong bond because overlapping in it takes place to large extent. The hybrid orbitals always from bond. a) s – s overlapping Molecular axis b)
s – p overlapping
c)
p – p overlapping
pz
2.
head on overlap
pz
p-p overlap
M.O.
Pi bond ( bond): The bond formed by the lateral overlapping of half filled atomic orbitals is known as pi bond. The sidewise overlapping takes place to less extent. Therefore, bond formed is a weak bond. bond overlapping takes place only at the sides of two lobes. A bond is formed when a bond already exists between the combining atoms.
p
p
p-p overlapping
M.O.
Example: In A – B molecule the bond formed is bond. B, molecule there are one and one bonds In A
In A
B, molecule there are one and two bonds
Thus, all the single bonds are bonds. Double bond has one and one bond. Triple bond has one and two bonds.
Hybridisation It is the mathematically fabricated concept that is introduced to explain the geometry/shapes of the covalent molecules of polyatomic ions containing covalent bonds. It is a process of intermixing of atomic orbitals with small difference in energy and belonging to the same atom, at the time of bonding so as to give another set of orbitals with equivalent shapes and energies. NARAYANA II T ACAD EMY- C P 3 ,
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CHEMICAL BONDING
sp3 Hybridisation: In ground state, the electronic configuration of carbon is 1s2, 2s2, 2p2. It is proposed that from 2s orbital, being quite near in energy to 2p orbitals, one electron may be promoted to the vacant 2pz orbital thus obtaining the excited atom. At this stage the carbon atom undoubtedly has four halffilled orbitals and can form four bonds. In the excited atom, all the four valence shell orbitals may mix up to give four identical sp3 hybrid orbitals. Each of these four sp3 orbital possesses one electron and overlaps with 1s orbitals of four H atoms thus forming four equivalent bonds in methane molecule. Due to the tetrahedral disposition of sp3 hybrid orbitals, the orbital are inclined at an angle of 109° 28’. Thus all the H– C– H angles are equal to 109° 28’
Energy
2p
2p
Promotion
sp3
of an electron
Hybridisation
2s
2s Ground State
Hybridised State
Excited State
H H 109.5° C
H
H
H
H H
H
Shape and formation of methane molecule
sp2 Hybridisation: When three out of the four valence obritals of carbon atom in excited state hybridize, we have three sp2 hybrid orbitals lying in a plane and inclined at an angle of 120°. If 2s and 2p, orbitals of the excited carbon atom are hybridized, the new orbitals lie in the xy plane while the fourth pure 2pz orbitals lies at right angles to the hybridized orbitals with its two lobes disposed above and below the plane of hybrid orbitals. Two such carbon atoms are involved in the formation of alkenes (compounds having double bonds). In the formation of ethene two carbon atoms (in sp2 hybridization state) form one sigma bond by ‘head-on’ overlap of two sp2 orbitals contributed one each by the two atoms. The remaining two sp2 orbitals of each carbon form bonds with H atoms. The unhybridized 2p, orbitals of the two carbon atoms undergo a side-wise overlap forming a bond. Thus the carbon to carbon double bond in ethene is made of one bond and one bond. Since the energy of a bond is less than that of a bond, the two bonds constituting the ethene molecule are not identical in strength. The molecule is a planar one. Pure p-orbital
Energy
2p
Promotion
2p
of an electron
2s Ground State
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sp2 Hybridisation
sp2 hybrid orbitals
2s Excited State
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CHEMICAL BONDING
pz
pz
H
sp 2 sp 2
H
H C
C sp 2
H
sp 2
sp 2 2 sp
H
H
H
H
Orbital model of ethane molecule
Different types of hybridization depend upon the type of atomic orbitals which are used for intermixing. Types of hybridization and spatial orientation of hybrid orbitals: The geometry and shapes of various species on the basis of VSEPR theory along with hybrid state of central atom is given below in tabular form. Types of atomic orbitals used 1.
one s + one p-orbital
Hybridisation Orientation of Examples hybrid orbitals sp Linear BeF2 , BeCl 2 , C 2 H 2 , HgCl 2
2.
one s + two p-orbitals
sp 2
Trigonal planar BF3 , C 2 H 4 , NO 3 , CO 32
3.
one s + three p-orbitals
sp 3
Tetrahedral CH 4 , CCl 4 , SiF4 , NH 4 , SO 24 , ClO 4
4.
one s + three p + d
sp 3 d
Trigonal bipyramidal
PF5 , PCl 5
5.
one s + three p + two d
sp 3 d 2
Octahedral
SF5 , [CrF6 ]3 , IF5
6.
one s + three p+three d
sp 3 d 3
Pentagonal Bipyramidal
7.
One d + one s + two p
dsp 2
Square planar
IF7
Only in complexes like [ Ni(CN) 4 ) 2 , [ PtCl 4 ] 2 etc. Note: i) ii) iii) iv) v) vi)
Orbitals participating in hybridization must have only small difference in their energies. Both half-filled and completely filled orbitals can get involved in hybridization. The number of hybrid orbitals is equal to the number of orbitals participating in hybridization. Hybrid orbital form more stronger bonds than pure atomic orbitals. Same atom can assume different hybrid states under different situations. Hybrid orbitals form sigma bonds.
Method of predicting the Hybrid state of the central atom in covalent molecules of polyatomic ions: The hybrid state of the central atom in similar covalent molecule or polyatomic ion can be predicted by using the generalized formula as described below : Simple Molecule Polyatomic Anion Polyatomic Cation 1 X [V G] 2
1 X [V G a ] 2
1 X [ V G c] 2
In the above formulae, V = Number of monovalent atoms or groups attached to the central atom G = Number of outer shell electrons in ground state of the central atom a = Magnitude of charge on anion c = Magnitude of charge on cation NARAYANA II T ACAD EMY- C P 3 ,
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CHEMICAL BONDING
Calculate the value of X and decide the hybrid state of central atom as follows : X 2 3 4 5 6 7 Hybrid state PF5
sp 2
sp
sp 3 d
sp 3 d 2 sp 3 d 3
ClO 4
COCl 2 NH 4
X 12 [5 5]
X 12 [2 4]
=5
=4
=3
sp 3
X 12 [4 5 1] X 12 [0 7 1]
=4
Hybrid state of P is sp 3 d
Hybrid state of C is sp 2
Hybrid state of N is sp 3
Hybrid state of Cl is sp 3
NO 3
IF5
CO 2
XeF4
X 12 [0 5 1] X 12 [5 7]
=3
=6
sp 2
sp 3 d 2 sp
PCl 6 PH3
=2 SF3
X 12 [0 4]
X 12 [ 4 8]
=6 sp 3 d 2
SF4
X 12 [6 5 1] X 12 [3 5]
X 12 [3 6 1] X 12 [4 6]
=6 =4 Hybrid state
=4 =5 Hybrid state
Hybrid state
Hybrid state
sp 3 d 2
sp 3
sp 3
sp 3 d
Molecular Orbital Theory Why He2 molecule does not exist and why O2 is paramagnetic? These questions cannot be explained by valence bond theory. In 1932 F. Hund and R.S. Mulliken put forward a theory known as Molecular Orbital Theory to explain above questions and many others. According to this theory, as the electrons of an atom are present in various atomic orbitals, electrons of a molecule are present in various molecular orbitals. Molecular orbitals are formed by the combination of atomic orbitals of comparable energy and proportional symmetry. While an electron in atomic orbital is influenced by one nucleus, in a molecular orbital, it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric. The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital (BMO) whereas other is anti-bonding molecular orbital (ABMO). BMO has lower energy and hence greater stability than the corresponding ABMO. First BMO are filled, then ABMO starts filling because BMO has lower energy than that of ABMO. Molecular orbitals like the atomic orbitals are filled in accordance with the Aufbau Principle obeying the Pauli’s Principle and the Hund’s rule. Order of energy of various molecular orbitals is as follows: For O2 and higher molecules 1s, *1s, 2s, *2s, 2px, [2py = 2pz], [*2py = *2pz], *2px For N2 and lower molecules 1s, *1s, 2s, *2s, [2py = 2pz], 2px, [*2py = *2pz], *2px
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CHEMICAL BONDING
Bond order: It may be defined as the half the difference between the number of electrons present in the bonding orbitals and the anti-bonding orbitals i.e. No. of electrons in BMO - No. of electrons in ABMO 2 A positive bonding order suggest a stable molecule while a negative bond order or zero bond order suggest an unstable molecule. Magnetic Behaviour: If all the molecular orbitals in species are spin paired, the substance is diamagnetic. However, if one or more molecular orbitals are singly occupied it is paramagnetic.
Bond order (B.O.) =
Illustration 2 Arrange the species O2, O2–, O22– and O2+ in the decreasing order of bond order and stability and also indicate their magnetic properties. Solution: The molecular orbital configuration of O2, O2–, O22– and O22+ are as follows: O2 = 1s2, *1s2, 2s2, *2s2, 2px2, 2py2, 2pz2, *2py1 = *2pz1 10 - 6 2 , No. of unpaired electrons = 2 2 paramagnetic O2– = 1s2, *1s2, 2s2, *2s2, 2px2, 2py2, 2pz2, *2py2 = *2pz1
Bond order =
10 - 5 2.5 , No. of unpaired electrons = 1 2 paramagnetic
Bond order =
O22– = 1s2, *1s2, 2s2, *2s2, 2px2, 2py2, 2pz2, *2py2 = *2pz2 Bond order =
10 - 8 1 , No. of unpaired electrons = 0 2
diamagnetic O2+ = 1s2, *1s2, 2s2, *2s2, 2px2, 2py2, 2pz2, *2py1 = *2pz0 10 - 5 2.5 , No. of unpaired electrons = 1 2 paramagnetic Now as the bond order decreases in the order O2+ O2 O2– O22– So, same will be the stability order of the above species because stability is directionally proportional to bond order. Note: Bond length is inversely proportional to bond order.
Bond order =
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CHEMICAL BONDING
PRAYAS - I Q.1 Q.2
Which of the following has zero dipole moment ? (A) NH3 (B) H2O (C) BCl3
(D) SO2
Which of the following has maximum bond energy ? (A) H – O – H (B) H – S – H (C) H – Te – H
(D) H – Se – H
Q.3
Which of the following statements is false ? (A) there will be 25% s character in hybrid orbitals if the angle between two hybrid orbital is 105° (B) in HClO4. HClO3 and HClO2 . number of valance electrons will be 32.26 and 20. respectively (C) bond length is inversely proportional to bond order (D) dipole moment of CH2Cl2 is zero
Q.4
The bond present in CuSO4.5H2O will be (A) ionic bond and hydrogen bond (C) coordinate bond
(B) covalent bond (D) all of the above
Ionic compound is (A) H2O
(C) Csl
Q.5 Q.6
(B) HCl
(D) NH3
If A (s) A+ (g) + e ................. 600 KJ mol–1/2 1 B (g) + e B– (g) .......... 260 KJ mol—1 2 2 1 A (s) + B2 (g) AB (s) ....... – 550 KJ mol–1 2
then lattice energy of AB (s) will be (A) – 160 (B) – 890
(C) – 420
(D) + 360
Q.7
Which of the following ions gives cyanide ion in aqueous solution ? (A) potassium ferrocyanide (B) potassium ferricyanide (C) potassium cyanide (D) potassium cobaltcyanide
Q.8
Nature of bond between two nonmetal atoms will be (A) Van der waals (B) covalent (C) ionic
Q.9
If the formula of a compound is X2 Y5, then what will be the numbers of electrons present in the valance shells of X and Y, respectively ? (A) 5 and 6 (B) 6 and 3 (C) 2 and 3 (D) 5 and 2
Q.10 The compound with two lone pairs and two bond electrons is (A) SCl2 (B) PH3 (C) NH3 Q.11
(D) all of the above
Which of the following has maximum melting point ? (A) SiC (B) Al4Cl3 (C) CO2
Q.12 Covalent compound is (A) K2O (B) Cl2O
(C) CaO
(D) HF (D) Cl2 (D) MgO
Q.13 A molecule MX3 has zero dipole moment. The orbitals used by M in the formation of bond will be(A) pure p (B) sp2 (C) sp (D) sp3d Q.14 Which of the following ions has an atom in a state of sp2 hybridisation ? (A) BeF3– (B) NF3 (C) OF2 (D) H3O+
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CHEMICAL BONDING
Q.15 Which of the following statements is false for BF3 ? (A) it is a Lewis acid (B) it fiorms an addition produced with ammonia (C) it is a planar molecule (D) it has ionic character Q.16 Maximum bond length is shown in (A) CO2 (B) CH4
(C) NH3
(D) H2O
Q.17 Which of the following does not have H - bond ? (A) water (B) phenol (C) liquefied HCl
(D) Liquefied NH3
Q.18 Coordinate bond is formed in (A) NaCl (B) NH4Cl
(D) BCl3
(C) Cl2
Q.19 Both the compounds of which pair have linear geometry ? (A) SO2, SnCl2 (B) CO2, BeH2 (C) SO2., CO2
(D) SO2, H2O
Q.20 XPO4 is a compound of a metal X, then the formula of the chloride of X will be (A) XCl3 (B) XCl (C) X2Cl3 (D) X2Cl2 Q.21 Which of the following has pyramidal geometry ? (A) PCl3 (B) SO3 (C) CO3–2
(D) NO3–
Q.22 Select the correct statement. (A) (CF3)3N is a base, whereas (CH3)3N is not (B) PbCl2 melts at 606° C. whereas PbCl4 melts at 114°C (C) NaCl is soluble in petrol (D) Dipole moment of BF3 is high Q.23 False statement is (A) SnCl2 is white and SnI2 is coloured (B) Boiling point of H2O is higher than H2S (C) Water is capable of extinguishing fire of petrol (D) Glucose forms H-bonds with water Q.24 Lewis structure of carbon suboxide (C3O2) in ground state is (A) O : C : : : C : C ::: O : (B) : O :: C : C : C :: O : (C) : :: C :: C :: C :: : (D) : O :: C :: C :: C :: O : Q.25 Which of the following hypothesis justifies that the bond angle of H2S is 92° ? (A) Lewis structure (B) Valence bond theory (C) Valance bond concept of hybrid orbitals (D) Octet rule Q.26 Geometry of a molecule in which the central atom has 50% p character, will be (A) linear (B) tetrahedral (C) trigonal (D) distorted tetrahedral Q.27 Configuration of the cation of which of the following metals is ns2, np6 , ndx type, and the total value of 2 + 6 + x is nine to eighteen ? (A) alkali metal (B) alkaline earth metal (C) inert metal (D) d block metal Q.28 Pb+4 is less stable than Pb+2, because of (A) inert pair effect (C) high ionisation potential
(B) small size (D) high electronegatively
Q.29 The element showing highest valency is – (A) Cl (B) I
(C) F
(D) Br
Q.30 Increasing order of dipole moment in H2O, NH3, NF3 and CCl4 is (A) CCl4 < NF3 < NH3 < H2O (B) CCl4 > NF3 > NH3 > H2O (C) NF3 > H2O > CCl4 > H2O (D) all the four have equal dipole moments NARAYANA II T ACAD EMY- C P 3 ,
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CHEMICAL BONDING
Q.31 Species in which peroxide ion is not present is (A) PbO2 (B) H2O2 (C) SrO2
(D) A and B both
Q.32 If s character decreases in hybrid orbital, then bond angle (A) decreases (B) increases (C) remains uncertain (D) all are wrong Q.33 Which of the following compounds does not have a coordinate bond ? (A) SO3 (B) H2SO4 (C) H2SO3 (D) HNO2 Q.34 Maximum covalency of an element is (A) the number of unpaired s electrons (C) number of unpaired d electrons
(B) number of unpaired s and p electrons (D) number of s and p electrons present in valance shell
Q.35 Which of the following molecules consists of only covalent bonds ? (A) KCl (B) PCl3 (C) NH4Cl
(D) BaClO4
Q.36 Number of hybrid orbitals is equivalent to (A) the atomic orbitals participating in hybridisation (B) electrons participating in the process (C) number of unpaired electrons in the valence shell (D) vacant orbitals in valence shell Q.37 When electron cloud of an anion is shared by the cation, then the polar bond so formed exhibits which of the following characters ? (A) covalent (B) metallic (C) coordinate (D) van der waals Q.38 Which of the following types of bond in an inorganic compound undergoes dissociation to gives ions ? (A) coordinate (B) covalent (C) electrovalent (D) hydrogen bond Q.39 What will be the ionic potential , when nature of the oxide of a metal is basic ? (A) > 2.2 and < 3.2 (B) > 2.2 (C) > 3.2 (D) < 2.2 Q.40 Which of the following does not have sp3 hybridisation ? (A) BF4– (B) H3O+ (C) OF2
(D) BF3
Q.41 Sulphur have maximum covalency – (A) 3 (B) 4
(D) 6
(C) 5
Q.42 Which of the following has least ionic character ? (A) Cu2Cl2 (B) KCl (C) CsCl
(D) BaCl2
Q.43 Which of the following are correctly matched ? Compound Hybridisation 1. Graphite – sp3 2. NH4+ – sp2 3. XeF2 – sp3d2 4. SF4 – sp3d (A) 4 (B) 3 (C) 2
(D) 1
Q.44 Formula of the phosphate of a metal is MHPO4. What will be the formula of its chloride (A) MCl (B) MCl2 (C) MCl3 (D) M2Cl2 Q.45 The central atom of which of the following compounds has four bond pairs and two lone pairs (A) SF4 (B) XeF4 (C) NH+4 (D) SF6
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CHEMICAL BONDING
Q.46 Three soluble salts of three metals A (Atomic weight 7), B (atomic weight 27) and C (atomic weigh 64) on electrolysis liberate 2.1, 2.7 and 9.6 grams of metals, respectively at the electrode. The valencies of the metals are, respectively , (A) 1, 3 and 2 (B) 3, 1 and 2 (C) 3, 1 and 3 (D) 2, 2 and 3 Q.47 Incorrect information about Cl2O is (A) angular structure (B) 110° bond angle
(C) four lone pairs
Q.48 Maximum energy is released from which process ? (A) F + F F2 (B) N + N N2 (C) Cl + Cl Cl2 Q.49 False statements about CO2 is that (A) its dipole moment is zero (C) it is diamagnetic
(D) two bonds (D) O + O O2
(B) its solid form is called dry ice (D) hybridisation state of C and O is same
Q.50 What will be the energy of the system on bringing x and y atoms closer upto bond distance ? (A) infinity (B) minimum (C) maximum (D) zero Q.51 Bond angle of HCN is similar to all of the following except (A) BeF2 (B) C2H2 (C) OF2 Q.52 When a molecule breaks, then (A) it absorbs energy (C) it neither transmit nor absorbs energy
(D) CO2
(B) it transmits energy (D) it transmit as well as absorbs energy
Q.53 Which of the following statements in true for hybridisation ? (A) sp3 hybridisation is found in SiF4. SnCl4. ClO4–, SO4–2, NH4+ and NH3 (B) sp hybridisation is found in BeCl2, BeF2 , BeH2, Hg Cl2. ZnCl2, CO2, C2H2 and HCN (C) sp2 hybridisation is found in BeF3–, BCl3, BH3, C2H4 AlCl3, NO3–1 (D) all of the above are correct. Q.54 Which of the following statements is true for [Cu(NH3)4]2+ ? (A) tetrahedral configuration and all paired electrons (B) square planar configuration and one unpaired electron (C) square planar configuration and all paired electrons (D) none of the above Q.55 What is the reason that ionic compounds do not show stereo isomerism (A) presence of ions (B) nondirectional nature of ionic bond (C) brittle nature (D) electrostatic forces of attraction between ions Q.56 In which of the following compounds, the coordinate bond is not present ? (A) CuSO4 (B) H2O (C) BF4– (D) AlCl4– Q.57 Bond length of which of the following types of bonds is maximum ? (A) sp2 – sp2 (B) sp – sp (C) sp3 – sp3 Q.58 Which of the following is correct for CO2 ? (A) nonpolar with polar bonds (C) polar with polar bonds
(D) sp3 – sp2
(B) polar with nonpolar bonds (D) all of the above are wrong
Q.59 Which of the following statements is true ? (A) there is H-bonding in aqueous hydrochloric acid (B) dipole moment of NF3 is more than the dipole moment of NH3 (C) dipole moment of CCl4 is less than the dipole moment of CHI3 (D) positions of atoms get changed in resonance
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CHEMICAL BONDING
Q.60 In a compound, 50% X (valency 1) and 50% Y (valency 1) are present, then the formula of the compound will be (A) X2Y (B) XY (C) XY2 (D) X2Y3 Q.61 Electronic configuration of an atom is ns2 np2 , which forms 2 and 2 bonds in a hydrocarbon, then hybridisation state and bond angle of the atom will be (A) sp, 180° (B) sp2, 120° (C) sp3, 109° 28 (D) dsp2, 90° Q.62 Which of the following compound is not known ? (A) OF2 (B) SF4 (C) SF6
(D) OF6
Q.63 Which of the following bonds should have higher polarity than the remaining three – (A) C – F (B) C – O (C) C – B (D) C – H Q.64 Which of the following has maximum melting point ? (A) ionic crystal (B) covalent crystal (C) metallic crystal Q.65 Inorganic graphite is – (A) boron nitrate (C) boron carbonate
(D) molecular crystal
(B) boron nitride (BN) (D) none of the above
Q.66 The type of bond between the layers in graphite is – (A) vander waals force (B) covalent (C) coordinate (D) ionic Q.67 Anhydrous HCl is – (A) an acid
(B) a base
(C) a salt
(D) a covalent compound
Q.68 Which of the following has ionic and covalent bonds ? (A) OF2 (B) KCl (C) AIN
(D) KCN
Q.69 Which of the following oxides is most acidic ? (A) MnO2 (B) Mn2O3 (C) Mn2O7
(D) MnO
Q.70 Which of the following has least bond energy – (A) C – O (B) C = O (C) C – C
(D) C = C
Q.71 Which of the following has minimum bond energy – (A) O – H (B) Cl – H (C) C – H
(D) N – H
Q.72 p – p overlapping is not possible in – (A) Cl2 (B) O2
(C) N2
(D) H2
Q.73 A bond is formed O2 by overlapping of – (A) 2py – 2pz (B) 2py – 2s
(C) 2pz – 2s
(D) 2pz – 2pz
Q.74 Highest bond angle will be in – (A) H2O (B) H2S
(C) H2Se
(D) H2Te
Q.75 Which of the following ions has tetrahedral geometry – (A) Na+ (B) NH4+ (C) Mg+2
(D) CO3–2
Q.76 Which of the followin molecules has hybridisdation on central atom different from that on the remaining three ? (A) SiH4 (B) CH4 (C) BF4– (D) [Ni(CN4)–2] Q.77 Which of following has square planar geometry ? (A) XeF4 (B) SiCl4 (C) NH4+ NARAYANA II T ACAD EMY- C P 3 ,
(D) BF4–1
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CHEMICAL BONDING
Q.78 Molecule having zero dipole moment is – (A) CH2Cl2 (B) NF3
(C) BF3
(D) ClO2–1
Q.79 Hybridisation state of carbon changes as follows. sp3 sp2 sp, then the angels of hybrid orbitals will – (A) remain unchanged (B) go on increasing (C) go on decreasing (D) first inct Q.80 CCl4 is a covalent compound, whereas LiCl is less covalent, because – (A) C – Cl bond is nonpolar (B) charge on CCl4 is more (C) Li – Cl bond is polar (D) moment of Li – Cl is not definite Q.81 Shape of xenon hexafluoride is – (A) tetrahderal (C) square planar
(B) distorted pentagonal bipyramidal (D) octahedral
Q.82 CO2 is isostructural with – (A) SnCl2 (B) HgCl2
(C) ZnI2
(D) OH2
Q.83 Which of the following statements is correct ? (A) bond order is a measure of the strength of the bond. (B) bond order is equal to the number of bonds present in a molecule (C) greater the bond order, more paramagnetic is the molecule (D) none of the above Q.84 Octahedral shape exists in hybridization – (A) sp3d (B) sp3d2
(C) sp3d3
(D) none of these
Q.85 Relative stabilities of O2, O2– and O2+, O22– are in the order – (A) O2 > O2+ > O2– > O22– (B) O2+ > O2 > O2– > O22– (C) O2+ > O22– > O2– > O2 (D) O22– > O2– > O2+ > O2 Q.86 Which of the following is not paramagnetic – (A) S2– (B) NO
(C) O2–
(D) N2–
Q.87 The dipole moment of a molecule of the type AX4 having a square planar geometry is – (A) O (B) 4D (C) 2D (D) none of the above Q.88 The nodle plane in the –bond of ethene is located in – (A) the molecular plane (B) a plane parallel to the molecular plane (C) a plane perpendicular to the molecular plane which bisect the carbon-carbon bond at right angle (D) a plane perpendicular to the molecular plane which contains the carbon-carbons bonds Q.89 The type of bonds present in CuSO4.5H2O are .............. only – (A) electrovalent and covalent (B) electrovalent and co-ordinate (C) electrovalent, covalent and co-ordinate (D) covalent and co-ordinate Q.90 The structure of XeF4 is – (A) planer (B) tetrahedral
(C) square planar
(D) pyramidal
Q.91 Which of the following is most stable – (A) Pb2+ (B) Ge2+
(C) Si2+
(D) Sn2+
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CHEMICAL BONDING
Q.92 Shape of molecules is decided by – (A) sigma bond (C) both sigma and -bonds
(B) -bond (D) neither sigma and -bonds
Q.93 The metallic lustre exhibited by sodium is explained by – (A) diffusion of sodium ions (B) excitation of free proton (C) oscillations of loose electrons (D) existence of body-centred cubic lattice Q.94 The molecule which has pyramidal shape is – (A) PCl3 (B) SO3
(C) CO32–
Q.95 Sulphuric acid molecule contains – (A) only covalent bonds (C) covalent and co-ordinate bonds
(B) covalent and ionic bonds (D) covalent, ionic and co-ordinate bonds
(D) NO3–
Q.96 NF3 is – (A) non-polar compound (B) electrovalent compound (C) having low value of dipole moment than NH3 (D) having more dipole moment than NH3 Q.97 The maximum possible number of hydrogen bonds in which a water (H2O) molecule can participate – (A) 4 (B) 3 (C) 2 (D) 6 Q.98 Variable valency is characteristic of – (A) noble gases (B) alkali metals
(C) transition metals
Q.99 PCl5 exists but NCl5 does not because – (A) nitrogen has no vacant 2d-orbitals (C) nitrogen atom is much smaller than p
(B) NCl5 is unstable (D) nitrogen is highly inert
Q.100 The type of bond formed between H+ and NH3 in NH4+ ion is – (A) ionic (B) covalent (C) dative
(D) non-metallic elements
(D) hydrogen
PRAYAS - II Q.1 Q.2
Q.3 Q.4
Q.5 Q.6 Q.7
Weakest bond is (A) ionic bond
(B) covalent bond
Which of the following statements is true ? (A) and bonds are weak bonds (C) both the bonds are equally strong
(C) coordinate bond
(B) bond is stronger than bond (D) all the above are wrong
Which of the following is nondirectional orbital ? (A) sp (B) sp3 (C) dsp2 orbital will formed by (A) overlapping of s-s orbitals (C) coaxial overlapping of p-p orbitals
(D) hydrogen bond
(D) s
(B) overlapping of sp-s orbitals (D) collateral overlampping of p-p orbitals
Which of the following compounds has linear geometry ? (A) CO2 (B) NO2 (C) SO2
(D) SnCl2
Which of the following has an atom having incomplete octet ? (A) BF3 (B) NH4+ (C) CCl4
(D) PH3
The compound having odd number of electron pairs is (A) XeF2 (B) NaCl (C) BF3
(D) ClO2
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Q.8
Electronic configuration of two elements X and Y are 2, 5 and 2, 7, respectively. If these elements form a covalent compound with each other, then the formula of the compound will be (A) XY3 (B) X2Y3 (C) XY2 (D) XY
Q.9
What is the oxidation number of carbon in diamond ? (A) 0 (B) 4 (C) 1
(D) 2
Q.10 The s character in sp3d hybrid orbital will be (A) Q.11
1 5
(B)
1 4
(C)
1 2
(D)
Which of the following is present in the lead pencil ? (A) Pb (B) Charcoal (C) Graphite
3 4
(D) Fe
Q.12 What is the type of hybridisation and nature of Ni in [Ni(CN)4–2 ion ? (A) dsp2 and diamagnetic (B) sp3d and paramagnetic (C) dsp3 and diamagnetic (D) sp3 and paramagnetic Q.13 Which of the following species as boron in sp3 hybridisation state ? (A) BF4– (B) BH3 (C) B2H6 Q.14 Hydrolysis of SiCl4 occurs very fast, because (A) Si is an electrophile due to vacant orbital in it (C) SiCl4 is covalent
(D) BCl3
(B) SiCl4 is ionic (D) Si is an element of V A group
Q.15 The factor responsible for insolubility of BaSO4 in water is (A) high dissociation energy (B) high covalent character (C) high hydration energy (D) high lattice energy Q.16 In hybridisation process (A) the hybrid orbitals formed are nondirectional (B) the hybrid orbitals of the atom have almost same energy (C) shapes of the hybrid orbitals are different (D) hybrid orbitals have high attractive forces between them Q.17 The compound soluble in ether is (A) NaCl (B) RbCl
(C) BaCl2
Q.18 Which of the following is not a covalent compound ? (A) NH3 (B) MgCl2 (C) SO2
(D) LiCl (D) CCl4
Q.19 Element X is strongly electropositive and Y is strongly electronegative. X and Y are monovalent, then the compound formed from them will be (A) X+ Y– (B) X - Y+ (C) X - Y (D) X Y Q.20 Which of the following statements is correct ? (A) all the covalent compounds are solids at room temperature (B) all the compounds of hydrogen are solids (C) all the halogen compounds are ionic (D) all the salts are generally ionic in nature Q.21 The valency of B in BCl3 is 3. This is justified on the basis of – (A) resonance (B) hybridisation (C) electronic configuration (D) shielding effect
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CHEMICAL BONDING
Q.22 Ionic bond is not a true bond, because (A) it is nondirectional (C) it is directional
(B) it is not strong (D) it has repulsion between ions
Q.23 The compound completing its octet by transfer of electrons is (A) MgO (B) H2S (C) PH3
(D) CCl4
Q.24 Ionic compounds are insoluble in nonpolar solvents, because (A) the dielectric constant of the solvents is high (B) their dipole moment is high (C) they get ionised very easily (D) they do not form ion-solvent complex Q.25 Transition metals show variable valency, because (A) more electrons can be lost from the outermost shell (B) they can lose electrons from the outermost and penultimate shells (C) they form weak metallic bonds (D) they have low ionisation potential Q.26 There is increase in polarisation, when – (A) the size of cation increases (C) the size of cation and anion decreases
(B) the size of cation decreases (D) the size of cation and anion increases
Q.27 Which of the following molecules does not have co-ordinate bond – (A) SO2 (B) O3 (C) CH3NC
(D) CO2
Q.28 Compound of nitrogen in which nitrogen is in the state of sp2 hybridisation, is (A) X– N = Y X
(B) N N
(C) N = Y
X
(D) NX3
Q.29 What is the reason of solubility of HCl in water ? (A) hydrogen bonding (B) ionisation (C) proton transfer (D) van der waals forces Q.30 An element y is in the first short period and its configuration is ns2p1. The formula and the nature of its oxide will be – (A) y2O3, acidic (B) yO2, amphotaric (C) y2O2, basic (D) yO3, acidic Q.31 Which of the following compounds is covalent and in which the extension of octet takes place during its formation ? (A) SF6 (B) NO (C) NH3 (D) HCl Q.32 The pair of ions having pseudo inert configuration, is – (A) Au+ and Ag+ (B) Na+ and Au+ (C) Cu+2 And Ag+
(D) Mg+2 and K+
Q.33 Increasing order of ionic character of the following metal halides is MgCl2, NaCl, CaCl2, SrCl2, BaCl2 (A) MgCl2 < CaCl2 < SrCl2 < BaCl2 < NaCl (B) NaCl < MgCl2 < CaCl2 < SrCl2 < BaCl2 (C) CaCl2 < MgCl2 < NaCl < SrCl2 < BaCl2 (D) BaCl2 < CaCl2 < MgCl2 < NaCl < BaCl2 Q.34 Bond length does not depend on – (A) electronegativity (B) electron affinity
(C) resonance
(D) hybridisation
Q.35 Presence of lone pair of electrons on an atom affects – (A) the bond angle (B) the geometry of the molecule (C) the physical properties (D) the geometry as well as bond angle Q.36 In which of the following types of hybridisation, all the bond angles are not equal ? (A) sp2 (B) sp3d (C) dsp2 (D) sp NARAYANA II T ACAD EMY- C P 3 ,
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CHEMICAL BONDING
Q.37 Dimethyl ether is a gas and the boiling point of ethanol is 78°, whereas molecular formula and molecular weight of both the compounds are same. What is the region ? (A) H-bonding in ethanol (B) presence of covalent bond in ethanol (C) ionic character of ethanol (D) basic nature of dimethyl ether Q.38 BeCl2 has which of the following types of orbital overlap ? (A) sp2 – p (B) sp – p (C) sp3 – p
(D) s – p
Q.39 Which of the following will show high covalent character ? (A) K2O (B) OF2 (C) Bi2O3
(D) none of the above
Q.40 In which of the following, the hybridisation has not taken place ? (A) C2H2 (B) SO2 (C) KCl
(D) NH4+
Q.41 Which of the following has octahedral geometry ? (A) SF6 (B) ClF2 (C) NH4+
(D) XeF4
Q.42 Dipole moment of which of the following is maximum ? (A) NH3 (B) NF3 (C) H2O
(D) OF2
Q.43 Atomic numbers of two atoms X and Y are 1 and 8, respectively, then what will be the formula of the compound formed from them ? (A) XY (B) X2Y & X2Y2 (C) XY3 (D) X2Y Q.44 Which of the following has minimum melting point ? (A) Cu2Cl2 (B) CuCl2 (C) CaCl2
(D) KCl
Q.45 Vander Waals forces are present in (A) inert gases (B) mixture of gases
(D) all of the above
(C) rare gases
Q.46 Iodine shows sp3d3 hybridisation in which of the following excited states ? (A) III (B) II (C) I (D) normal state Q.47 According to Sudgen, the compound having a one electron bond is (A) NCl3 (B) PCl5 (C) B2H6
(D) AlCl3
Q.48 What is the reason of high reactivity of fluorine ? (A) low bond energy of F – F bond (B) F2 is liquid at room temperature (C) high bond energy of F – F bond (D) electronegativity of F is maximum Q.49 Decomposition temperature is maximum for (A) NH3 (B) PH3
(C) AsH3
(D) SbH3
Q.50 Most acidic compound is (A) N2O3 (B) P2O3
(C) As2O3
(D) Sb2O3
Q.51 In which of the following compounds, dipole-dipole attraction is present ? (A) H2O (l) (B) H2(g) (C) N2(g) (D) HCl (g) Q.52 In which of the following, the bond order is 2.5 ? (A) HCl (B) KCl (C) CO
(D) NO
Q.53 Supposing that, an atom can be stabilised by 6 electrons and not by an octet, then the most stable structure of N will be (A) N+ (B) N– (C) N2– (D) N+2 Q.54 Correct order of boiling points of H2, Cl2 and Br2 is (A) H2 < Cl2 < Br2 (B) Cl2 > Br2 > H2 (C) H2 > Cl2 < Br2 NARAYANA II T ACAD EMY- C P 3 ,
(D) equal in all the three
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CHEMICAL BONDING
Q.55 The cation having ns2, np6, ndx configuration in which 2 + 6 + x = 9 to 18, are generally formed by– (A) alkali metals (B) alkaline earth metals (C) transition elements (D) inner transition elements Q.56 Total number of lone pairs in N2H4 should be – (A) 3 (B) 2 (C) 1
(D) 5
Q.57 Which of the following is angular molecule ? (A) NO2 (B) HgCl2
(D) SO3
(C) BeF2
Q.58 A covalent compound is good conductor of electricity. Which of the following is true for this compound ? (A) single macromolecule (B) molecule formed by separate layers (C) ionic molecule (D) all of the above Q.59 Which of the following starred atoms does not have a lone pair of electrons ? (A)
(B)
(C)
(D)
Q.60 Which of the following cations is capable of maximum polarising the chloride ion ? (A) Ca+2 (B) Zn+2 (C) Sr+2 (D) K+ Q.61 Most stable carbonate is – (A) CaCO3 (B) BaCO3
(C) SrCO3
(D) MgCO3
Q.62 Which of the following has highest bond energy – (A) C – Cl (B) C – Br (C) C – I
(D) C – F
Q.63 Which of the following molecules is bent – (A) CO2 (B) O3
(D) none of these
(C) N2O
Q.64 In hybridisation process – (A) hybrid orbitals have strong attraction between them. (B) shape of hybrid orbitals is different (C) orbitals of almost equal energy hybridise in a single atoms (D) hybrid orbitals formed are nondirectional Q.65 Molecular orbital is formed by theoverlap of two atomic orbitals. It will be called – (A) ionic bond (B) covalent bond (C) coordinate bond (D) hydrogen bond Q.66 Number of coordinate bonds in SO3 molecule, will be – (A) 1 (B) 2 (C) 3
(D) 4
Q.67 Which of the following has minimum energy ? (A) bond (B) bond (C) ionic bond
(D) hydrogen bond
Q.68 Which of the following compound is covalent ? (A) H2 (B) CaO (C) KCl
(D) Na2S
Q.69 Strongest bond in between – (A) NaCl (B) CsCl
(C) both
(D) none
Q.70 Which of the following has a linear shape ? (A) B2H6 (B) BeF2
(C) AlCl3
(D) SO2
Q.71 In which of the following the angle between the two covalent bonds is greatest ? (A) H2O (B) NH3 (C) CO2 (D) CH4
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CHEMICAL BONDING
Q.72 AsF5 molecule is trigonal by pyramidal. The orbitals of As atom involved in hybridization are – (A) d x 2 y 2 , d z 2 , s, px, py (B) dxy, s, px, py, pz (D) d x 2 y 2 , s, p x , p y , p z
(C) s, px, py, pz, d z 2
Q.73 Which of the following species has bond order of zero ? (A) Ne2 (B) He2+ (C)O2+
(D) N2+
Q.74 The compound which does not obey the octet rule is – (A) OF2 (B) SO2 (C) PCl3
(D) SnCl4
Q.75 Which of the following molecular species has unpaired electrons ? (A) N2 (B) F2 (C) O2–
(D) O22–
Q.76 Which of the following statements is not correct for sigma and pi bonds formed between two carbon atoms? (A) Sigma-bond determines the direction between carbon atoms but a pi-bond has no primary effect in this regard. (B) Sigma-bond is stronger than a pi-bond (C) Bond energies of sigma-and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol, respectively (D) Free rotation of atoms about a sigma-bond is allowed but not in case of a pi-bond Q.77 Which contains both polar and non-polar covalent bonds – (A) NH4Cl (B) HCN (C) H2O2
(D) CH4
Q.78 A sp3-hybrid orbitals contains – (A) 1/4 s-character (B) 1/2 s-character
(C) 2/3 s-character
(D) 3/4 s-character
Q.79 Ionic reaction take place in – (A) liquid state (B) solid state
(C) solution state
(D) gaseous state
Q.80 The lowest bond energy exist in the following bonds for – (A) C – C (B) N – N (C) H – H
(D) O – O
Q.81 The total number of valency electrons in PH4+ ion is – (A) 8 (B) 9 (C) 6
(D) 14
Q.82 HCl molecule in the vapour state in an example of – (A) non-polar bond (B) ionic bond (C) polar covalent bond (D) pure covalent bond Q.83 The nature of bonding in diamond is – (A) ionic (B) covalent
(C) metallic
(D) co-ordinate
Q.84 Among LiCl, BeCl2, BCl3 and CCl4, the covalent bond character follows the order – (A) LiCl > BeCl2 > BCl3 > CCl4 (B) LiCl < BeCl2 < BCl3 < CCl4 (C) LiCl > BeCl2 > CCl4 > BCl3 (D) LiCl < BeCl2 < BCl3 > CCl4 Q.85 The example of the p-p orbital overlapping is the formation of – (A) H2 molecule (B) Cl2 molecule (C) Hydrogen chloride (D) hydrogen bromide molecule Q.86 Which statement is not correct – (A) double bond is shorter than a single bond (B) sigma bond is weaker than pi-bond (C) double bond is stronger than a sigma bond (D) covalent bond is stronger than hydrogen bond
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CHEMICAL BONDING
Q.87 Which atomic orbital is always involved in sigma bonding only – (A) s (B) p (C) d
(D) f
Q.88 Which involves breaking of covalent bond – (A) boiling H2S (B) melting KCN
(D) boiling CF4
(C) melting SiO2
Q.89 Fluorine molecule is formed by the overlapping of – (A) s-p orbitals (B) s-s orbitals (C) p-p orbitals by end to end manner (D) p-p orbitals by sides to sides manner Q.90 Which does not apply to metallic bond – (A) overlapping (C) delocalised electrons
(B) mobile valency electrons (D) none
Q.91 Most covalent halide of aluminium is – (A) AlCl3 (B) AlI3
(C) AlBr3
(D) AlF3
Q.92 The number of lone pairs is same in PCl3 and – (A) BCl3 (B) NCl3 (C) CCl4
(D) PCl5
Q.93 Which does not show hydrogen bonding – (A) C2H5OH (B) liquid NH3
(D) liquid HBr
(C) H2O
Q.94 Which of the following has unchanged valency – (A) H (B) Na (C) Fe
(D) O
Q.95 A molecule in which sp2-hybrid orbitals are used by the central atom in forming covalent bond is – (A) He2 (B) SO2 (C) PCl5 (D) N2 Q.96 Iron is tougher than sodium because – (A) iron atom is smaller (C) metallic bonds are stronger in iron
(B) iron atoms are more closely packed (D) none of these
Q.97 The ratio of and -bonds in benzene is – (A) 2 (B) 6
(C) 4
(D) 8
Q.98 How many bonded electron pairs are present in IF7 molecule – (A) 6 (B) 7 (C) 5
(D) 8
Q.99 The octet rule is not followed in – (A) F2 (B) NaF
(D) BF3
(C) CaF2
Q.100 In a crystal, cations and anions are held together by – (A) electrons (B) electrostatic forces (C) nuclear forces
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(D) covalent bonds
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CHEMICAL BONDING
CHEMICAL BONDING ANSWER KEY PRAYAS-I Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans
1 C 11 A 21 A 31 A 41 D 51 C 61 A 71 B 81 B 91 A
2 A 12 B 22 B 32 A 42 A 52 A 62 D 72 D 82 C 92 A
3 D 13 B 23 A 33 D 43 A 53 D 63 B 73 D 83 A 93 C
4 D 14 A 24 C 34 D 44 B 54 B 64 B 74 A 84 B 94 A
5 C 15 D 25 C 35 B 45 B 55 B 65 B 75 B 85 B 95 C
6 B 16 A 26 A 36 A 46 A 56 B 66 A 76 C 86 A 96 C
7 C 17 C 27 D 37 A 47 C 57 C 67 D 77 A 87 A 97 A
8 B 18 B 28 A 38 C 48 B 58 A 68 D 78 C 88 A 98 C
9 A 19 B 29 A, B 39 D 49 D 59 C 69 C 79 C 89 C 99 A
10 A 20 A 30 A 40 D 50 B 60 B 70 C 80 A 90 C 10 0 C
7 D 17 D 27 D 37 A 47 B 57 D 67 D 77 C 87 A 97 C
8 A 18 B 28 A 38 B 48 A, D 58 B 68 A 78 A 88 C 98 B
9 A 19 A 29 B 39 B 49 A 59 C 69 B 79 C 89 C 99 D
10 A 20 D 30 A 40 C 50 A 60 B 70 B 80 D 90 A 100 B
PRAYAS-II Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans Q ue . Ans
1 D 11 C 21 B 31 A 41 A 51 D 61 B 71 C 81 A 91 B
2 B 12 A 22 A 32 A 42 C 52 D 62 D 72 C 82 C 92 B
NARAYANA II T ACAD EMY- C P 3 ,
3 D 13 A 23 A 33 A 43 B 53 B 63 B 73 A 83 B 93 D
4 D 14 A 24 D 34 B 44 B 54 A 64 C 74 B 84 B 94 B
5 A 15 D 25 B 35 D 45 D 55 C 65 B 75 C 85 B 95 B
6 A 16 B 26 B 36 B 46 A 56 B 66 B 76 C 86 B 96 C
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CHEMICAL BONDING
PARIKRAMA - I Q.1
Why the atoms of noble gases do not react with other atom to form compounds ?
Q.2
Why ? The ionic bond is called a non directional but covalent bond is called a direction bond.
Q.3
Why MgO is more stable than NaCl ?
Q.4
Why the NaCl is more stable than CsCl ?
Q.5
Why ionic compound do not conduct electricity when they are in the solid state ? or
Q.6
Why the ionic solid conduct electricity when they are in water solution or in molten state ? or
Q.7
Why the ionic solids are soluble in polar solvent like H2O. ? or
Q.8
Explain the solubility of ionic solid in a polar solvent like water ? or
Q.9
Why ? An ionic solid dissolve in polar solvent ? or
Q.10 Explain why ionic solids are not soluble in non polar solvents ? Q.11
Why elements show variable covalency ?
Q.12 Why the covalent compounds are consist of discrete molecules ? Q.13 Why some covalent compounds ore good conductor of electricity ? Q.14 Why covalent solid are soluble in non-polar solvents ? Q.15 Why covalent solids having the giant molecules are insoluble in all the solvents ? Q.16 Why the covalent compound neither hard not brittle ? Q.17 Why the reaction of covalent compound are slow ? Q.18 Why ionic compounds are not show isomerism but covalent compounds show this phenomenone ? Q.19 Why an alloy is hard then pure metal ? Q.20 Why metals are highly tensile ? Q.21 Why metallic bond is weaker than a covalent bond ? Q.22 Why only O, N and F are capable of forming H – bonds ? Q.23 Why NH3 is better electron donner than PH3, H2S & H2O ? Q.24 Why metals have high density ? Q.25 Why the alkaline earth metals are harder than alkali metals ? Q.26 Explain the hardness sequence Li > Na > K ? Q.27 In ordinary conditions water exist as liquid but H2Se and H2Te as gases why ? NARAYANA II T ACAD EMY- C P 3 ,
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Q.28 Explain the following sequence for boiling point ? He < Ne < Ar < Kr < Xe < Rn Q.29 Explain how can a molecule has less energy and high stability than isolated atoms ? Q.30 Why – bond is weak than – bond ? Q.31 Why a molecule is more stable in terms of energy than the uncombined atoms ? Q.32 Why sodium chloride does not conduct electricity in solid state but does so in molten state ? Q.33 Why H2O is liquid while H2S is a gas at ordinary temperature ? Q.34 Why BCl3 and BF3 are non polar ? Q.35 Why BeF2 and BF3 are stable though Be and B have less than 8 electrons ? Which one is more stable? Q.36 Why the repusion between non bonded orbitals is greater than between the bonded orbitals ? Q.37 Why Valency of oxygen is generally two whereas sulphur shows valency of two, four and six ? Q.38 Why anhydrous HCl is bad conductor but in aqueous medium, it is good conductor of current ? Q.39 Why diamond is a hard solid with a very high melting point but a non-conductor of electricity ? Q.40 Why AlCl3 is largely covalent while AlF3 is largely ionic ? Q.41 Arrange the following bonds in increasing order of strength C – C, Ge – Ge, Si – Si Q.42 F2 and Cl2 are gases, Br2 is liquid and I2 is solid why ? Q.43 CO2 is gas but SiO2 is solid. Why ? Q.44 NH3 can be liquefied easily but PH3 is not. Why ? Q.45 What is the increasing order of ionic character in HF, HCl, HBr and HI, Explain ? Q.46 What is the increasing order of ionic character in H2O, H2S and H2Se, Explain ? Q.47 What is the decreasing order of ionic character in NCl3 and PCl3, Explain ? Q.48 NF3 is less stable than NCl3. Why ? Q.49 Why is PCl5 unstable ?
PARIKRAMA -II Q.1
Why KHF2 is exist but KHCl2 is not ?
Q.2
Why Co+3 ion is form with great difficulty but Fe+3 is not ?
Q.3
Why an ionic solid do not exist as individual neutral independent molecules ?
Q.4
Why NaCl have less boiling point than SiCl4 ?
Q.5
Explain why Cu+ cation polarises the anions more strongly than Na+ cation even both have +1 charge?
Q.6
Why Ag+ (cation) 4s2p6d10 has greater polarising power than K+ (cation) 3s2p6 ?
Q.7
Why Li+ compounds are soluble in non polar solvent but not in polar solvents ?
Q.8
Explain the following sequences of solubility ? AlF3 > AlCl3 > AlBr3 > Al I3
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Q.9
CHEMICAL BONDING
Why co-ordinate bond is called semi polar bond ?
Q.10 Why H – atom alone is not capable of forming hydrogen bond ? Q.11
Why metals have intermediate melting and boiling point as compared to those of covalent and ionic compounds ?
Q.12 Explain why melting point of o-nitrophenol is lesser than m and p-isomers ? or o-nitrophenol is volatile in steam and less soluble in water than the other two isomers ? Q.13 Boiling point of hydrides of VA, VIA & VIIA decrease from top to bottom but NH3, H2O & HF show a sudden increase why ? Q.14 Explain why glycerol (CH2(OH) – CH(OH) – CH2(OH)) have more viscosity, high heat of vaporisation and high dielectric constant ? Q.15 The density of ice is less then that of water or ice floats over water. Explain ? Q.16 Methanoic acid, HCOOH, has one carbon-oxygen bond of length 123 pm, and another of 136 pm. Which bond has which length ? Q.17 Both carbon-oxygen bonds in the methanoate ion, HCOO–, have the same length (127 pm). What does this tell about bonding ? Q.18 Water can react with hydrogen ion to make the oxonium ion, H3O+. What is present in water molecule that allows it to react with a hydrogen ion ? Describe the bonding in the oxonium ion. Draw a dot and cross diagram for the molecule ? Q.19 Why MgCl2 is linear but SnCl2 is angular ? Q.20 Why NCl5 does not exist while PCl5 does ? Q.21 Why Lead prefers to form divalent compounds ? Q.22 Why CH4, NH3 and H2O contain same number of electron but their shapes are different ? Q.23 Why Calcium fluoride is more ionic than CaI2 ? Q.24 Why Ammonium salts are much more soluble in water than the corresponding sodium salts ? Q.25 Why Carbon has two electrons unpaired in the outer most, but it forms tetravalent in organic compounds ? Q.26 The electronegativities of nitrogen and chlorine are same but NH3 exists as liquid whereas HCl as gas. Why ? Q.27
Although CO2 has no dipole moment, SO2 and H2O have considerable dipole moments. Why?
Q.28 Nitrogen trifluoride (NF3) and ammonia (NH3) have identical shape and a lone pair of electrons on nitrogen and further the electronegativity difference between the elements is nearly the same but the dipole moment of NH3 is very high in comparison to NF3. Why ? Q.29 Why Three carbon-oxygen bonds are equal in carbonate ion ? Q.30 Why NF3 is weaker base than NH3, NCl3, NBr3 and NI3 ? Q.31 Why BaSO4 is insoluble in water ? Q.32 Why compounds of normal elements are diamagnetic ? Q.33 Why IF7 is exist but ClF7 is not. NARAYANA II T ACAD EMY- C P 3 ,
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Q.34 Which among the following has highest boiling point and why ? H2, He, Ne, Xe, CH4 Q.35 At room temperature, NO2 gas is paramagnetic in nature. When it is cooled below 0ºC acquires diamagnetic behaviour. Explain this observation on the basis of structure ? Q.36 Explain why the super oxides KO2, RbO2 and CsO2 are paramagnetic ? Q.37 NH4+ has bond angle identical to CH4 but NH3 has different bond angle; explain with proper reasoning ? Q.38 Explain why ClF2– is linear but ClF2+ is bent molecular ion ? Q.39 Ether
and water
have same hybridization at oxygen but they have not same
bond angle. Why ? Q.40 In trimethylamine, the nitrogen has a pyramidal geometry whereas in trisilylamine N(SiH3)3 it has a planar geometry. Account for this fact. Q.41 Carbon has maximum covalency of four. Explain the covalency of carbon in given compound.
Q.42 Which among the following will have the highest melting point ? PH3, NH3, (CH3)3N Q.43 Discuss the hybridization and shape of the molecule
. Indicate whether the
molecule is polar or non-polar. Q.44 The dielectric constant of H2O2 is more than that of H2O, but H2O2 is not a good solvent. Why ? Q.45 What is the increasing order of bond angle in OF2 and Cl2O ? Q.46 Why in hydrogen & vander waal interaction bonding atoms do not lose their identity ? Q.47 AgCl is white, AgBr is yellow & AgI is deep yellow. Why ? Q.48 Why CCl4 does not get hydrolised but SiCl4 get hydrolised. Q.49 Why SF6 is exist but OF6 is not ?
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SOLUTIONS (Parikrama– I) Ans.1 Because the outer most electron configuration of the atoms of noble gas is a stable configuration of 8 electron which is also called octet. Ans.2 Because in ionic bond, an ion can attract other opposite charged ions from any direction and extends equally in all directions so the nature of ionic bond is non-directional, but on the other band a covalent bond is formed by the proper overlapping of orbitals so it have directional character. Ans.3 A/c to the force of attraction between constituent ions.
F
q1q 2 qq 1 2– 2 2 d (rA rB )
The electric charge on doubly charged Mg+2 and doubly charged O–2 ion in Mg+2O–2 ionic crystal is 4 times greater than between monovalent Na+ and Cl– ions in Na+ Cl– Ans.5 Because Na+ is smaller than Cs+ ion so the force of attraction between Na+Cl– ion in NaCl is higher than attraction between Cs+Cl– in CsCl. So NaCl have high lattice energy then CsCl and have greater stability. Ans.6 Due to the following two reasons (i) Unstable configuration of the core (Kernal) (ii) Inert electron pair effect. Ans.7 Because in the crystal lattice the cation and anions are tightly held together with each other so the ions therefore ; can not move freely to any large extent when an electric current is passed through the ionic solids. Ans.8 Because when the ionic compound goes in to the molten state. The kinetic energy of ions becomes so high that the attractive forces acting between the ions are over come so the arrangement of ions in crystal is destroyed and the ions become free to move in liquid medium. Ans.9 The electrostatic force between cation and anions is reduced by the high dielectric constant value of polar solvent. Ions are move freely and interact with solvent molecule to form the solvated ions. Ans.10 The water molecule is a dipole and hence the positive end of water dipole interact with the negative ion of the ionic solids and the negative ion of the ionic solids and the negative end of the dipole interacts with the positive ion of the same ionic crystals. Ans.11 For an ionic solid to dissolve in a polar solvent the solvation energy of the solvent must be greater than the lattice energy of the ionic solid so that the solvation energy may overcome the lattice energy. Ans.12 Because the force of attraction between adjacent covalent molecules is weak. Ans.13 Because these compounds having the layer lattices (graphite) and they have a free electrons which are good conductors of electricity. In such solids electrons can pass from one layer to the other and thus current can be carried. Ans.14 Due to the similarity in covalent nature of the molecule of the solute and solvent their solubility is based on the principle like dissolves like. Ans.15 This is due to the fact because due to their big size, are not able to interact with the solvent molecule. Ans.16 Because there are weak forces holding the molecules in solid crystal. A molecular layer in the crystal easily slips relative to other adjacent layer and there are no forces of repulsion between the other layers like those in ionic compound. NARAYANA II T ACAD EMY- C P 3 ,
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Ans.17 Because there are no strong electrical force is present to speed up the reaction between molecules. Ans.18 Because ionic bonds are not rigid and non-directional but covalent bond are region and directional. Ans.19 Because if an alloying metal is added to a metal. The structural homogeneity of the principal metal is disturbed and hence the resulting allow is hard and brittle. Ans.20 It is due to the great attraction between the positive metal ion and the mobile electron. Ans.21 Due to the presence of the delocalised, electron (mobile electron) the attraction force between valence electron and the nuclei is weak but on the other hand is covalent bond due to the presence of localised electron there is more attraction between valence electron and nuclei is develop and thus covalent bond is stronger than metallic bond. Ans.22 It is due to their high electronegativity and small atomic size. Ans.23 It is due to the presence of vacant d-orbitals in PH3 and H2S. In H2O due to more electronegativity of O than N. Electron losing capacity of H2O will be less than NH3. Ans.24 This is due to fact that the metal bond keeps the metal atoms closely packed in the metallic crystal. Ans.25 Because the strength of metallic bond increases with increase of valence electrons and we know that alkaline earth metals have two and alkali metals have one valence electron. Ans.26 Because the size of Li+, Na+ and K+ ion is in the order. Li+ < Na+ < K+ Ans.27 Due to the high electronegativity of O-atom than S, Se and Te. H2O molecule associate to form a polymerised molecule while H2S, H2Se and H2Te cannot do so. Ans.28 With increase of vander wall forces the boiling point or melting point of a substance also increase. Ans.29 Because the overlapping of orbital involves a release of energy. This show that the formation of covalent bond is always accompanied by evolution of energy and thus stablilises the molecules. Ans.30 Because -bond is formed by the side to side overlapping between two p-orbital which much less effective than -bond which formed by the head to head overlapping between p-orbitals. Ans.31 When the atoms combined together to form molecule there is always release of energy. Thus potential energy of molecule is less than that of uncombined atoms and therefore, the molecule is more stable. Ans.32 Because solid sodium chloride has crystalline structure in which the ions are not free to move the ion become mobile when it is in molten state and thus the electricity can be conduct. Ans.33 Because oxygen has high electronegativity than sulphur. As a result, H2S forms hydrogen bonding so molecule of water come nearer to each other through hydrogen bonding. This results in higher boiling point of water and hence it is a liquid. Ans.34 Both are triangular and symmetrical molecules their dipole moment are zero. Ans.35 The stability is explained by symmetrical linear structure of BeF2 and triangular planer structure of BF3. BeF2 is more stable because of its greater bond angle (180º). Ans.36 The non-bonded orbitals relatively occupy space compared to the bonded orbitals and thus repulsion are greater. Ans.37 Because there are no d-orbitals in oxygen and hence it can not extend it/s covalency but vacant dorbitals are present in S are paired orbitals can be unpaired by shifting electrons to d-orbital so either making four orbitals singly occupied or six orbitals singly occupied showing valency 4 or 5 besides ? NARAYANA II T ACAD EMY- C P 3 ,
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Ans.38 HCl is a covalent compound and in gaseous state it does not conduct current. In water however it react to form ions, HCl + H2O H3+O + Cl– and thus conduct current. Ans.39 Because diamond is a giant molecule-three dimensional figure as every carbon is sp3 hybridised so one carbon atom is linked to four other carbon atoms tetrahedrally. Due to continuous covalent bonding all the atoms are held very closely are strongly as a result diamond is a very hard solid with a very high MP. It is non conductor as free electrons are not available. Ans.40 Due to large size of Cl– ion than F– ions there is greater degree of distortion in AlCl3. Ans.41 Ge – Ge < Si – Si < C – C Bigger is the atomic size lower is the bond strength. Ans.42 Intermolecular distance is high in gaseous state, low in solid state and inter mediate in liquid state. When molecular weight increases, vander waal's forces also increases. So F2 and Cl2 are gases due to the weak vander waals forces and I2 is solid. Ans.43 CO2 has only weak vander waals forces but in SiO2, one silicon atom is surrounded by four oxygen atoms tetrahedrally forming macromolecular string So SiO2 is solid. Ans.44 Because H-bonding present in NH3, while PH3 is non polar due to same value of electronegativity of P and H. Ans.45 HI < HBr NCl3, Because electronegativity difference of P and Cl is greater than N and Cl. Ans.48 NH3 is more polar than NCl3 – SO bond energy of N – F will be higher. Ans.49 String of PCl5 is unsymmetrical because the value of bond angles are different.
SOLUTIONS (Parikrama – II) Ans.1 HF2– is formed by H-bonding which formes KHF2 with K+ while H-bond is not formed due to low electronegativity of Cl– thus KHCl2 is not exist Ans.2 Increasing the nuclear charge in an atom tend to prevent the removal of electron from a lower energy level. As Co atom forms Co+3 ion with great difficulty because 3d electrons are more firmly retained by cobalt nucleus with positive charge equal to +27 which one unit higher than that on iron nucleus (= +26) Fe(26) 3s2p6d6, 4s2, Fe+2(24) 3s2p6d6, Fe+3(23) 3s2p6d5 Co(27) 3s2p6d7, 4s2, Co+2(25) 3s2p6d7, Co+3(24) 3s2p6d6 Ans.3 Because in crystal lattice a large number of cation and anion attract each other due to electro static force of attraction which extends in all directions. So in order to occupy minimum space the ions arrange themselves systemetically in an alternating cation-anion pattern thus we see that an ionic crystal consist of three dimensional solid string. Ans.4 NaCl is ionic and SiCl4 is covalent bond. Ans.5 According to their electronic configuration Cu+(3s2p6d18)have 18 electron in its valence shell but Na+(2s2p6) have 8 electron in its valence shell so the d-electron of the 18 electron shell shield the nuclear charge of cation less effectively. NARAYANA II T ACAD EMY- C P 3 ,
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Ans.6 same reason (Q. 5) Ans.7 Because Li+ cation has a maximum polarising power due to its small size. Ans.8 So AlF3 > AlCl3 > AlBr3 > AlI3. Increase the covalent character lower the solubility. The AlF3 have maximum ionic character thus it have maximum solubility. Ans.9 Because it is formed by these two steps. Ist step – In this the doner (A) transfer 1 electron of it lone pair to the acceptor (B) so A develop unit positive & B develop unit negative charge. This step is similar to the formation of ionic bond.
A
+B
–
A + B
IInd step – In this the two electrons one each with A+ & B– are shared by both the ion in this step is similar to the formation of covalent bond.
–
A + B
A
B
Thus we say that a co-ordinate bond is equivalent to a combination of an electrovalent bond (polar bond) and a covalent bond (non polar bond) so it is called semi polar bond. Ans.10 Because H-atom has small size with only one electron in its Ist energy level, when this electron is taken away, the proton (H+) left behind which can easily manage between two electronegative atoms bringing them closer together by creating electrostatic force. Ans.11 This is due to the reason that the attractive forces (metallic bond) in metallic crystal are intermediate between those in covalent and ionic compound. Ans.12 Because in o-nitrophenol a six membered ring (chelation) is formed by intramoleculer H-bond so it have different properties from those of other isomers. Because this type of bonding is not possible in m & p-isomer because of the size of the ring that would result.
Ans.13 The abnormal high boiling point of NH3, H2O and HF are due to the fact that the molecule are capable of forming the associated molecule as (NH3)x, (H2O)x, and (HF)x by H-bonding between them. Ans.14 This is because of this fact that glycerol with three –OH group in its molecule can for any H-bond per molecule. This result in greater attraction among the molecule of glycerol. Thus it is more viscous, have high heat of vapourisation and high dielectric constant. Ans.15 In ice, water molecule is associated with four other water molecule through hydrogen bonding in a tetrahedral manner and form open-cage like structure. When ice melts the molecule come closer to each other so the density of water in liquid state is more than in solid state thus ice floats on water.
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CHEMICAL BONDING
O
Ans.16 The structure of molecule is CH3 – C
. O H The double bonded carbon oxygen bond is shorter than the other. A double bond between two atoms is always stronger and shorter than a single bond between the same atoms and HCOO– have same bond length.
Ans.17 This is due to the presence of resonance in molecule
Each bond has part of the character of single bond and part of character of a double bond. The electron are delocalised over the three atoms.
Ans.18 There are two lone pairs, but only one of them make a co-ordinate bond with an empty orbital on hydrogen atom. There are three bond in the H3O+ ion all the three are identical but two of them are covalent and one coordinate is the dot and cross diagram is –
+
H × O × H
H
Ans.19 In MgCl2, Mg is sp hybridised while in SnCl2, Sn is sp2 hybridised. Ans.20 In N–atom d-orbital are not present while in P-atom d-orbital are present and ns electron can be shifted to nd orbital. Ans.21 Due to inert pair effect lead prefer's form divalent compound.
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CHEMICAL BONDING
Ans.22 Because the central atom in each of the three molecules CH4, NH3 and H2O undergoes sp3 hybridisation. In CH4 no lone pair is present the. In NH3 one lone pair is present, while in water molecule two lone pair are present. The lone pair and bond pair electrons repulsion changed the molecules shapes
Ans.23 Because the size of the I– ion is bigger than F– ion. In CaI2 therefor more polarisation is present thus CaF2 is more ionic than CaI2. Ans.24 Na+ ion is solvated by ion-dipole interaction while NH4+ ion is solvated by H–bonding which is a stronger attraction force. Ans.25 Since all the four valencies of carbon are identical, 2s electron is shifted to one of the vacant p-orbital with the result four unpaired orbitals are present, these undergo hybridisation and form four hybrid orbitals. Ans.26 Because the size of nitrogen is less then the size of chlorine, therefore, electron density in nitrogen is more than that of chlorine, so nitrogen forms H-bonding leading to association of molecules. Hence NH3 is a liquid. H-bonding is not possible with chlorine. Ans.27 CO2 has linear structure while SO2 and H2S have V-shaped string.
O
H
= 1.84
H
Ans.28 It is due to different directions of moments of the N – H and N – F bond.
N F
F F So according the structure the net dipole moment of NH3 is nearly zero by vector rule.
Ans.29 The carbonate ion exhibits resonance and all the three C – O bonds are equal. Ans.30 It is due to larger electronegativity of F – atoms of NF3. The lone pair is most difficult to be donated. Ans.31 In BaSO4, lattice energy is more than it hydration energy. Ans.32 Because normal element acquire inert gas configuration in compounds so all the orbitals are double occupied and thus compound are diamagnetic. Ans.33 IF7 is exist because electro negative difference & size of iodine is big which is help to expansion of iodine orbital is possible but not in Cl. Ans.34 Xe, it has highest vander waal's force because of having more electron.
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CHEMICAL BONDING
O
Ans.35 At room temp.
N
; there is unpaired electron at nitrogen. On cooling dimer exists O
; and there is no unpaired electron with it. Ans.36 KO2 has O2– ion; there exist odd electron bond as shown in structure as shown in structure. Odd electron bond makes is it paramagnetic. Ans.37 In NH4+ there are four bond pairs and no lone pair like NH3. However, NH3 has only three bond pair and one lone pair, hence its bond angle is less due to lone pair-bond pair repulsion. Ans.38 Chlorine atoms lies in sp3d hybrid state. Three lone pairs are oriented along the corners of triangular plane.
Chlorine atom lies in sp3 hybrid state. Two lone pairs are oriented along two corners of tetrahedra. [ClF2+] Ans.39 In H2O, bond angle is less than 109º28 due to lone pair and bond pair repulsion but in ether due to strong mutual between two alkyl groups bond angle become greater than 109º28.
Ans.40 In N(CH3)3; there is sp3 hybridisation at nitrogen but due to lone pair bond pair repulsion shape becomes pyramidal.
In tri silyl ammonia, there is vacant d-orbital at silicon hence, formation p – d back bonding take place and geometry becomes planar.
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CHEMICAL BONDING
Ans.41 This molecule has bridged carbon with three centred bonding. Thus, covalency of carbon is maintained. Ans.42 NH3, due to the presence of H – bonding.
Ans.43 All fluorine atom do not lie in plane molecule is not symmetrical thus it will be polar. Ans.44 Because H2O2 decomposes readily to H2O at room temperature 2H2O2 2H2O + O2 Ans.45 OF2 > ClO2 Because electronegativity difference in F – O is greater is greater than in O – Cl. Ans.46 Because both the bonds, are physical bonds. Ans.47 I– is largest, so polarisation of I– by Ag+ is maximum and the shared electrons are easily transferred to higher energy level in visible light. So AgI is deep yellow, but in chloride; electrons are not transferrer easily so AgCl is colourless. Ans.48 In CCl4 (i) Octet of C is complete (ii) there is no electron pair on C (iii) there is no vacant d-orbital. so CCl4 neither donates nor accepts electron and thus does not react with water. In SiCl4 due to presence of vacant d-orbital in Si. It accept electrons and thus reacts with OH– of water and forms Si(OH)4 (unstable) which gets converted to more stable SiO2 by loss of two water molecule. Ans.49 Because oxygen is a second period element, so due to absence of 3d orbitals the valency of oxygen will not be six. But in sulphur due to presence of 3d orbitals the valency of sulphur may be six
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CHEMICAL BONDING
[ IIT JEE & REE 1987 to 2006 ] 1987 Q 1. Q 2. Q 3.
The first IP in eV of N & O are respectively given by : (A) 14.6, 13.6 (B) 13.6, 14.6 (C) 13.6, 13.6
(D) 14.6,14.6
Atomic radii of F & Ne in Angstrom are respectively given by (A) 0.72, 1.60 (B)1.60, 1.60 (C) 0.72, 0.72
(D) none
The electronegativity of the following elements increases in the order (A) C, N, Si, P (B) N, Si., C, P (C) Si, P, C, N
(D) P, Si, N, C
Q 4.
The bond between carbon atom (1) & carbon atom (2) in compound N C-CH = CH2 involves the hybrids as: 1 2 (A) sp2 & sp2 (B) sp3 & sp (C) sp & sp2 (D) sp & sp
Q 5.
Amongst the trihalides of nitrogen, which one is least basic? (A) NF3 (B) NCl3 (C) NBr3
(D) NI3
Hydrogen bonding is maximum in : (A) Ethanol (B) Diethylether
(D) Triethylamine
Q 6.
(C) Ethyl chloride
Q 7.
Bromine can be liberated from potassium bromide solution by action of : (A) Iodine solution (B) Chlorine water (C) sodium chloride (D) Potassium iodide
Q 8.
Silver chloride is sparingly soluble in water because its lattice energy is greater than _______ energy.
Q 9.
_________ phosphorus is reactive because of its highly strained tetrahedral structure.
Q 10. In benzene carbon uses all the three p-orbitals for hybridisation. Q 11. sp2 hybrid orbitals have equal S & P character. Q 12. In group I A of alkali metals, the ionisation potential decreases down the group. Therefore lithium is a poor reducing agent. Q 13. Explain the molecule of magnesium chloride is linear whereas that of stannous chloride is angular.
1988 Q 14. Give reason carbon oxygen bond lengths in formic acid are 1.23 A0 & 1.36 A0 and both the carbon oxygen bonds in sodium formate have the same value i.e. 1.27 A0. Q 15. Give reason that valency of oxygen is generally two whereas sulphur shows of 2, 4 & 6. Q 16. The first Ionization potential of Na, Mg, Al & Si are in the order (A) Na < Mg > Al < Si (B) Na > Mg > Al > Si (C) Na < Mg Si (D) Na > Mg > Al < Si Q 17. The species which the central atom uses sp2 hybrid orbitals in its bonding is (A) PH3 (B) NH3 (C) CH3+ (D) SbH3 Q 18. The molecule that has linear structure is (A) CO2 (B) NO2
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(C) SO2
(D) SiO2
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Q 19. The statements that are true for the long form of the P.T.are (a) It reflects the sequence of filling the electrons in the order of sub-energy levels s, p, d & f (b) It helps to predict the stable valency states of the elements (c) It reflects trends in physical & chemical properties of elements (d) It helps to predict the relative ionicity of the bond between any two elements. Q 20. Arrange N2, O2, F2, Cl2 in increasing order of bond dissociation energy. Q 21. Arrange CO2, N2O5, SiO2, SO3 is the increasing order of acidic character. Q 22. Arrange HOCl, HOClO2, HOClO3, HOClO in increasing order of thermal stability.
1989 Q 23. Explain the first I.E. of carbon atom is greater than that of boron atom whereas the reverse is true for the second I.E. Q 24. The compound which has zero dipole moment is (A) CH2Cl2 (B) BF3 (C) NF3
(D) ClO2
Q 25. Which of the following is paramagnetic (A) O2(B) CN-
(C) CO
(D) NO+
Q 26. The molecule which has pyramidal shape is (A) PCl3 (B) SO3
(C) CO32-
(D) NO3-
Q 27. The compound in which *C uses its sp3 hybrid orbitals for bond formation is : (A) H*COOH (B) (H2N)*CO (C) (CH3)3*COH (D) CH3*CHO Q 28. The C - H bond distance is the longest in (A) C2H2 (B) C2H4
(C) C2H6
(D) C2H2Br2
Q 29. Which one of the following is the smallest in size (A) N3(B) O2(C) F–
(D) Na+
Q 30. The number of sigma and pi bonds in 1-butene-3-yne are (A) 5 sigma 5 pi (B) 7 sigma 3 pi (C) 8 sigma 2 pi
(D) 6 sigma 4 pi
Q 31. Sodium sulphate is soluble in water whereas Barium sulphate is sparingly soluble because (A) the hydration energy of sodium sulphate is more than its lattice energy (B) the lattice energy of barium sulphate is more than its hydration energy (C) the lattice energy has no role to play in solubility (D) the hydration energy of sodium sulphate is less than its lattice energy Q 32. All the Al - Cl bond in Al2Cl6 are equivalent Q 33. Both potassium ferrocyanide & potassium ferricyanide are diamagnetic
1990 Q 34. Amongst the following the one having highest I.E. is (A) [Ne] 3 s2 3 p1 (B) [Ne] 3 s2 3 p3 (C) [Ne] 3 s2 3 p2
(D) [Ar] 3 d0 4 s2 4 p3
Q 35. The valence atomic orbitals on C in silver acetylide is ________ hybridised Q 36. The presence of polar bonds in a polyatomic molecule suggests that the molecule has non-zero dipole moment. Q 37. Write the two resonance structures of N2O that satisfy the octet rule. NARAYANA II T ACAD EMY- C P 3 ,
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1991 Q 38. The hybridisation of C atoms in C-C single bond of HC C - CH = CH2 is (A) sp3 - sp3 (B) sp2 - sp3 (C) sp- sp2 (D) sp3 - sp Q 39. The linear structure is assumed by : (A) SnCl2 (B) NCO-
(D) NO2+
(C) CS2
(E) SO2
Q 40. Nitric oxide, though an odd electron molecule , is diamagnetic is liquid state Q 41. Arrange the following as stated. (i) Increasing order of ionic size : N3- , Na + , F- , O2- , Mg2+ (ii) Increasing order of basic character : MgO, SrO, K2O, NiO, Cs2O (iii) Increasing strength of H - bonding . (X...H - X) O, S, F, Cl, N (iv) Increasing order of extent of hydrolysis CCl4, MgCl2, AlCl3, PCl5, SiCl4 Q 42. Write two resonance structures of ozone which satisfy the octet rule.
1992 Q 43. The statement that is not correct for the periodic classification of element is : (A) The properties of elements are periodic functions of their atomic number (B) Non metallic elements are lesser in number than metallic (C) The IE1 of elements along a period do not vary in a regular manner with increase in atomic number (D) For transition elements the d-subshells are filled with electrons monotonically with increase in atomic numbers Q 44. The type of hybrid orbitals used by the chlorine atom in ClO2- is (A) sp3 (B) sp2 (C) sp
(D) none
Q 45. The CN- & N2 are isoelectronic. But in contrast to CN- , N2 is chemically inert because of (A) Low bond energy (B) Absence of bond polarity (C) Unsymmetrical electron distribution (D) Presence of more number of electron in bonding orbitals Q 46. The maximum possible number of hydrogen bonds a water molecule can form is (A) 2 (B) 4 (C) 3 (D) 1 Q 47. The species that do not contains peroxide ions are (A) PbO2 (B) H2O2 (C) SrO2
(D) BaO2
Q 48. Which of the following have identical bond order (A) CN(B) O2(C) NO+
(D) CN+
Q 49. The molecules that will have dipole moment are (A) 2 , 2 - dimethyl propane (C) trans - 2 - pentene Q50.
(B) cis - 3 -Hexene (D) 2 , 2 ,3 , 3-tetra methyl butane
Amongst the three isomers of nitrophenol, the one that is least soluble in water is _______ .
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1993 Q51.
The decreasing order of E A of F, Cl, Br is F > Cl > Br.
Q52.
Diamond is harder than graphite.
Q53.
The basic nature of hydroxides of group 13 (IIIB) decreases progressively down the group.
Q54.
The tendency for catenation is much higher for C than Si.
Q55.
HBr is stronger than HI because of H-bonding.
Q56.
The dipole moment of CH3F is greater than CH3Cl.
Q57.
The D.M. of KCl is 3.336 × 10-29 Coulomb-metre which indicates that it is highly polar molecule. The interatomic distance between K+ and Cl– in this molecule is 2.6 × 10-10m. Calculate the dipole moment of KCl molecule if there were opposite charges of one fundamental unit located at each nucleus. Calculate the % ionic character of KCl.
1994 Q58.
The kind of delocalization involving sigma bond orbitals are called___________.
Q59.
The two types of bonds present in B2H6 are covalent and _________.
Q60.
Explain PCl5 is formed but NCl5 cannot.
Q61.
Using VSEPR theory, identify the type of hybridisation and draw the structure of OF2. What are oxidation states of O and F.
1995 Q62.
Explain why the dipolemoment of NH3 is more than that of NF3.
Q63.
The experimentally determined N-F bond length in NF3 is greater than the sum of single bond covalent redii of N and F. Explain.
1996 Q64.
Explain why the dipole moment of NH3 is more than that of NF3.
Q65.
The order of increasing thermal stabilities of K2CO3, MgCO3, CaCO3, BaCO3 is _____
Q66.
Identify isostructural pairs from NF3, NO3–, BF3, H3O+, HN3.
Q67.
(i) The number and type of bonds between two C-atom in CaC2 are (A) 1 sigma 1 pi (B) 1 sigma 2 pi (C) 1 sigma 1 and half pi (D) 1 sigma (ii) Which is correct for CsBr3. (A) it is a covalent compound (C) it contains Cs+ and Br3- ions
(B) it contains Cs3+, Br– ions (D) it contains Cs+, Br– and lattice Br2 molecule
Q68.
Arrange in increasing order of dipole moment. Toluene, m-dichlorobenzene, O-dichlorobenzene, P-dichlorobenzene.
Q69.
When N2 goes to N2+, the N-N distance ________ and when O2 goes to O2+, the O-O bond distance __________.
Q70.
Explain the difference in the nature of bonding in LiF and LiI.
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Q71.
Compare qualitatively the Ist and IInd IP of Cu and Zn. Explain the observation.
Q72.
Which is least stable and has doubtful existence? (A) CI4 (B) GeI4 (C) SnI4
Q73. Q74.
Which one of the following oxides is neutral. (A) CO (B) SnO2
(C) ZnO
(D) PbI4 (D) SiO2
The decreasing order of acid strength of ClOH, BrOH, IOH.
1997 Q75.
The incorrect statement among the following is (A) the IE1 of A1 is less than IE1 of Mg (B) the IE2 of Mg is greater than IE2 of Na (C) the IE1 of Na is less than IE1 of Mg (D) the IE3 of Mg is greater than IE3 of Al
Q76.
Among KO2, AlO2-, BaO2 and NO2+ unpaired electron is present in (A) NO2+ and BaO2 (B) KO2 and AlO2(C) KO2 only (D) BaO2 only
Q77.
Property of the alkaline earth metals that increases with their atomic number is (A) IE (B) solubility of their hydroxides (C) solubility of their sulphates (D) electronegativity
Q78.
Among N2O, SO2, I3+ and I3-, the linear species are ________ and _________.
Q79.
Among PCl3, CH3+, NH2- and NF3, __________ is least relative towards water.
Q80.
The P–P–P angle in P4 molecule is _________.
Q81.
Among the following compounds the one that is polar and has the central atom with sp2 hybridisation is (A) H2CO3 (B) SiF4 (C) BF3 (D) HClO2
Q82.
Which contains both polar and non polar bonds (A) NH4Cl (B) HCN (C) H2O2
(D) CH4
Q83.
The critical temp. of water is higher than that of O2, because the H2O molecule has: (A) Fewer electron than O2 (B) 2 covalent bonds (C) V– shape (D) dipole moment
Q84.
Which one has sp2 hybridisation (A) CO2 (B) SO2
(C) N2O
(D) CO
Q85.
Compounds that formally contain Pb4+ are easily reduced to Pb2+. The stability of lower oxidation state is due to ______________.
Q86.
Arrange in order of increasing radii, Li+, Mg2+, K+, Al3+.
Q87.
Arrange BeSO4, MgSO4, CaSO4, SrSO4 in order of decreasing thermal stability.
Q88.
Anhydrous AlCl3 is covalent. From the data given below predict whether it would remain covalent or become ionic in aqueous solution. [IE for Al = 5137 kJ/mol] H hydration for Al3+ = –4665 kJ/mol; Hhydra for Cl- = -381kJ/mol.
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CHEMICAL BONDING
1998 Q89.
The geometry and the type of hybrid orbitals present about the central atom in BF3 is (A) linear, sp (B) trigonal planar, sp2 (C) tetrahedra sp3 (D) pyramidal, sp3
Q90.
State whether True of False: (i) F atom has less negative E A than 1 atom. (ii) LiCl is predominantly a covalent compound. (iii) Al(OH)3 is amphoteric in nature.
Q91.
Interpret the non-linear shape of H2S molecule and non planar shape of PCl3 using VSEPR theory.
Q92.
Write electronic structure of ozone.
1999 Q93. (i)
(ii)
The correct order of increasing C-O bond length of, CO, CO32-, CO2 is (A) CO32- < CO2 < CO (B) CO2 < CO32- < CO 2(C) CO < CO3 < CO2 (D) CO < CO2 < CO32In the dichromate anion (A) 4 Cr - O bonds are equivalent (C) all Cr - O bonds are equivalent
(B) 6 Cr - O bonds are equivalent (D) all Cr - O bonds are non equivalent
(iii)
The geometry of H2S and its dipole moment are (A) angular and non zero (B) angular and zero (C) linear and non zero (D) linear and zero
(iv)
In compounds type E Cl3, where E = B, P, As or Bi, the angles Cl-E-Cl for different E are in the order (A) B >P = As = Bi (B) B > P >As > Bi (C) B < P = As = Bi (D) B < P < As Cl-
(D) P3+ > P5+
(vii)
Give reasons for the following in one or two sentences only. (a) BeCl2 can be easily hydrolyed (b) CrO3 is an acid anhydride
(viii)
Discuss the hybridisation of C - atoms in allene (C3H4) and show the - orbital overlaps.
(xi)
Explain why o-hydroxybenzaldehyde is a liquid at room temperature, while p-hydroxybenzaldehyde is a high melting solid.
2000 Q94. (i) (ii)
The correct order of radii is (A) N < Be < B (B) F- < O2- < N3-
(C) Na < Li < K
The number of P-O-P bonds in cyclic metaphosphoric acid is (A) zero (B) two (C) three
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(D) Fe3+ < Fe2+ < Fe4+ (D) four
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CHEMICAL BONDING
(iii)
The hybridisation of atomic orbitals of N in NO2+, NO3– and NH4+ are respectively. (A) sp, sp3, sp2 (B) sp, sp2, sp3 (C) sp2, sp, sp3 (D) sp2, sp3, sp
(iv)
The correct order of acid strength is (A) Cl2O7 > SO2 > P4O10 (C) Na2O > MgO > Al2O3
(B) CO2 > N2O5 > SO3 (D) K2O > CaO > MgO
(v)
Molecular shapes of SF4, CF4 and XeF4 are respectively (A) the same with 2, 0 and 1 lp (B) the same with 1, 1 and 1 lp (C) different with 0, 1, 2 lp (D) different with 1, 0, 2 lp
Q95.
The IE1 or Be is greater than that of B.
Q96.
Amongst H2O, H2Se and H2Te the one with the highest boiling point is (A) H2O of H-bonding (B) H2Te of higher molecular weight (C) H2S of H-bonding (D) H2Se of lower molecular weight
Q97.
Write the M.O. electron distribution of O2. Specify its bond order and magnetic property.
Q98.
Draw the molecular structures of XeF2, XeF4 and XeO2F2, indicating the location of l.p. of electrons.
Q99.
Give reason why elemental nitrogen exists as a diatomic molecules where as elemental phosphorus is a tetratomic molecule.
Q100. The set with correct order of acidity (A) HClO < HClO2 < HClO3 < HClO4 (C) HClO < HClO4 < HClO3 < HClO2
(B) HClO4 < HClO3 < HClO2 < HClO (D) HClO4 < HClO2 < HClO3 < HClO
Q101. The no. of S-S bonds in S3O9 is (A) 3 (B) 2
(C) 1
(D) zero
Q102. The common features of the species CN–, CO, NO+ are (A) bond order three and isoelectric (B) B.O. = 3 and weak field ligand (C) B.O. 2 and acceptor (D) isoelectric and weak field ligands. Q103. Between SiCl4 and CCl4 only SiCl4 reacts with H2O. Q104. SiCl4 is ionic and CCl4 is covalent. Q105. The complex ion which has no. “d” e- in the central atom is (A) [MnO4–] (B) [Co(NH3)6]3+ (C) [Fe(CN)6]3–
(D) [Cr(H2O)6]3+
Q106. The set representing correct order of IP1 is (A) K>Na>Li (B) Be>Mg>Ca
(D) Fe>Si>C
(C) B>C>N
Q107. The correct order of hybridisation of the central atom in the following species NH3, [PtCl4]2-, PCl5 and BCl3 is (A) dsp2, dsp3, sp2, sp3 (B) sp3, dsp2, dsp3, sp2 2 2 3 3 (C) dsp , sp , sp , dsp (D) dsp2, sp3, sp2, dsp3 Q108. The correct order of radii is (A) N MgO
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CHEMICAL BONDING
Q 112. Molecular shapes of SF4 , CF4, and XeF4 are (A) the same, with 2, 0 and 1 l.p. (B) the same, with 1, 1 and 1 l.p. (C) different, with 0, 1 and 2 l.p. (D) different, with 1, 0 and 2 l.p. Q 113. The first I E of Be is greater than that of B because 2 p orbital is lower in energy than 2S. Q 114. Amongst H2O, H2S, H2Se and H2Te, the one with the highest Boiling point is (A) H2O; of H- bonding (B) H2Te; of higher molecular weight (C) H2S; of H- bonding (D) H2Se; of lower molecular weight Q 115. Draw the structures of [Co(NH3)6]3+ , [Ni(CN)4]2+ and Ni(CO)4. Write the hybridisation of atomic orbitals of the transitional metal in each case. Q 116. Write the M.O. electron distribution of O2, Specify its bond order and magnetic property. Q 117. Draw the molecular structures of XeF2, XeF4 and XeO2F2, indicating the location of lone pair(s) of electrons. Q 118. Give reason(s) why elemental nitrogen exists as a diatomic molecule whereas elemental phosphorus is a tetraatomic molecule.
2001 Q 119. Draw the structure of Diborane. Q 120. State hybridisation of Cr in [Cr(NH3)4Cl2]Br and [Cr(NH3)4ClBr] Cl. Calculate their magnetic moments (spin-only). Q 121. Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BF3 & NH3. (A) N : tetrahedral sp3, B: tetrahedral sp3 (B) N : pyramidal sp3, B: pyarmidal sp3 (C) N : planar sp3, B: planar sp3 (D) N : pyramidal sp3, B: tetrahedral sp3 Q 122. Identify the correct order to solubility of Na2S, CuS and ZnS in aq. medium (A) CuS > ZnS > Na2S (B) ZnS > Na2S > CuS (C) Na2S > CuS > ZnS (D) Na2S > ZnS > CuS Q 123. Identify the correct order of boiling point of CH3 CH2 CH2 CH2 OH; CH3 CH2 CH2 CHO; (A) 1> 2> 3 (B) 3> 1> 2 (C) 1> 3 > 2
CH3 CH2 CH2 COOH (D) 3> 2 >1
Q 124. Identify the correct order of acidic acid strengths , CO2 CuO, CaO, H2O (A) CaO > CuO > H2O > CO2 (B) H2O < CuO < CaO < CO2 (C) CaO < H2O < CuO < CO2 (D) H2O < CO2 < CaO M - Cl > M - Br > M - I BiH3 < SbH3 < AsH3 < PH3 < NH3 HOCl < HOClO < HOClO2 < HOClO3
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(b) (d) (f) (h)
HF < HCl < HBr < HI Ga2O3 < As2O3 < GeO2 < ClO2 H2O < H2S < H2Se < H2Te HOI < HOBr < HOCl
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31.
CHEMICAL BONDING
(a) Cl2O 2x- 2 = 0 Oxiation state of Cl in Cl2O = + 1 (b) ClO2 x + 2 (-2) = 0 Oxiation state of Cl in ClO2 = + 4 (c) KBrO3 +1 + x + 3 (-2) = 0 x-5=0 Oxiation state of Br in KBrO3 = + 1 (d) NaClO4 +1 + x + 4 (-2) = 0 x-7=0 Oxiation state of Cl in NaClO4 = + 7
x=+1 x = +4
x = +5
x = +7
32.
Catenation : The peroperty of forming chains of identical atoms is called catenation. The property of catenation depends upon the strength of atom atom bodn. the greater the strength of atom-atom bond, greater is the extent of catenation. In the elements of the 16th group, s shows strongest tendency of catenation because of s-s bond strength. Polysulphide showsing catenation are H—S— S— H H— S— S— S— H (polysulphides or H— S— S— S— S— H polysulphanes) Besides these polysulphanes (H — Sn — H) polysulphuric acids HO3S — Sn — SO3H and various allotropic forms of the elements containing different sized Sn rings and chains are also known. The s-s bond is also found in number of compounds of biological importance such as cysteine, proteins, enzymes etc.
33.
(a) Shape of SiF4 : Tetrahedral Si : Shows sp3 hybridization (b) SiF62- hybridizatio sp3d2 shape octahedral (c) PF5P : Hybridization sp3d P = 3s23p3 Geometry : Trigonal bipyramid Two axial bonds 15 pm longer than 3 equatorial bonds.
NARAYANA II T ACAD EMY- C P 3 ,
e = equatorial bonds, a = axial bonds
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CHEMICAL BONDING
XII NCERT Question & Solution 1.
Draw the shape (bondary surfaces) of the following orbitals ; (a) 2py (b) 3dz2 (c) 3dx2-y2 (Show coordinate axes in you sketches.)
2.
Discuss the similarities and differences between a 1s and 1 2s orbital.
3.
For each of the following pair of hydrogen orbitals, indicated which is higher in energy ; (a) 1s, 2s ; (b) 2p, 3p (c) 3dxy, 3dyz, (d) 3s, 3d (e) 4f, 5s
4.
Which orbital in each of the following pairs is lower in energy in amny-electron atom ? (a) 2s, 2p (b) 3p, 3d (c) 3s, 4s (d) 4d, 5f
5.
Explain the meaning of the symbol 4d6.
6.
The ground state electron configurations listed here are incorrect. Explain what mixtakes have been made in each and write the correct electron configuration. Al : 1s22s22p43s23p3 B : 1s22s22p5 C : 1s22s22p6
7.
Draw orbital diagrams for atoms with the following electronic configuraiton : (a) 1s22s22p5 (b) 1s22s22p6 (c) 1s22s22p63s23p64s23d7
8.
Two p orbitals from one atom and two p orbitals from another atom are combined to form molecuar orbitals. How many MOs will result from this combination ? Explain.
9.
Show the shapes of bonding and antibonding MOs formed by combination of (a) two s orbital (b) two p orbitals (side to side)
10.
How do the bonding and antibonding MOs formed from a given pari to AOs compare to each other with respect to (a) energy (b) presence of nodes (c) internuclear electron density ?
11.
Arrange the following species in order of increasieng stability : Li2, Li2+, Li2Justify your choice with a molecular orbitla energy level diagram.
12.
Use molecular orbitla thoery to explain why the Be2 molecule cues not exist.
13.
Explain why the bond order of N2 is greater than N2+, but the bodn order of O2 is less than that of O2+.
14.
Compare the relative stability fo the followign species and indicating their magnetic properties (diamagnetic or paramagnetic) ; O2, O2+, O2- (superoxide), O22- (peroxide ion)
15.
Explain the significance of bond order. Can bond order be used for quantiative compoarisons of the strengths of chemcial bonds ?
16.
What is the energy gap in band theory ? compare its size in conductors, semiconductors and insulators.
17.
Which of the following substances exhibit H-bonding ? Draw the H bonds between two molecules of the substnace where appropriate :
(A) CH3CH2OH
O || (B) CH 3 C OH
O || (C) CH 3 C CH 3
O || (D) CH 3 C NH 2
18.
How can one nonpolar molecule inducea dipole in a nearby nonpolar molecule ?
19.
What type(s) of intermolecualr forces exist between the following pairs ? (a) HBr and H2S (b) Cl2 and CBr4, (c) I2 and NO3- and (d) NH3 and C6H6 ?
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CHEMICAL BONDING
(Hint : To identify intermolecular forces, it is useful to classify the species being considered as (1) nonpolar molecules, (2) polar molecules, and (3) ions. Keep in mind that dispersion forces exist between all species.)
SOLUTIONS
1.
(a) 2py
(b) 3dz2
(c) 3dx2–y2
2.
Similarities between 1s and 2s (1) both are non-directional. (2) spherical in shape (3) the density of charge cloud is maximum at the nucleus and decrease with increase in distance from the nucleus both in 1s and 2s. Difference : (1) The principal quantum no. for 1s orbital is 1 but in 2s is 2. (2) the 2s have 1 node or nodal plane but 1s have no any node. (3) the energy of 2s orbital is higher than 1s orbital.
3.
(a) 2s (b) 3p (c) 3dxy & 3dz have equal energy. (d) for hydrogen 3s and 3d orbital have identical energy. (e) 5s has higher energy than 4f orbital for hydrogen.
4.
According to Aufbau rule, energy of the orbital increases 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s ........... In multi-electron systems, the relative order of energies can also be predicted with the help of ‘n + 1’ rule or BOHR-BURY RULE
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CHEMICAL BONDING
103
According to this (i) in neutral atoms, a subshell with lower value of (n + 1), has lower energy (ii) If two subshell have equal value of (n + 1), the subshell with lower value of n has lower energy (a) out of 2s and 2p, 2s is filled first ; its n + 1 is lower or 2s has lower energy than 2p. (b) Energy of 3p is lower than 3d 3p 3d n=3 n=3 =1 =2 n+1=4 n+1=5 (c) 3s is lower in energy than 4s 3s 4s n=3 n=4 =0 =0 n+1=3 n+1=4 (d) 4d is lower in energy than 5f 4d 5f n=4 n=5 =2 =3 n+1=6 n+1=8 5.
4d6 Principal energy level = 4 Number of electrons = 6 Subshell = d ‘4’ indicates the principal energy level. ‘4d’ indicates the d-subshell of the 4th valence shell. 6 indicated the number of electrons present in ‘4d’ subshell.
6.
Al = 1s2 2s2 2p6 3s2 3p1 Here 3s2 3p3 is filled first before completing 2p subshell, which as to attain 6 electrons before electron start filling 3s or 3p. B = 1s2 2s2 2p1 Atomic no. of B = 5 F = 1s2 2s2 2px2 2py2 2p1 [ Atomic no. of F = 9]
7.
(a) F = 9 = 2, 7
(b) P = 15 = 2, 8, 5
(c) Co = 2, 8., 15, 2
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8.
CHEMICAL BONDING
For example if we take oxygen molecule both the oxygen atoms contains two p–orbitals. If we take one p–orbital from one oxygen atom and another p–orbital from another oxygen atom then
If we take another set of p-orbitals each from each oxygen atom then
So we get 4 molecular orbitals from the combination of two p-orbitals from two atoms. 9.
(a) From two s–orbitals The formation of molecular orbitals by linear combination of two s orbital is as shown below
(b) From two p–orbital (side–side) When two 2p orbitals which have mutually parallel axis interact to give rise molecular orbitals as shown below
side wise approach 10.
In combine state
(A + B)2 = A2 + B2 + 2AB NARAYANA II T ACAD EMY- C P 3 ,
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CHEMICAL BONDING
Uncombined state
(A – B)2 = A2 + B2 – 2AB Nuclear – nuclear repulsion will be more energy in more stability is less. (a) Energy - Energy of the antibonding orbital is raised above the energy of the parent atomic orbitals that have combined and the energy of the bonding orbital has lowered than the parent orbitals. The total energy of two molecular orbitals however remains the same as that of the two original atomic orbitals. (b) presence of node & (c) internuclear electron density - bonding molecular orbital most of the electron density is located between the nuclei of the bonded atoms and hence the repulsion between the nuclei is very low while in an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei, as a matter of fact there is a nodal plane (i.e., plane in which the electron density is zero.) 11.
The electronic configuration of Li is 1s22s1. The M. O. energy level diagram of Li2 is as follows For Li2
Bond order
=
1 (Nb – Na) 2
=
1 2 (4 – 2) = = 1 2 2
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CHEMICAL BONDING
For Li2+ one electron should be removed from the bonding molecular orbital so bond order will be =
1 (Nb – Na) 2
1 1 (3 – 2) = = 0.5 2 2 – for Li2 one electron should be added to the antibonding molecular orbital so bond order will be
=
1 1 (4 – 3) = = 0.5 2 2 The greater the bond order of a diatomic molecule the more stable it will be So in between Li2, Li2+, Li2– Li2 is more stable than both Li2+, Li2– stability order is Li2, Li2+, Li2– In Li2+, the positive charge is more that is the nucleus will attract the electrons so size will be less than Li2– & stability is more.
=
12.
The electronic configuration of Be is 1s22s2 so the M. O. diagram is as follows
Here the Bond order
=
1 (Nb – Na) 2
1 (4 – 4) = 0 2 If the bond order is zero i.e., the molecule doesn’t exist.
=
NARAYANA II T ACAD EMY- C P 3 ,
INDRA VIHAR, TALWANDI, KOTA-324005 (Raj.), Ph.N o.-0744-3200119,website : www.narayanaiitkota.com
107
CHEMICAL BONDING
13.
The electronic configuration of N2 molecule is (1s)2 (*1s)2 (2s)2 ( 2s)2 (2px)2 (y2py)2 So the bond order is =
(z2pz)2
1 6 (10 – 4) = = 3 2 2
1 5 (9 – 4) = = 2.5 2 2 (one electron is removed from bonding molecular orbital) So the bond order of N2+ is less than N2. The electronic configuration of O2 molecule is (1s)2 (1s*)2 (2s)2 (2s*)2 (2px)2 (py2py)2 (pz2pz)2
For N2+ the bond order is =
Bond order of O2 is =
(*y2py)1
(*z2pz1)
1 4 (10 – 6) = = 2 2 2
1 5 (10 –5) = = 2.5 2 2 (i.e., One electron is removed from antibonding orbital i.e., p*) so bond order of O2+ is greater than O2
bond order of O2+ is =
14.
The electronic configuration of O2 is as follows O2 (1s)2 (1s*)2 (2s)2 (*2s)2 (2pz)2 (2px2 = py2) The bond order of O2, O2+, O2–, O22– are as follows for O2 :
1 4 (10 – 6) = = 2 2 2
For O2+
1 5 (10 – 5) = = 2.5 2 2
For O2– :
(2px1 = *2py1)
1 3 (10 – 7) = = 1.5 2 2
1 2 (10 – 8) = = 1 2 2 As the bond order increases stability increase so the stability order is O2+ > O2 > O2– > O22– Magnetic Properties for O2 there are two unpaired electron so it is paramagnetic in nature. For O2+ it also contain one unpaired electron so it is also paramagnetic in nature. For O2– it also has one unpaired electron so it is also paramagnetic in nature. but O22– is diamagnetic due to absence of unpaired electron.
For O22– :
1 (Nb – Na) 2 A positive bond order means a stable molecule while a negative (or) zero bond order means an unstable molecule. Bond length decreases bond order increases. Yes bond order can be used for quantitative comparison of the strength of the chemical bonds.
15.
Bond order =
16.
The highest occupied energy bond is valence bond while the lowest unoccupied energy bond is conduction band. The energy gap between the top of the valence bond and the bottom of the conduction band is called the energy gap (Eg). In case of insulators the energy gap is very large and in case of semiconductor the energy gap is very small and in case of conductors the energy gap is zero.
NARAYANA II T ACAD EMY- C P 3 ,
INDRA VIHAR, TALWANDI, KOTA-324005 (Raj.), Ph.N o.-0744-3200119,website : www.narayanaiitkota.com
108
CHEMICAL BONDING
Formation of energy bands in (a) Sodium crystal (b) Magnesium crystal
17.
(A)
(intermolecule hydrogen bonding)
(B) No H–bonding is present in the molecule The energy of a hydrogen bond is of the order 40 kJ/mole.
(C)
(intermolecule hydrogen bonding)
(D)
(intermolecule hydrogen bonding)
18.
If in an atom the electrons are moving at some distance from the nucleus. At any instant it is likely that the atom has a dipole moment created by the specific position of electrons. It is instantaneous dipole which lasts for just a fleeting moment and in the next instant the electron are in different positions and the atom has a new instantaneous dipole and so on.
19.
(a) (b) (c) (d)
HBr & H2S Cl2 & CBr4 I2 & NO3NH3 & C6H6
NARAYANA II T ACAD EMY- C P 3 ,
dipole-dipole interaction (keesen force) Instaneous dipole-instaneous induced dipole (London force) dipole induce dipole dipole induce dipole INDRA VIHAR, TALWANDI, KOTA-324005 (Raj.), Ph.N o.-0744-3200119,website : www.narayanaiitkota.com
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