D-Block Elemnets Theory_E

CHEMISTRY

Transition Elements s (d-Block Elements) INTRODUCTION : Four series of elements are formed by filling the 3d, 4d, 5d and 6d subshells of electrons. Collectively these comprise the d-block elements. They are often called as ‘transition elements because their position in periodic table is between the s-block and p-block elements. Their properties are transitional between the highly reactive metallic elements of s-block (which form ionic compounds) and the elements of p-block (which are largely covalent). Typically the transition elements have an incompletely filled d level. A transition element may be defined as the element whose atom in ground state or ion in one of common oxidation states, has partly filled d-sub shell i.e. having electrons between 1 to 9. Group 12 (the zinc group) elements have completely filled d-orbitals in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. The general electronic configuration of d-block elements is (n–1) d1–10 ns1–2, where n is the outer most shell. However, palladium does not follow this general electronic configuration. It has electron configuration [Kr]36 4d10 5s0 in order to have stability. 

st transition series or 3d series corresponding to filling of 3d sub-shell consists of the following 10 elements of 4th period. 21



Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and 30Zn.

nd transition series or 4d series corresponding to filling of 4d sub-shell consists of the following 10 elements of 5th period. 39



Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag and 48Cd.

rd transition series or 5d series corresponding to filling of 5d sub-shell consists of following 10 elements of 6th period. 57



La, 72Hf, Ta, W, Re, Os, Ir, Pt, Au and 80Hg.

IVth transition series or 6d series corresponding to the filling of 6d sub-shell starts with 89Ac followed by elements with atomic number 104 onwards. There are greater horizontal similarities in the properties of the transition elements. However, some group similarities also exist.

Example-1

On what ground you can say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

Solution

On the basis of incompletely filled 3d orbitals of scandium, [Ar]18 3d1 4s2 , and completely filled 3d orbitals of Zn, [Ar]18 3d10 4s2, they are considered transition and non-transition elements respectively.

Example-2

The element with the electronic configuration [Xe]54 4f14 5d1 6s2 is a : (A) representative element

(B) d-block element

(C) lanthanoid Solution

After achieving 4f

(D) actinoid 14

0

2

5d 6s configuration, the next electron goes to 5d and this is the case of

Lu(Z = 71) which is the last element of lanthanoid series. Therefore, (C) option is correct. Example-3

The number of transition series is : (A) two

Solution

(B) three

(C) four

(D) five

There are four transition series ; 3d, 4d, 5d and 6d series. Therefore, (C) option is correct.

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CHEMISTRY GENERAL TRENDS IN THE CHEMISTRY OF TRANSITION ELEMENTS. Metallic character : In d-block elements the last but one (i.e. the penultimate) shell of electrons is expanding. Thus they have many physical and chemical properties in common. Hence nearly all the transition elements display typical metallic properties such as high tensile strength, ductility, malleability, high thermal and electrical conductivity and metallic lustre. With the exceptions of Zn,Cd, Hg and Mn, they have one or more typical metallic structures at normal temperatures. Lattice Structures of Transition Metals (Table-1) Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

hcp (bcc) Y Hcp (bcc) La hcp (ccp, bcc)

hcp (bcc) Zr hcp (bcc) Hf hcp (bcc)

bcc

bcc (bcc,ccp) Mo bcc

X (hcp) Tc hcp

bcc (hcp) Ru hcp

ccp

ccp

ccp

X (hcp) Cd

W bcc

Re hcp

Os hcp

Ir ccp

Nb bcc Ta bcc

Rh

Pd ccp

Ag ccp

Pt ccp

ccp Au ccp

X (hcp) Hg X

bcc = body centred cubic ; hcp = hexagonal close packed ccp = cubic close packed ; X = a typical metal structure The transition elements (with the exception of copper) are very much hard and have low volatility. They have high enthalpies of atomisation which are shown in figure given below. The maxima at about the middle of each series indicate that one unpaired electron per d orbital is particularly favourable of strong interatomic interaction. In general, greater the number of valence electrons, stronger is the resultant bonding. Cr, Mo and W have maximum number of unpaired electrons and therefore, these are very hard metals and have maximum enthalpies of atomization. Hence the metals with very high enthalpy of atomisation (i.e., very high boiling point) tend to be noble in their reactions. The metals of the second, 4d and third, 5d series have greater enthalpies of atomisation than the corresponding elements of the first series, 3d and this is an important factor indicating for the occurrence of much more frequent metal-metal bonding in compounds of the heavy transition metals.

Graph showing Trends in enthalpies of atomisation of transition elements Size of atoms and ions : The atomic radii of the transition metals lie in-between those of s- and p-block elements. The covalent radii of the elements decreases from left to right across a row in the transition series, until near the end when the size increases slightly. The decrease in size is small after mid way. In the beginning, the atomic radius decreases with the increase in nuclear charge (as atomic number increases), where as the shielding effect of d-electrons is small. After mid way as the electrons enters the last but one shell, the added d-electron shields the outer most electrons. Hence with the increase in the d-electrons screening effect increases. This counter balances the increased nuclear charge. As a result, the atomic radii remain practically same after chromium.

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CHEMISTRY Near the end of the series, the increased electron repulsions between the added electrons in the same orbitals are greater than the attractive forces. This results in the expansion of the electron cloud and thus atomic radius increases. The atomic radii, in general, increase down the group. The atomic radii of second series are larger than those of first transition series. In the atoms of the second transition series, the number of shells are more than those of the Ist transition series. As a result, the atoms of IInd transition series are larger than those of the elements of the first transition series. But the atomic radii of the second and third transition series are almost the same. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction which essentially compensates for the expected increase in atomic size with increasing atomic number. The net result of the lanthanoid contraction is that the second and the third d-series exhibit similar radii (e.g., Zr 160 pm, Hf 159 pm) and have very similar physical and chemical properties much more than that expected on the basis of usual family relationship. 

The trend followed by the ionic radii is same as that followed by atomic radii.



The ionic radii of transition metals are different in different oxidation states.

Melting and boiling points : The melting and boiling points of the transition series elements are generally very high. The melting points of the transition elements rise to a maximum and then fall as the atomic number increases. Manganese and technetium have abnormally low melting points. Strong metallic bonds between the atoms of these elements attribute to their high melting and boiling points. In a particular series, the metallic strength increases upto the middle with increasing number of unpaired electrons i.e up to d5. After chromium, the number of unpaired electrons goes on decreasing. Accordingly, the melting points decrease after middle (Cr) because of increasing pairing of electrons.  The dip in melting points of Mn and Tc can be attributed to their stable electronic configurations (half filled 3d5 and fully filled 4s2 ). Due to this stable electronic configuration, the delocalisation of electrons may be less and thus the metallic bond is much weaker than preceding elements. Example-4

Why do the transition elements have higher boiling & melting points ?

Solution

Because of having larger number of unpaired electrons in their atoms, they have stronger interatomic interaction and hence stronger bonding between atoms. Hence strong metallic bonds between the atoms of these elements attribute to their high melting and boiling points.

Example-5

The atomic radii of transition elements in a period are (A) smaller than those of s-block as well as p-block elements. (B) larger than those of s-block as well as p-block elements. (C) smaller than those of s-block but larger than those of p-block elements. (D) larger than those of s-block but smaller than those of p-block elements. Across the period the atomic radii generally decrease with increasing atomic number as effective nuclear charge increase. Therefore, (C) option is correct.

Solution

Density : The atomic volumes of the transition elements are low compared with the elements of group 1 and 2. This is because the increased nuclear charge is poorly screened and so attracts all the electrons more strongly. In addition, the extra electrons added occupy inner orbitals. Consequently the densities of the transition metals are high. The densities of the second row are high and third row values are even higher. Elements with the highest densities are osmium 22.57 g cm–3 and irridium 22.61 g cm–3.  Across a period from left to right atomic volumes decrease and atomic masses increase. Hence the densities also increase across a period.  The last element zinc is an exception, having large atomic volume and hence lower density.

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CHEMISTRY Ionisation energies or Ionisation enthalpies : 

The first ionisation energies of d-block elements are intermediate between those of the s- and pblocks. This suggests that the transition elements are less electropositive than groups 1 and 2 and may form either ionic or covalent bonds depending on the conditions. Generally, the lower valent states are ionic and the higher valent states are covalent. Across a period from left to right ionisation energies gradually increase with increase in atomic number. This is because the nuclear charge increases and the atomic size decreases with increase in atomic number along the period. Consequently making the removal of outer electron difficult.



In a given series, the difference in the ionisation energies between any two successive d-block elements is very much less than the difference in case of successive s-block or p-block elements. The effect of increased nuclear charge and screening effect of the added d-electrons tend to oppose each other. Hence due to these two counter effects, the difference in the ionisation energies of two successive transition elements is very small on moving across a period.



The first ionisation energy of Zn, Cd, and Hg are very high because of their fully filled (n–1) d10 ns2 configuration.



IInd and IIIrd ionisation energies also increase along a period. IInd ionisation energy of Cr > Mn and Cu > Zn This is because after the removal of Ist electron Cr and Cu acquire stable configurations (d5 & d10) and the removal of IInd electron thus become very difficult. Similarly the E2 of 23V < 24Cr > 25Mn and 28Ni < 29Cu > 30Zn

The third ionization energy of Mn is very high because the third electron has to be removed from the stable half-filled 3d orbital. 

The first ionisation energies of the 5d elements are higher as compared to those of 3d and 4d elements. This is because the weak shielding of nucleus by 4f electrons in 5d elements results in higher effective nuclear charge acting on the outer valence electrons.

Graph Showing Trends in Ionisation Energies Example-6

Zn forms only Zn2+ and not Zn3+, why?

Solution

In the formation of Zn3+, an electron from the d10 configuration has to be removed which requires abnormally higher amount of energy.

Oxidation states : The transition metals exhibit a large number of oxidation states. With the exception of a few elements, most of these show variable oxidation states. These different oxidation states are related to the electronic configuration of their atoms. The existence of the transition elements in different oxidation states means that their atoms can lose different number of electrons. This is due to the participation of inner (n – 1) d-electrons in addition to outer nselectrons because, the energies of the ns and (n – 1) d-sub-shells are nearly same. For example, scandium has the outer electronic configuration 3d14s2. It exhibits an oxidation state of +2 when it uses both of its 4s-electrons for bonding but it can also show oxidation state of +3 when it uses its two s-electrons and one d-electron. Similarly, the other atoms can show oxidation states equal to ns-and (n – 1) d-electrons.

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CHEMISTRY Table-2 : Different oxidation states of first transition series.

Element

Outer electronic configuration

Sc

3d14s2

Ti

2

3d 4s

V

3d34s2

Cr

5

3d 4s

Mn

3d54s2

+2, +3, +4, (+5), +6, +7

Fe

3d64s2

+2, +3, (+4), (+5), (+6)

Co

7

3d 4s

Ni

3d84s2

Cu

10

3d 4s

1

+1, +2

Zn

10

2

+2

Oxidation states +3

2

+2, +3, +4 +2, +3, +4, +5

1

+2, +3, (+4), (+5), +6

2

3d 4s

+2, +3, (+4) +2, +3, +4

Oxidation states given in parenthesis are unstable. Table-3 :Different oxidation states of elements of second transition series. Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

+3

(+3)

(+2)

+2

+2

+2

+2

+2

+1

+2

+4

(+3)

+3

(+4)

+3

+3

(+3)

(+2)

(+4)

+4

(+5)

+4

+4

+4

(+3)

+5

+5

(+5)

(+6)

+6

(+6) (+7) (+8)

Table-4 : Different oxidation states of elements of third transition series. La

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

+3

(+3)

(+2)

+2

(-1)

+2

+2

+2

+1

+1

+4

(+3)

(+3)

(+1)

+3

+3

(+3)

+3

+2

(+4)

+4

(+2)

+4

+4

+4

+5

+5

+3

+6

(+6)

(+5)

+6

+4

+8

(+6)

+5 (+6) +7

It may be noted that the stability of a given oxidation state depends upon the nature of the elements with which the metal is combined. The highest oxidation states are found in compounds of fluorides and oxides because fluorine and oxygen are most electronegative elements. The examination of the common oxidation states exhibited by different transition metals reveals the following facts :

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CHEMISTRY (i)

The variable oxidation states of transition metals are due to participation of inner (n – 1)d and outer nselectrons. The lowest oxidation state corresponds to the number of ns-electrons. For example, in the first transition series, the lowest oxidation states of Cr (3d54s1) and Cu(3d104s1) are +1 while for others, it is +2 (3d1 – 104s2).

(ii)

Except scandium, the most common oxidation state of the first row transition elements is +2 which arises due to loss of two 4s-electrons. This means that after scandium 3d-orbitals become more stable and, therefore, are lower in energy than the 4s-orbitals. As a result, electrons are first removed from 4s-orbitals.

(iii)

The elements which show the greater number of oxidation states occur in or near the middle of the series. For example, in the first transition series, manganese exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states in the beginning of series can be due to the presence of smaller number of electrons to lose or share (Sc, Ti). On the other hand, at the extreme right hand side end (Cu, Zn), lesser number of oxidation state is due to large number of d electrons so that only a fewer orbitals are available in which the electron can share with other for higher valence. The highest oxidation state shown by any transition metal is +8.

(iv)

In the +2 and +3 oxidation states, the bonds formed are mostly ionic. In the compounds of higher oxidation states (generally formed with oxygen and fluorine), the bonds are essentially covalent. Thus, the bonds in +2 and +3 oxidation states are generally formed by the loss of two or three electrons respectively while the bonds in higher oxidation states are formed by sharing of d-electrons. For example, in MnO4– (Mn in +7 state) all the bonds are covalent.

(v)

Within a group, the maximum oxidation state increase with atomic number. For example, iron (group 8) shows common oxidation states of +2 and +3 but ruthenium and osmium in the same group form compounds in the +4, +6 and +8 oxidation states.

(vi)

Transition metals also form compounds in low oxidation states such as +1 and 0 or negative. The common examples are [Ni(CO)4], [Fe(CO)5] in which nickel and iron are in zero oxidation state.

(vii)

The variability of oxidation states in transition elements arises because of incomplete filling of the d-orbitals in such a way that their oxidation states differ by unity such as VII, VIII, VIV and VV. This behaviour is in contrast with the variability of oxidation states of non-transition elements (p-block elements), where oxidation states normally differ by a unit of two such as Sn2+, Sn4+, In+, In3+, etc.

(viii)

Unlike p-block elements where the lower oxidation states are favoured by heavier members (due to inert pair effect), the higher oxidation states are more stable in heavier transition elements. For example, in group 6, Mo (VI) and W (VI) are found to be more stable than Cr (VI). Therefore, Cr (VI) in the form of dichromate in acidic medium is a strong oxidising agent whereas MoO3 and WO3 are not.

Standard electrode potentials : The magnitude of ionization enthalpy gives the amount of energy required to remove electrons to form a particular oxidation state of the metal in a compound. Thus, the value of ionisation enthalpies gives information regarding the thermodynamic stability of the transition metal compounds in different oxidation states. Smaller the ionisation enthalpy of the metal, the stable is its compound. For example, the first four ionisation enthalpies of nickel and platinum are given below : Table-5 : Ionisation enthalpies

Ni

Pt

IE1 + IE2

2.49 × 103 kJ mol–1

2.66 × 103 kJ mol–1

IE3 + IE4

8.80 × 103 kJ mol–1

6.70 × 103 kJ mol–1

Total

11.29 × 103 kJ mol–1

9.36 × 103 kJ mol–1

It is clear form the above table that the sum of first two ionization enthalpies is less for nickel than for platinum. Ni  Ni2+ + 2e– I.E. = 2.49 × 103 kJ mol–1 Pt  Pt2+ + 2e– I.E. = 2.66 × 103 kJ mol–1 2+ Therefore, ionization of nickel to Ni is energetically favourable as compared to that of platinum. Thus, the nickel (II) compounds are thermodynamically more stable than platinum (II) compounds. On the other hand, the sum of first four ionisation enthalpies is less for platinum than for nickel as : Ni  Ni4+ + 2e– I.E. = 11.29 × 103 kJ mol–1 Pt  Pt4+ + 2e– I.E. = 9.36 × 103 kJ mol–1

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CHEMISTRY Thus, the platinum (IV) compounds are relatively more stable than nickel (IV) compounds. Therefore, K2PtCl6 [having Pt (IV)] is a well-known compound whereas the corresponding nickel compound is not known. However, in solutions the stability of the compounds depends upon electrode potentials. Electrode potentials : In addition to ionisation enthalpy, the other factors such as enthalpy of sublimation, hydration enthalpy, ionisation enthalpy etc. determine the stability of a particular oxidation state in solution. This can be explained in terms of their electrode potential values. The oxidation potential of a metal involves the following process: M(s)  M+(aq) + e– This process actually takes place in the following three steps as given in following flowchart :

(i)

In the first step, the atoms get isolated from one another and become independent in the gaseous state. This converts solid metal to the gaseous state. The energy needed for this step is known as enthalpy of sublimation. M(s)  M+(g) Enthalpy of sublimation, subH

(ii)

In the second step, the outer electron is removed from the isolated atom. The energy required for this change is ionisation enthalpy. M(s)  M+(g) + e– Ionisation enthalpy, IE

(iii)

In the third step the gaseous ion gets hydrated. In this process, energy known as hydration enthalpy, is liberated. M+(g) + nH2O  M+(aq)

Enthalpy of hydration, hydH

The oxidation potential which gives the tendency of the overall change to occur, depends upon the net effect of these three steps. The overall energy change is H = subH + IE + hydH Thus, H gives the enthalpy change required to bring the solid metal, M to the monovalent ion in aqueous medium, M+ (aq). An exactly similar cycle may be constructed for the formation of an anion in solution except that the ionization enthalpy may be replaced by electron gain enthalpy when the gaseous atom goes to gaseous anion. H helps to predict the stability of a particular oxidation state. The smaller the values of total energy change for a particular oxidation state in aqueous solution, greater will be the stability of that oxidation state. The electrode potentials are a measure of total energy change. Qualitative, the stability of the transition metal ions in different oxidation states can be determined on the basis of electrode potential data. The lower the electrode potential i.e., more negative the standard reduction potential of the electrode, the more stable is the oxidation state of the transition metal in the aqueous solution. The electrode potentials of different metals can also be measured by forming the cell with standard hydrogen electrode. For the measurement of electrode potential of M2+ | 1M, the e.m.f. of the cell in which the following reaction occurs is measured : 2H+(aq) + M(s)

M2+ (aq) + H2(g)

Knowing the potential of 2H+(aq) | H2(g), it is possible to determine the potential of M2+ (aq) | M. For the first transition series, the E values of M2+(aq) | M are given below :

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CHEMISTRY The observed values of E and those calculated using the data are compared in the following figure.

Thermochemical data (kJ mol–1) for the first row Transition Elements and the Standard Electrode potentials for the Reduction of MII to M Table-6 :  aHq (M)  f H1

Element (M)

469 515 398 279 418 427 431 339 130

Ti V Cr Mn Fe Co Ni Cu Zn

(i)

1H2

661 648 653 716 762 757 736 745 908

hydH (M )

E /V

-1866 -1895 -1925 -1862 -1998 -2079 -2121 -2121 -2059

-1.63 -1.18 -0.90 -1.18 -0.44 -0.28 -0.25 0.34 -0.76



1310 1370 1590 1510 1560 1640 1750 1960 1730

2+



The results lead to the following conclusions : There is no regular trend in these values. This is attributed to the irregular variation of ionisation enthalpies (IE1 + IE2) and the sublimation energies in the period.

(ii)

It may be noted that the electrode potentials of transition metals are low in comparison to elements of group 2 (e.g., Ca = –2.87 V). Compared to group 2 elements, the transition elements have fairly large ionisation enthalpies and very large enthalpies of atomisation. These reduce their electrode potentials though their hydration enthalpies are large.

(iii)

Zinc has low enthalpy of atomisation and fairly large hydration energy. But it has also low electrode potential (–0.76 V) because of its very high ionisation enthalpy (IE1 + IE2).

(iv)

It is clear from above table and figure that copper has positive reduction potential, E (0.34 V) and this shows that copper is least reactive metal out of the first transition series. This unique behaviour (+ve) E value of copper) also accounts for its inability to liberate H2 from acids. It has been observed that only oxidizing acids (such as nitric acid and hot concentrated sulphuric acid) react with copper in which the acids are reduced. The high energy required to convert Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy.

(v)

In general, the value becomes, less negative across the series. This is related to the general increase in the sum of first and second ionisation enthalpies. It is interesting to note that the values of Eº of Mn, Ni and Zn are more negative than expected from the general trend. The relatively more negative values of E for Mn and Zn are due to stability of half filled d-sub-shell in Mn2+ (3d5) and the completely filled (3d10) configuration in Zn2+. The exceptionally high E value of Ni from regular trend is related to the highest negative enthalpy of hydration of Ni2+ ion. Trends in the M3+ | M2+ Standard Electrode Potentials Except copper and zinc, all other elements of first transition series show +3 oxidation states also to form M3+ ions in aqueous solutions. The standard reduction potentials for M3+ | M2+ redox couple are given below : Table-7 :

E  (M3+(aq) | M2+(aq) (in Volt)

Ti

V

Cr

Mn

Fe

Co

–0.37

–0.26

–0.41

+1.57

+0.77

+1.97

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CHEMISTRY These values reveal the following facts : The low value of scandium reflects the stability of Sc3+ which has a noble gas configuration.

(i) (ii)

The comparatively high value for Mn shows that Mn2+ (d5 configuration) is particularly stable. On the other hand comparatively low value for Fe shows the extra stability of Fe3+ (d5 configuration).

(iii)

The comparatively low value of V is related to the stability of V2+ (due to half filled t2g3 energy level of 3d orbitals in octahedral crystal field spitting).

(iv)

The E value for Mn3+ /Mn2+ couple much more positive than for Cr3+/Cr2+ or Fe3+ /Fe2+. This is because of the much larger IIIrd ionisation energy of Mn (removal of electron from d5 configuration). Trends in Stability of Higher Oxidation States Standard electrode potential data provide valuable information about the stabilities of different oxidation states shown by an element. The highest oxidation states are shown generally among halides and oxides.

1.

In metal halides. The transition elements react with halogens at high temperatures to form transition metal halides. These reactions have very high heat of reaction. But once the reaction starts, the heat of reaction is sufficient to continue the reaction. The halogens react in the following decreasing order ; F2 > Cl2 > Br2 > I2 Table-8 : Halides of first transition series Oxidation Number

Sc

Ti

V

Cr

VF5

+5 +4

+2

Fe

Co

Ni

Cu

Zn

NiX2

CuX2b

CrF6

+6

+3

Mn

ScX3

CrF5

TiX4

VX4

a

CrF4

MnF4

TiX3

VX3

CrF3

MnF3

FeX3a

CoF3

TiX2

c

CrF2

MnX2

FeX2

CoX2

VX2

+1

CuX a

b

ZnX2

c

c

where X = F, Cl, Br, I, X = F, Cl, Br, X = F, Cl, X = Cl, Br, I

(i)

Within each of the transition groups 3 - 12, there is a difference in the stability of the various oxidation states. In general, the second and third transition series elements exhibit higher coordination number and their higher oxidation states are more stable than the corresponding first transition series elements. The following trends are observed from table regarding transition metal halides : In general, the elements of first transition series tend to exist in low oxidation states. Chromium to zinc form stable difluorides and the other chlorides are also known.

(ii)

Since fluorine is the most electronegative element, the transition metals show highest oxidation states with fluorine. The highest oxidation states are found in TiX4 (tetrahalides, X = F, Cl, Br and I), VF5 and CrF6.

(iii)

The +7 oxidation state for Mn is not shown by simple halides. However, MnO3F is known in which the oxidation state of Mn is +7.

(iv)

After Mn, the tendency to show higher oxidation states with halogens are uncommon. Iron and cobalt form trihalides FeX3 (X = F, Cl or Br) and CoF3.

(v)

The tendency of fluorine to stabilise the highest oxidation state is due to either higher lattice enthalpy as in case of CoF3 or higher bond enthalpy due to higher covalent bonds e.g., VF5 and CrF6.

(vi)

V(V) is shown by VF5 only. However, the other halides undergo hydrolysis to form oxohalides, VOX3.

(vii)

Fluorides are relatively unstable in their low oxidation states. For example, vanadium form only VX2 (X = Cl, Br or I) and copper can form CuX (X = Cl, I). All copper (II) halides are known except the iodide. This is because, Cu2+ oxidises I– to I2. 2Cu2+ + 4I–  Cu2I2(s) + I2 It has been observed that many copper (I) compounds are unstable in aqueous solution and they undergo disproportionation to Cu(II) and Cu(0) as : 2Cu+  Cu2+ + Cu Copper in +2 oxidation state is more stable than in +1 oxidation state. This can be explained on the basis of much larger negative hydration enthalpy (hydH) of Cu2+ (aq) than Cu+, which is much more than compensates for the large energy required to remove the second electron i.e., second ionisation enthalpy of copper.

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CHEMISTRY 2.

In metal oxides and oxocations. Table-9 : Oxides of first transition series Oxidation Number

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

CoO

NiO

CuO

ZnO

Mn2O7

+7 CrO3

+6 V2O5

+5 +4 Sc2O3

+3 +2

TiO2

V2O4

CrO2

MnO2

Ti2O3

V2O3

Cr2O3

Mn2O3

Fe2O3

TiO

VO

(CrO)

MnO

FeO

Cu2O

+1 Mixed oxides

Mn3O4

Fe3O4

Co3O4

The ability of oxygen to stabilize the highest oxidation state is demonstrated in their oxides. The highest oxidation states in their oxides concides with the group number. For example, the highest oxidation state of scandium of group 3 is +3 in its oxides, Sc2O3 whereas the highest oxidation state of manganese of group 7 is +7, in Mn2O7. However beyond group 7, no higher oxides of iron above Fe2O3 are known. Although higher oxidation state such as +6 is shown in ferrates such as FeO42– in alkaline medium, but they readily decompose to Fe2O3 and O2. Besides the oxides, oxocation of the metals also stabilise higher oxidation states. For example, VV as VO2+, VIV as VO2+ and TiIV as TiO2+. It may be noted that the ability of oxygen to stabilise these high oxidation states exceeds that of fluorine. For example, manganese forms highest fluoride as MnF4 whereas the highest oxide is Mn2O7. This is due to the fact that oxygen has great ability to form multiple bonds to metals. In the covalent oxide. Mn2O7, each Mn is tetrahedrally surrounded by oxygen atoms and has Mn–O–Mn bridge. The tetrahedral [MO4]n– ions are also known for vanadium (V), chromium (VI), manganese (VI) and manganese (VII). The transition elements in the +2 and +3 oxidation states mostly form ionic bonds whereas with higher oxidation states, the bonds are essentially covalent e.g., in MnO4– all bonds are covalent. As the oxidation number of a metal increases, the ionic character of their oxides decrease. For example, in case of Mn, Mn2O7 is a covalent green oil. In these higher oxides the acidic character is predominant. Thus CrO3 gives H2CrO4 and H2Cr2O7 and Mn2O7 gives HMnO4. V2O5 is, however amphoteric though mainly acidic and with alkalies as well as acids gives VO43– and VO2+ respectively. Example-7

Solution Example-8 Solution

Example9 Solution

For the first series of transition metals the E values are E V Cr Mn Fe Co Ni Cu (M2+/M+1) – 1.18 – 0.91 – 1.18 – 0.44 – 0.28 – 0.25 + 0.34 Explain the irregularity in the above values. This is because of irregular variation of ionization energies (E1 + E2) and also the sublimation energies which are much less for manganese and vanadium Write the formulae of different oxides of manganese. What is the oxidation state of manganese in each of them ? Arrange them in order of their decreasing acidic character. Mn+2O, Mn23+ O3, Mn4+O2, Mn27+O7, Mn3O4 (mixed oxide of MnO and Mn2O3) As the oxidation state of Mn increases, the electronegativity values of the central metal ions increase. As a result of this the difference in the electronegativity values between metals and oxygen decrease. So the acidic character increases as given below. MnO < Mn2O3 < MnO2 < Mn2O7 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. In Cr2O72– and CrO42– the oxidation state of Cr is +6 (2x + 7(–2) = –2 and x + 4 (– 2) = –2) . The group number is equal to the number of electrons in (n-1) d and ns sub-shells for d-block elements. So group no. = 5 + 1 = 6. In MnO4– the oxidation state of Mn is +7 (2x + 7(–2) = –1). The group number is equal to the number of electrons in (n-1) d and ns sub-shells for d-block elements. So group no. = 5 + 2 = 7

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CHEMISTRY Example-10 Solution

Zinc does not show variable valency because of : (A) complete ‘d’ sub-shell (B) inert pair effect (C) 4s2 sub-shell (D) none. 18 10 2 Zn = [Ar] 3d 4s . Zinc has completely filled d-sub-shell so removal of electron from completely 30 filled 4d sub-shell would be quite difficult. Thus it does not show variable valency. Therefore, (A) option is correct.

Formation of Coloured Ions : Most of the compounds of transition metals are coloured in the solid form or solution form. The colour of the compounds of transition metals may be attributed to the presence of incomplete (n – 1) d-sub-shell. In the case of compounds of transition metals, the energies of the five d-orbitals in the same sub-shell do not remain equal. Under the influence of approaching ions towards the central metal ion, the d-orbitals of the central metal split into different energy levels. This phenomenon is called crystal field splitting. For example, when the six ions or molecules approach the metal ion (called octahedral field), the five d-orbitals split up into two sets : one set consisting of two d-orbitals ( d x 2  y 2 , dz 2 ) of higher energies and the other set consisting of three d-orbitals (dxy, dyz, dzx) of lower energies. In the case of the transition metal ions, the electron can be easily promoted from one energy level to another in the same d-sub-shell. These are called d-d transitions. The amount of energy required to excite some of the electrons to higher energy states within the same d-sub-shell corresponds to energy of certain colours of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain colour, is absorbed and the electron gets raised from lower energy set of orbitals to higher energy set of orbitals as shown below :

Figure showing electronic transition from t2g to eg orbitals. The excess of other colours constituting white light are transmitted and the compound appears coloured. The observed colour of a substance is always complementary colour of the colour which is absorbed by the substance. For example, Ti3+ compounds contain one electron in d-sub-shell (d1). It absorbs green and yellow portions from the white light and blue and red portions are emitted. Therefore, Ti3+ ions appear purple. Similarly, hydrated cupric compounds absorb radiations corresponding to red light and the transmitted colour is greenish blue (which is complementary colour to red colour). Thus, cupric compounds have greenish-blue colour. Table-10 : Colour of different hydrated transition metal ions. Ion Sc (III), Ti (IV) Ti (III) V (IV) V (III) Cr (III) Mn (III) Cr (II) Mn (II) Fe (III) Fe (II) Co (III) Co (II) Ni (II) Cu (II) Cu (I) Zn (II)

Outer Configuration 3d0 3d1 3d1 3d2 3d3 3d4 3d4 3d5 3d5 3d6 3d6 3d7 3d8 3d9 3d10 3d10

Colour of the ion Colourless Purple Blue Green Green Violet Blue Pink Yellow Green Blue Pink Green Blue Colourless Colourless

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CHEMISTRY Example-11

Explain the blue colour of CuSO4.5H2O.

Solution

Cu2+ ion (3d9) absorbs red light from the visible region, for the promotion of 3d electrons, the ions reflect blue light and appear blue.

Magnetic Properties It is interesting to note that when the various substances are placed in a magnetic field, they do not behave in a similar way i.e., they show different behaviour which are known as magnetic behaviour. These are classified as : (i)

Paramagnetic substances. The substances which are attracted by magnetic field are called paramagnetic substances and this character arises due to the presence of unpaired electrons in the atomic orbitals.

(ii)

Diamagnetic substances. The substances which are repelled by magnetic field are called diamagnetic substances and this character arises due to the presence of paired electrons in the atomic orbitals. Most of the compounds of transition elements are paramagnetic in nature and are attracted by the magnetic field. The transition elements involve the partial filling of d-sub-shells. Most of the transition metal ions or their compounds have unpaired electrons in d-sub-shell (from configuration d1 to d9) and therefore, they give rise to paramagnetic character. The magnetic character is expressed in terms of magnetic moment. The larger the number of unpaired electrons in a substance, the greater is the paramagnetic character and larger is the magnetic moment. The magnetic moment is expressed in Bohr magnetons abbreviated as B.M. Table-11 : Calculated and observed magnetic moment of ions of first transition series.

Ion

Outer Configuration

Number of unpaired electrons

Magnetic moment (B.M.) Calculated

observed

3+

3d

0

0

0

0

Ti

3+

3d

1

1

1.73

1.75

Ti

2+

3d

2

2

2.84

2.76

3d

3

3

3.87

3.86

3d

4

4

4.90

4.80

Sc

2+

V

Cr

2+

3d

5

5

5.92

5.96

2+

3d

6

4

4.90

5.3 - 5.5

2+

3d

7

3

3.87

4.4 - 5.2

Mn Fe

2+

Co

2+

3d

8

2

2.84

2.9 - 3.4

2+

3d

9

1

1.73

1.8 - 2.2

2+

3d

10

0

0

0

Ni

Cu Zn

The experimental data are mainly for hydrated ions in solution or in the solid state. Each unpaired electron have magnetic moment associated with its spin angular momentum and orbital angular momentum. In the compounds of first transition series, the orbital angular momentum does not contribute much and thus has no significance. Therefore, for the first transition series elements, the magnetic moment arise only from the spin of the electrons. This can be calculated from the relation :  = n (n  2) B.M. where n is the number of unpaired electrons and  is magnetic moment in Bohr magneton (BM) units. The paramagnetism first increases in any transition series and then decreases. The maximum paramagnetism is observed around the middle of the series (as contains maximum number of unpaired electrons). In addition to paramagnetic and diamagnetic substances, there are a few substances such as iron metal, iron oxide which are highly magnetic (about 1000 times more than ordinary metals). These are very strongly attracted by applied magnetic field and retained their magnetism when removed from the field are called ferromagnetic substances.

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CHEMISTRY Example-12

Determine the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

Solution

The metal having atomic number 25 has electron configuration [Ar]18 3d5 4s2. So its divalent ion in aqueous solution will have electron configuration [Ar]18 3d5 4s0. Thus it has five unpaired electrons. So

(spin) =

5(5  2) = 5.92 BM.

Formation of complexes : In contrast to representative elements, the transition elements form a large number of coordination complexes. The transition metal ions bind to a number of anions or neutral molecules in these complexes. The common examples are [Ni(NH3)6]2+, [Co(NH3)6]3+, [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+, etc. The great tendency of transition metal ions to form complexes is due to : (i) small size of the atoms and ions, (ii) high nuclear charge and (iii) availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons donated by ligands. Formation of Interstitial Compounds : Transition metals form interstitial compounds with elements such as hydrogen, boron, carbon and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) get trapped in vacant spaces of the lattices of the transition metal atoms as shown below.

Figure showing formation of interstitial compounds They are generally non-stoichiometric and are neither typically ionic nor covalent. The common examples of interstitial compounds of transition metals are TiC, Mn4N, Fe3H, TiH2 etc. It may be noted that these formula do not correspond to any normal oxidation state of the metal. Generally, the nonstoichiometric materials are obtained having the composition as TiH1.7, VH0.56, etc. Because of the nature of their composition, these compounds are referred to as interstitial compounds. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard. These interstitial compounds have similar chemical properties as the parent metals but differ significantly in their physical properties particularly, density, hardness and conductivity. For example, steel and cast iron are hard because they form interstitial compounds with carbon. The transition metals can easily accommodate the small non-metallic atoms because of void sites between the packed atoms of the crystalline metal. These spaces are present because of defects in their structures and existence of variable oxidation states. The general characteristic physical and chemical properties of these compounds are : (i)

They have high melting points which are higher than those of pure metals.

(ii)

They retain metallic conductivity i.e. of pure metals.

(iii)

They are very hard and some borides have hardness as that of diamond.

(iv)

They are chemically inert. Catalytic properties Many transition metals and their compounds act as good catalysts for various reactions. Of these, the use of Fe, Co, Ni, V, Cr, Mn, Pt, etc. are very common. For example,

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CHEMISTRY (i)

Iron-molybdenum is used as catalyst in the synthesis of ammonia by Haber’s process.

(ii)

Nickel is used in hydrogenation reactions in organic chemistry.

(iii)

V2O5 is used for the oxidation of SO2 to SO3 in the Contact process for the manufacture of H2SO4.

(iv)

MnO2 is used to catalyse the decomposition of H2O2 solution.

(v)

Cobalt salts catalyse the decomposition of bleaching powder. The catalytic property of transition metals is due to their tendency to form reaction intermediates with suitable reactants. These intermediates give reaction paths of lower activation energy and, therefore increase the rate of the reaction. These reaction intermediates readily decompose yielding the products and regenerating the original substance. The transition metals form these reaction intermediates due to the presence of vacant orbitals or their tendency to form variable oxidation states.

(i)

In some cases, the transition metal catalysts provide a suitable large surface area for the adsorption of the reactant. This increases the concentration of the reactants at the catalyst surface and also weakens the bonds in the reactant molecules. Consequently, the activation energy gets lowered. For example, during the conversion of SO2 to SO3, V2O5 is used as a catalyst. Solid V2O5 adsorbs a molecule of SO2 on the surface to form V2O4 and the oxygen is given to SO2 to form SO3. The divanadium tetroxide is then converted to V2O5 by reaction with oxygen : V2O5 + SO2 (catalyst)  SO3 + V2O4 (divanadium tetroxide) 2V2O4 + O2  2V2O5

(ii)

In some cases, the transition metal ions can change their oxidation states and become more effective as catalysts. For example, cobalt salts catalyse decomposition of bleaching powder as cobalt can easily change oxidation state from +2 to +3 as : Co2+ + OCl– + H2O  Co3+ + Cl– + 2OH– 2Co3+ + 2OH–  2Co2+ + H2O + ½O2 Iron (III) also catalyses the reaction between iodide and persulphate ions (S2O82–).

Example-13 Solution

How iron (III) catalyses the reaction between iodide & persulphate? 2Fe3+ + 2–  2Fe2+ + 2 2Fe2+ + S2O82–  2Fe3+ + 2SO42– ________________________________________________________

Fe (  ) 2– + S2O82–   2 + 2 SO42–

Alloy Formation : Alloys are homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other metal. The alloys are generally formed by those atoms which have metallic radii within about 15% of each other. Transition metals form a large number of alloys. The transition metals are quite similar in size and, therefore, the atoms of one metal can substitute the atoms of other metal in its crystal lattice. Thus, on cooling a mixture solution of two or more transition metals, solid alloys are formed. Such alloys are hard, have high melting points and are more resistant to corrosion than parent metals. For example, the most common known alloys are ferrous alloys. Chromium, manganese, vanadium, tungsten, molybdenum etc. are used to produce variety of steels and stainless steel. Alloys of transition metals with non-transition metals such as bronze (copper-tin), brass (copper-zinc) are also industrially important alloys.

Figure showing formation of alloys

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CHEMISTRY Example-14 Solution

Which of the following form an alloy ? (A) Ag + Pb (B) Fe + Hg

(C) Pt + Hg

(D) Fe + C

Argentiferrous lead is an alloy contains Ag & Pb. 'Fe' and 'C' form an alloy called cementite, Fe3C. Therefore, (A) & (D) option are correct.



PREPARATIONS AND PROPERTIES OF SOME IMPORTANT d-BLOCK METAL COMPOUNDS

[A]

COMPOUNDS OF IRON :



Ferrous Sulphate, FeSO4.7H2O (Green vitriol) It is commonly known as harakasis 

(iii)

Preparation By iron scrap : Fe + H2SO4  FeSO4 + H2 From kipp's waste : 3FeS2 + 2H2O + 11O2  FeSO4 + Fe2(SO4)3 + 2H2SO4 Fe2(SO4)3 + Fe  3FeSO4 FeCO3 + H2SO4  FeSO4 + H2O + CO2



Properties

(i) (ii)

(a) Physical :

Hydrated ferrous sulphate is a green crystalline compound, effloresces on exposure to air. Anhydrous FeSO4 is colourless.

(b) Chemical : (i) On exposure to atmosphere, it turns brownish-yellow due to the formation of basic ferric sulphate. 4FeSO4 + 2H2O + O2  4Fe(OH) . SO4 (basic ferric sulphate) (ii) Aqueous solution is acidic due to hydrolysis. Fe2+ + 2H2O Fe(OH)2 + 2H+ (iii)

140 C ºC  FeSO4. H2O 300   2FeSO4 FeSO4 . 7H2O  

(iv)

FeSO4 + 6KCN  K4[Fe(CN)6] + K2SO4

(v)

It acts as reducing agent. AuCl3 + 3FeSO4  Au + Fe2(SO4)3 + FeCl3 6HgCl2 + 6FeSO4  3Hg2Cl2 + 2Fe2(SO4)3 + 2FeCl3 MnO4– + 8H+ + 5Fe2+  5Fe3+ + Mn2+ + 4H2O Cr2O72– + 14H+ + 6Fe2+  6Fe3+ + 2Cr3+ + 7H2O

Uses :



High  Temp.

Fe2O3 + SO2 + SO3

It is used : (i) for making Blue - Black ink. (ii) as mordant in dyeing (iii) as insecticide in agriculture (iv) for making laboratory reagents like Mohr's salt etc. FeSO4 + (NH4)2 SO4 + 6H2O)  FeSO4 (NH4)2SO4.6H2O (Mohr's salt) (v) FeSO4 + H2O2 known as Fenton's reagent is used as catalyst.

Ferric Oxide, Fe2O3 : 

Preparation :

(i) (ii)

 Fe2O3 + 3H2O 2Fe(OH)3     2FeSO4 Fe2O3 + SO2 + SO3

(iii)

4FeS + 7O2  2Fe2O3 + 4SO2



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CHEMISTRY 

Properties :

(a) Physical : It is a deep red powder and is insoluble in water. (b) Chemical : (i) Amphoteric nature of iron (III) oxide. Fe2O3 + 6HCl  2FeCl3 + 3H2O Fe2O3 + 2NaOH fusion  H2O + 2NaFeO2 Fe2O3 + Na2CO3 fusion  2NaFeO2 + CO2 NaOCl and chlorine oxidises the NaFeO2 in to Na2FeO4. NaFeO2 + Cl2 + 4NaOH  2Na2FeO4 + 2NaCl + 2H2 (ii)

ºC    4Fe3O4 + O2 6Fe2O3 1300

(iii)

Fe2O3 + 3H2  2Fe + 3H2O

Uses :



It is used as (i) a red pigment, (ii) an abrasive polishing powder and (iii) a catalyst.

Ferric Chloride, FeCl3 :  Preparation : (a) Anhydrous FeCl3 :  (i) 12FeCl2 (anhydrous) + 3O2  2Fe2O3 + 8FeCl3 

(ii) (iii)

2Fe + 3Cl2(dry)  2FeCl3 FeCl3 . 6H2O + 6SOCl2  FeCl3 + 12HCl + 6SO2

(iv)

OCH3 | FeCl3.6H2O + 6CH3 — C — CH3  FeCl3 + 12CH3OH + 6CH3COCH3 | OCH3

(b) (i) (ii) (iii) (iv)

Hydrated FeCl3 : Fe(OH)3 + 3HCl  FeCl3 + 3H2O Fe2O3 + 6HCl  2FeCl3 + 3H2O Fe2(CO3)3 + 6HCl  2FeCl3 + 3H2O + 3CO2 2Fe + 4HCl + Cl2  2FeCl3 + 2H2  All solutions of FeCl3 obtained in above chemical reactions on crystallisation gives hydrated FeCl3.6H2O.

 Properties

(a) Physical : Anhydrous ferric chloride is dark black solid while hydrated salt, FeCl3 . 6H2O is yellowish-brown deliquescent crystalline solid. Both are soluble in water as well as in ether forming solvated species, [Fe(H2O)4Cl2]Cl . 2H2O and

O  FeCl3 respectively..

It is sublimed at 300°C giving a dimeric gas,

.

(b) Chemical : (i) Action of heat :  (a) 2FeCl3 (anhydrous)  2FeCl2 + Cl2  (b) 2[FeCl3. 6H2O]  Fe2O3 + 6HCl + 9H2O (ii) Aqueous solution is acidic due to hydrolysis. [Fe(H2O)6]3+ + H2O [Fe(H2O)5(OH)]2+ + H3O+ acid base base acid (iii) As oxidising agent : 2FeCl3 + SnCl2  2FeCl2 + SnCl4 2FeCl3 + H2S  2FeCl2 + 2HCl + S 2FeCl3 + SO2 + 2H2O  2FeCl2 + H2SO4+ 2HCl 2FeCl3 + 2KI  2FeCl2 + 2KCl + 2

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CHEMISTRY (iv) (v)

Formation of addition compounds : FeCl3 + 6NH3  FeCl3. 6NH3 ; FeCl3 + NOCl  FeCl3.NOCl 2FeCl3 + 3Na2CO3 + 3H2O  2Fe(OH)3 + 6NaCl + 3CO2

Uses :

(i) (ii) (iii) (iv)

It is used : as a medicine (its alcoholic solution named as tincture is ferri perchloride). for detection of acetates and phenols. for making prussian blue dye. as an oxidising agent.

[B] COMPOUNDS OF ZINC :



Zinc oxide, ZnO (Chinese white or philosopher's wool) It is found in nature as zincite or red zinc ore.  Preparation :

(i)

2Zn + O2  2ZnO

(ii)

 ZnCO3  ZnO + CO2

(iii)

 2Zn(NO3)2  2ZnO + 4NO2 + O2

(iv)

 Zn(OH)2  ZnO + H2O

 Properties

(a) Physical : It is a white powder which becomes yellow on heating due to change in the structure of lattice but again turns white on cooling. it is insoluble in water and sublimes at 400°C. (b) Chemical : (i) It is amphoteric in nature. ZnO + H2SO4  ZnSO4 + H2O ZnO + 2NaOH  Na2ZnO2 + H2O (ii)



ZnO + H2   Zn + H2O 400 º C

Uses :

(i) (ii) (iii) (iv) (v) Example-15

It is used : as a white paint. It does not get tarnished even in presence of H2S because ZnS is also white. for preparing Rinmann's green (Green paint - ZnCoO2) as catalyst for preparation of methyl alcohol for making soft rubber for making cosmetic powders, creams and in medicine 

Zn (OH)2  [X]. Select the correct statement (s) for the compound X. (A) X on heating with cobalt nitrate gives green mass (B) X on heating alone, becomes yellow but turns white on cooling. (C) Solution of X in dilute HCl gives bluish white/white precipitate with excess potassium ferrocyanide. (D) X is insoluble in aqueous sodium hydroxide.

Solution

(A) X is ZnO which on heating with cobalt nitrate gives ZnO. CoO, the Rinmann's green. (B) It turns yellow on heating and becomes white on cooling. (C) ZnCl2 forms bluish white/white precipitate. Zn3K2[Fe(CN)6]2. 3 Zn2+ + 2 K+ + 2 [Fe(CN)6]4–  K2Zn3[Fe(CN)6]2  (D) ZnO + 2NaOH  Na2ZnO2 (soluble complex) + H2O. So options A, B & C are correct and (D) is incorrect.

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CHEMISTRY 

Zinc Sulphate, ZnSO4 . 7H2O (White vitriol)  Preparation : (i) (ii) (iii)

Zn + H2SO4  ZnSO4 + H2 ZnO + H2SO4  ZnSO4 + H2O ZnCO3 + H2SO4  ZnSO4 + H2O + CO2

 Properties :

(a) Physical :

It is a colourless, crystalline solid soluble in water. It slowly effloresces when exposed to air. It is isomorphous with Epsom salt (MgSO4 . 7H2O)

(b) Chemical : (i) ZnSO4 + 2NaOH  Zn(OH)2 (white) + Na2SO4 Zn(OH)2 + 2NaOH  Na2ZnO2 (soluble complex) + 2H2O (ii) ZnSO4 + 2NaHCO3  ZnCO3 + Na2SO4 + H2O + CO2 (iii)

ºC ºC ºC   ZnSO4 . 6H2O 280   ZnSO4 800   ZnO + SO3 ZnSO4.7H2O 100

Uses :

It is used : (i) as eye lotion, (ii) for making lithophone - mixture of BaS + ZnSO4 (white paint) and (iii) as mordant in dyeing



Zinc Chloride ZnCl2.2H2O  Preparation :

It is prepared by dissolving zinc oxide, carbonate or hydroxide in dilute hydrochloric acid ZnO + 2HCl  ZnCl2 + H2O The solution so obtained on concentration and cooling gives the crystals of ZnCl2 . 2H2O. Anhydrous zinc chloride cannot be prepared merely by heating the crystals of ZnCl2 . 2H2O because a basic chloride Zn(OH)Cl is formed during decomposition which on further heating gives ZnO. ZnCl2 . 2H2O  Zn(OH)Cl + HCl + H2O Zn(OH)Cl  ZnO + HCl Hence anhydrous zinc chloride is obtained either by heating zinc in a current of chlorine gas or by distilling a mixture of zinc powder and mercuric chloride  Zn + Cl2  ZnCl2 ; Zn + HgCl2  ZnCl2 + Hg  Properties :

It is a white crystalline solid, deliquescent and soluble in water. On heating it forms the oxychloride and finally the oxide. Its concentrated solution sets to a hard mass when mixed with zinc oxide and the product is used as a dental filling. [C]

COMPOUNDS OF COPPER



Copper Sulphate, CuSO4 . 5H2O (Blue Vitriol) It is also called 'Nilathotha'. 

(i) (ii)

Preparation : CuO + H2SO4  CuSO4 + H2O Cu(OH)2 + H2SO4  CuSO4 + 2H2O

(iii)

Cu + H2SO4 +

(iv)

CuCO3.Cu(OH)2 (malachite ore) + 2H2SO4  2CuSO4 + 3H2O + CO2 

1 O  CuSO4 + H2O 2 2

 Properties

(a) Physical : It is a blue crystalline compound and is soluble in water. (b) Chemical : (i)

CuSO4 + 2NaOH  Cu(OH)2  (blue) + Na2SO4 2CuSO4 + Na2CO3 + H2O  Cu(OH)2  (blue) + Na2SO4 + CO2  .

(ii)

2CuSO4 + SO2 + 2H2O + 2KI  CuI2 + 2H2SO4 + K2SO4 2CuSO4 + 2FeSO4 + 2KI  Cu2 I2 + Fe2(SO4)3 + K2SO4

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CHEMISTRY (iii)

CuSO4 + Fe  Cu + FeSO4 CuSO4 + Zn  Cu + ZnSO4

(iv)

2CuSO4 + 2KSCN + SO2 + 2H2O  2CuSCN  (white) + K2SO4 + 2H2SO4

100 º C 250 º C 750 º C (v) CuSO4.5H2O air CuSO4.4H2O    CuSO4 . H2O   CuSO4    CuO + SO2 + O2

effloresce s



Uses : It is used : (i) for making other copper compounds. (ii) for electroplating, electrotyping, as mordant in dyeing. (iii) in making Bordeaux mixture which is used in agriculture as fungicide and germicide. (iv) in making Fehling's solution. (v) in medicine as antiseptic. (vi) in electric batteries. Example-16

Solution

Anhydrous white solid (A) on addition of potassium iodide solution gave a brown precipitate which turned white (B) on addition of excess of hypo solution. When potassium cyanide is added to an aqueous solution of (A) a white precipitate is formed which then dissolves in excess forming (C). A solution (1%) of (A) on adding to a solution of white portion of egg produced violet colouration in alkaline medium (i.e. in presence of NaOH). Identify compound (A) and explain the reactions. As 1% solution of (A) produced violet colouration with white portion of egg (Biuret test) and (A) with potassium iodide gives brown precipitate which turned white on adding hypo. The (A) may be anhydrous CuSO4 (White). This is further confirmed by the reaction of (A) with potassium cyanide. 2CuSO4 + 4K  Cu22  (white) + 2  (yellow or brown) + 2K2SO4 (A) (B) 2 + 2Na2S2O3  Na2S4O6 + 2Na 2CuSO4 + 4KCN  2 Cu(CN)2  (yellow) + 2K2SO4 ; 2Cu(CN)2  2CuCN  (white) + (CN)2  2 CuCN + 6 KCN  2 K3 [Cu(CN)4] (colourless soluble complex). (C)

Example-17

100 º C 250 º C CuSO4.5H2O    [X] (s)    [Y] (s) (A) X and Y are CuSO4. 3H2O and CuSO4 (B) X and Y are CuSO4. 3H2O and CuSO4 . H2O (C) X and Y are CuSO4. H2O and CuSO4 (D) X and Y are CuSO4 and CuO

Solution

100 º C 250 º C  CuSO4.H2O(s)    CuSO4 (s). CuSO4.5H2O   So, (C) option is correct.



Cupric Oxide, CuO It is called black oxide of copper and is found in nature as tenorite.  Preparation : It is prepared (i) By heating Cu2O in air or by heating copper for a long time in air (the temperature should not exceed above 1100°C) 2Cu2O + O2  4CuO 2Cu + O2  2CuO  (ii) Cu(OH)2  CuO + H2O  (iii) 2Cu(NO3)2  2CuO + 4NO2 + O2 (iv) On a commercial scale, it is obtained by heating/calcination of malachite which is found in nature. CuCO3 . Cu(OH)2  2CuO + CO2 + H2O  Properties

(a) Physical : It is black powder insoluble in water and stable to moderate heating.

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CHEMISTRY (b) Chemical : (i) The oxide dissolves in acids HCl, H2SO4 of HNO3 forming corresponding salts. CuO + 2H+  Cu2+ + H2O (ii) When heated to 1100 – 1200°C, it is converted into cuprous oxide with evolution of oxygen. 4CuO  2Cu2O + O2 (iii) It is reduced to metallic copper by reducing agents like hydrogen, carbon and carbon monoxide. CuO + H2  Cu + H2O



Cupric Chloride, CuCl2 . 2H2O  Preparation : (i) The metal or cupric oxide or cupric hydroxide or copper carbonate is dissolved in concentrated HCl. The resulting solution on crystallisation gives green crystals of hydrated cupric chloride. 2Cu + 4HCl + O2  2CuCl2 + 2H2O CuO + 2HCl  CuCl2 + H2O Cu(OH)2CuCO3 + 4HCl  2CuCl2 + 3H2O + CO2 (ii)

Anhydrous cupric chloride is obtained as a dark brown mass when copper metal is heated in excess of chlorine gas or by heating hydrated cupric chloride in HCl gas at 150°C. Cu + Cl2  CuCl2 ºC   CuCl2 + 2H2O CuCl2 . 2H2O 150 HCl gas

 Properties

(a) Physical : It is deliquescent compound and is readily soluble in water. The dilute solution is blue but concentrated solution is, however, green. It changes to yellow when concentrated HCl is added. The blue colour is due to complex cation [Cu(H2O)4]2+ and yellow colour due to complex anion [CuCl4]2– and green when both are present. (b) Chemical : (i) The aqueous solution is acidic due to its hydrolysis CuCl2 + 2H2O Cu(OH)2 + 2HCl (ii) The anhydrous salt on heating forms Cu2Cl2 and Cl2 2CuCl2  Cu2Cl2 + Cl2 

3CuCl2 . 2H2O  CuO + Cu2Cl2 + 2HCl + Cl2 + 5H2O Strong

(iii)

It is readily reduced to Cu 2Cl 2 by copper turnings or sulphur dioxide gas or nascent hydrogen (obtained by the action of HCl on Zn) or SnCl2. CuCl2 + Cu  Cu2Cl2 2CuCl2 + SO2 + 2H2O  Cu2Cl2 + 2HCl + H2SO4 2CuCl2 + 2H  Cu2Cl2 + 2HCl 2CuCl2 + SnCl2  Cu2Cl2 + SnCl4

(iv)

A pale blue precipitate of basic cupric chloride CuCl2 . 3Cu(OH)2 is obtained when NaOH is added. CuCl2 + 2NaOH  Cu(OH)2 + 2NaCl CuCl2 + 3Cu(OH)2  CuCl2 . 3Cu(OH)2  It dissolves in ammonium hydroxide forming a deep blue solution. On evaporation of this solution deep blue crystals of tetraammine cupric chloride are obtained. CuCl2 + 4NH4OH  [Cu(NH3)4]Cl2 . H2O + 3H2O

[D]

COMPOUNDS OF SILVER :



Silver Nitrate, AgNO3 (Lunar caustic) : 

Preparation : It is prepared by heating silver with dilute nitric acid. The solution is concentrated and cooled when the crystals of silver nitrate separate out. heat 3Ag + 4HNO3 (dilute)    3AgNO3 + NO + 2H2O

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CHEMISTRY 

Properties

(a) Physical : It is a colourless crystalline compound, soluble in water and alcohol. It melts at 212°C. (b) Chemical : (i) It possesses powerful corrosive action on organic tissues, which it turns black especially in presence of light. The blackening is due to finely divided metallic silver, reduced by organic tissue. It is, therefore, stored in coloured bottles. (ii) On heating above its melting point, it decomposes to silver nitrite and oxygen. 2AgNO3  2AgNO2 + O2 When heated red hot, it gives metallic silver 2AgNO3  2Ag + 2NO2 + O2 (iii) Solutions of halides, phosphates, sulphides, chromates, thiocyanates, sulphates and thiosulphates, all give a precipitate of the corresponding silver salt with silver nitrate solution (for reactions refer qualitative analysis sheet). (iv)

Solid AgNO3 absorbs ammonia gas with the formation of an addition compound, AgNO3. 3NH3. When NH4OH is added to silver nitrate solution, a brown precipitate of silver oxide appears which dissolves in excess of ammonia forming a complex salt. 2AgNO3 + 2NH4OH  Ag2O + 2NH4NO3 + H2O Ag2O + 2NH4NO3 + 2NH4OH  2[Ag(NH3)2]NO3 + 3H2O

(v)

The ammonical solution of AgNO3 react with acetylene to form a white precipitate. 2AgNO3 + 2NH4OH + C2H2  Ag2C2 (silver acetylide) + 2NH4NO3 + 2H2O

(vi)

Ammonical silver nitrate is called Tollen’s reagent and is used to identify reducing sugars and aldehydes.  RCHO + 2Ag+ + 3OH–  RCOO– + 2Ag + 2H2O It is known as “silver mirror” test of aldehydes and reducing sugars. Ag2O + HCHO  2Ag + HCOOH Ag2O + C6H12O6  2Ag + C6H12O7

(vii)

Reaction with iodine : (a) 6AgNO3 (excess) + 3I2 + 3H2O  AgIO3 + 5AgI + 6HNO3 (b) 5AgNO3 + 3I2 (excess) + 3H2O  HIO3 + 5AgI+ 5HNO3

(viii)

Silver is readily displaced from an aqueous silver nitrate solution by the base metals, particularly, if the solution is some what acidic. 2AgNO3 + Cu  2Ag + Cu(NO3)2 2AgNO3 + Zn  2Ag + Zn(NO3)2

(ix)

Phosphine, arsine and stibine all precipitate silver from silver nitrate solution PH3 + 6AgNO3 + 3H2O  6Ag + 6HNO3 + H3PO3 AsH3 + 6AgNO3 + 3H2O  6Ag + 6HNO3 + H3AsO3

Uses :

(i)

It is used as a laboratory reagent for the identification of various acidic radicals especially for chloride, bromide and iodide. The ammonical silver nitrate solution i.e., Tollen's reagent is used in organic chemistry for testing aldehydes, reducing sugars, etc.

(ii)

Silver nitrate is used for making silver bromide which is used in photography.

(iii)

It is used in the preparation of making inks and hair dyes.

(iv)

It is used extensively for the preparation of silver mirrors. The process of depositing a thin and uniform layer of silver on a clean glass surface is known as silvering of mirrors. It is employed for making looking glasses, concave mirrors and reflecting surfaces. The process is based on the reduction of ammonical silver nitrate solution by some reducing agent like HCHO. The silver film deposited on the glass is first coated with a varnish and finally painted with red lead to prevent its being scraped off.

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CHEMISTRY 

Silver oxide (Ag2O) :  Preparation : It is prepared as brown solid by adding sodium hydroxide solution to silver nitrate solution. 2AgNO3 + 2NaOH  Ag2O + 2NaNO3 + H2O  Properties : It is a brown solid sparingly soluble in water. (i) On heating, it is decomposed to black metallic silver with evolution of oxygen at 160°C. (ii) (iii)

(iv)

 2Ag2O  4Ag + O2  It is reduced to the metal by hydrogen on heating  Ag2O + H2  2Ag + H2O The aqueous solution is strongly basic, solution behaving as AgOH. Moist silver oxide hydrolyses the alkyl halide to alcohol. C2H5Br + AgOH  C2H5OH + AgBr Dry distillation of a mixture of alkyl halide with silver oxide produces ether.  2C2H5Br + Ag2O  C2H5 – O – C2H5 + 2AgBr

Uses :  

It is used : As Tollen’s reagent for the detection of aldehydes, formic acid and terminal alkynes. In the manufacture of mirrors.

Photography : A photographic film consists of a light sensitive emulsion of fine particles (grains) of silver salts in gelatine spread on a clear celluloid strip or a glass plate. The grain size is very important to photographers, as this affects the quality of the pictures produced. AgBr is mainly used as the light sensitive material. Some AgI is used in ‘fast’ emulsions. The film is placed in a camera. When the photograph is exposed, light from the subject enters the camera and is focussed by the lens to give a sharp image on the film. The light starts a photochemical reaction by exciting a halide ion, which loses an electron. The electron moves in a conduction band to the surface of the grain, where it reduces a Ag+ ion to metallic silver. light 2AgBr(s)   2Ag + Br2 In modern photography only a short exposure of perhaps 1/100th of a second is used. In this short time, only a few atoms of silver (perhaps 10–50) are produced in each grain exposed to light. Parts of the film which have been exposed to the bright parts of the subject contain a lot of grains with some silver. Parts exposed to paler parts of the subject contains a few grains with some silver, whilst parts not exposed contain none. Thus the film contains a latent image of the subject. However, the number of silver atoms produced is so small that the image is not visible to the eye. Next the film is placed a developer solution. This is a mild reducing agent, usually containing quinol. Its purpose is to reduce more silver halide to Ag metal. Ag is deposited mainly where there are already some Ag atoms. Thus the developing process intensifies the latent image on the film so it becomes visible. The correct conditions for processing must be used to obtain an image of the required blackness.

2AgBr(s) + 2OH–(aq) + HO

OH (aq)  2Ag(s) + 2H2O() + O

O + 2Br–(aq)

hydroquinol hydroquinone If the film was brought out into daylight at this stage, the unexposed parts of the emulsion would turn black and thus destroy the picture. To prevent this happening any unchanged silver halides are removed by placing the film in a fixer solution. A solution of sodium thiosulphate is used as fixer. It forms a soluble complex with silver halides. AgBr + 2Na2S2O3  Na3[Ag(S2O3)2] + NaBr After fixing, the film can safely be brought out into daylight. Parts blackened by silver represent the light parts of the original picture. This is therefore a negative. To obtain an image with light and dark the right way round, a print must be made. Light is passed through the negative onto a piece of paper coated with AgBr emulsion. This is then developed and fixed in the same way as before.

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CHEMISTRY [E]

POTASSIUM PERMANGANATE (KMnO4) :  Preparation

This is the most important and well known salt of permanganic acid and is prepared from the pyrolusite ore. It is prepared by fusing pyrolusite ore either with KOH or K2CO3 in presence of atmospheric oxygen or any other oxidising agent such as KNO3. The fused mass turns green with the formation of potassium manganate, K2MnO4. 2MnO2 + 4KOH + O2  2K2MnO4 + 2H2O 2MnO2 + 2K2CO3 + O2  2K2MnO4 + 2CO2 The fused mass is extracted with water and the solution is now treated with a current of chlorine or ozone or carbon dioxide to convert manganate into permanganate. 2K2MnO4 + Cl2  2KMnO4 + 2KCl 2K2MnO4 + H2O + O3  2KMnO4 + 2KOH + O2 3K2MnO4 + 2CO2  2KMnO4 + MnO2 + 2K2CO3 Commercially it is prepared by fusion of MnO2 with KOH followed by electrolytic oxidation of manganate. MnO42– (green)  MnO4– (purple) + e– In the laboratory, a manganese(II) ion salt is oxidised by peroxodisulphate to permanganate 2Mn2+ + 5S2O82– + 8H2O  2MnO4– + 10SO42– + 16H+  Properties

(a) Physical : It is purple coloured crystalline compound. It is moderately soluble in water at room temperature. (b) Chemical : (i) Effect of heating When heated alone or with an alkali, it decomposes evolving oxygen. K   K2MnO4 + MnO2 + O2 2KMnO4 750 Re d  2K MnO + O 2K2MnO4  2 3 2 Heat

  4KMnO4 + 4KOH  4K2MnO4 + 2H2O + O2 or   4MnO4– + 4OH–  4MnO42– + O2 + 2H2O.

MnO42– in dilute alkaline, water and acidic solutions is unstable and disproportionates to give MnO4– and MnO2 . 3MnO42– + 4H+  2MnO4– + MnO2  + 2H2O 3MnO42– + 2H2O  2MnO4– + MnO2  + 4OH– or 2– 3MnO4 + 3H2O  2MnO4– + MnO (OH)2  + 4OH– On heating in current of H2 , solid KMnO4 gives MnO   2KMnO4 + 5H2  2KOH + 2MnO + 4H2O

(ii)

On treatment with concentrated H2SO4 (KMnO4 is taken in excess), it forms manganese heptoxide via permanganyl sulphate which decomposes explosively on heating. 2KMnO4 + 3H2SO4  2KHSO4 + (MnO3)2SO4 + 2H2O (MnO3)2SO4 + H2O  Mn2O7 + H2SO4 3 Mn2O7  2MnO2 + O2 2 KMnO4 + 3H2SO4 (conc.)  K+ + MnO3+ (green) + 3HSO4– + H3O+.

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CHEMISTRY (iii)

 

Potassium permanganate is a powerful oxidising agent. Potassium permanganate acts as an oxidising agent in alkaline, neutral or acidic solutions. A mixture of sulphur, charcoal and KMnO4 forms an explosive powder. A mixture of oxalic acid and KMnO4 catches fire spontaneous after a few seconds. The same thing happens when glycerine is poured over powdered KMnO4. In alkaline & neutral medium : In strongly alkaline medium KMnO4 is reduced to manganate. 2KMnO4 + 2KOH (conc.)  2K2 MnO4 + H2O + [O] or e– + MnO4–  MnO42– However if solution is dilute then K2MnO4 is converted in to MnO2 which appears as a brownish precipitate. 2K2MnO4 + 2H2O  2MnO2 + 4KOH + 2[O] or 2e– + 2H2O + MnO42–  MnO2 + 4OH– This type of behaviour is shown by KMnO4 itself in neutral medium 3e– + 2H2O + MnO4–  MnO2 + 4OH– In alkaline or neutral medium KMnO4 shows following oxidising properties. (a) It oxidises ethene to glycol.



In alkaline medium KMnO4 solution is also known as Bayer’s reagent (1% alkaline KMnO4 solution). (b) It oxidises iodide into iodate. 3e– + 2H2O + MnO4–  MnO2 + 4OH– × [2] 6OH– + –  O3– + 3H2O + 6e– ––––––––––––––––––––––––––––––––––––––––––– 2MnO4– + – + H2O  2MnO2 + O3– + 2OH– (c) H2S is oxidised into sulphur : 2MnO4– + 3H2S  2MnO2 + 2OH– +2H2O + 3S (d) Manganese sulphate is oxidised to MnO2 : 2MnO4– + 3MnSO4 + 2H2O  K2SO4 + 5MnO2 + 2H2SO4 (e) In neutral/ faintly alkaline solution thiosulphate is quantitatively oxidised to sulphate. 8MnO4– + 3S2O32– + H2O  8MnO2 + 6SO42– + 2OH– (f) In the presence of sodium hydroxide, sodium sulphite is oxidised in to sodium sulphate. 2MnO4– + 3SO32– + 3H2O  2MnO(OH)2  + 3SO42– + 2OH–.



In acidic medium (in presence of dilute H2SO4) : Manganous sulphate is formed. The solution becomes colourless. 2KMnO4 + 3H2SO4  K2SO4 + 2MnSO4 + 3H2O + 5[O] or MnO4¯ + 8H+ + 5e–  Mn2+ + 4H2O This medium is used in quantitative (volumetric) estimations. The equivalent mass of KMnO4 in acidic medium is =

Molecular mass . 5

(a) Ferrous salts are oxidised to ferric salts. MnO4¯ + 5Fe2+ + 8H+  5Fe3+ + Mn2+ + 4H2O (b) Iodine is evolved from potassium iodide. 2MnO4¯ + 10I¯ + 16H+  2Mn2+ + 5I2 + 8H2O

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CHEMISTRY (c) H2S is oxidised to sulphur : 2MnO4– + 16H+ + 5S2–  2Mn2+ + 8H2O + 5S (d) SO2 is oxidised to H2SO4 : 2MnO4– + 5SO2 + 2H2O  5SO42– + 2Mn2+ + 4H+ (e) Nitrites are oxidised to nitrates : 2MnO4– + 5NO2– + 6H+  2Mn2+ + 3H2O + 5NO3– (f) Oxalic acid is oxidised to CO2 : This reaction is slow at room temperature, but is rapid at 60ºC. Mn(II) ions produced catalyse the reaction; thus the reaction is autocatalytic. 60 º C 2MnO4– + 16H+ + 5C2O42–  2Mn2+ + 8H2O + 10CO2

(g)

HCHO is oxidised to HCOOH 2MnO4– + 5HCHO + 6H+  2Mn2+ + 5HCOOH + 3H2O.

(h)

It oxidises hydrogen halides (HCl, HBr or HI) into X2 (halogen) 2MnO4– + 16H+ + 10X–  2Mn2+ + 8H2O + 5X2 H2O2 is oxidised to O2. 2MnO4– + 6H+ + 5H2O2  5O2 + Mn2+ + 8H2O.

(i)

 Uses :

(i) (ii)

(iii)

KMnO4 is used as an oxidising agent in laboratory and industry. Alkaline potassium permanganate is called Bayer's reagent. This reagent is used in organic chemistry for the test of unsaturation. KMnO4 is used in the manufacture of saccharin, benzoic acid, acetaldehyde etc. KMnO4 is used in qualitative analysis for detecting halides, sulphites, oxalates, etc.

Example-18

Potassium permanganate acts as an oxidant in neutral, alkaline as well as acidic media. The final products obtained from it in three conditions are respectively : (A) MnO42– , Mn3+ and Mn2+ (B) MnO2 , MnO2 and Mn2+ + 3+ (C) MnO2 , MnO2 and Mn (D) MnO , MnO2+ and Mn2+

Solution

3e– + 2H2O + MnO4–  MnO2 + 4OH– (neutral medium) e– +MnO4–  MnO4–2 (dilute alkaline medium) MnO4¯ + 8H+ + 5e–  Mn2+ + 4H2O (acidic medium) Therefore, (B) option is correct.

[F]

POTASSIUM DICHROMATE (K2Cr2O7) :  Preparation : The chromite ore is roasted with sodium carbonate in presence of air in a reverberatory furnace Roasting  8Na CrO + 2Fe O + 8CO 4FeO. Cr2O3 (chromite ore) + 8Na2CO3 + 7O2  2 4 2 3 2 in air

The roasted mass is extracted with water when Na 2CrO 4 goes into the solution leaving behind insoluble Fe2O3. The solution is then treated with calculated amount of H2SO4. 2Na2CrO4 + H2SO4  Na2Cr2O7 + Na2SO4 + H2O The solution is concentrated when less soluble Na2SO4 crystallises out. The solution is further concentrated when crystals of Na2Cr2O7 are obtained. Hot saturated solution of Na2Cr2O7 is then treated with KCl when orange red crystals of K2Cr2O7 are obtained on crystallisation. Na2Cr2O7 + 2KCI  K2Cr2O7 + 2 NaCl

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CHEMISTRY



K2Cr2O7 is preferred over Na2Cr2O7 as a primary standard in volumetric estimation because Na2Cr2O7 is hygroscopic in nature but K2Cr2O7 is not.

 Properties

(a) Physical : It is orange-red coloured crystalline compound. It is moderately soluble in cold water but freely soluble in hot water. It melts at 398°C. (b) Chemical : (i)

Effect of heating : On heating strongly, it decomposes liberating oxygen. 2K2Cr2O7  2K2CrO4 + Cr2O3 +

3 O 2 2

On heating with alkalies, it is converted to chromate, i.e., the colour changes from orange to yellow. On acidifying, yellow colour again changes to orange. K2Cr2O7 + 2KOH  2K2CrO4 + H2O  2CrO 2 Cr2O 2 7 + 2OH¯ 4 + H2 O

Orange

Yellow

+  Cr2O 2 2CrO 2 4 + 2H 7 + H2 O

Yellow Orange Thus CrO42– and Cr2O72– exist in equilibrium and are interconvertable by altering the pH of solution. + 2CrO 2 4 + 2H

2HCrO4–

Cr2O 2 7 + H2 O



In alkaline solution, chromate ions are present while in acidic solution, dichromate ions are present.

(ii)

K2Cr2O7 + 2H2SO4 (conc. & cold)  2CrO3  (bright orange/red) + 2KHSO4 + H2O

2K Cr O 2

2

7

+ 8H2SO4 (conc. & Hot)  2K2SO4 + 8H2O + 2Cr2(SO4)3 + 3O2

(iii)

Acidified K2Cr2O7 solution reacts with H2O2 to give a deep blue solution due to the formation of CrO5. Cr2O72– + 2H+ + 4H2O2  2CrO5 + 5H2O Blue colour in aqueous solution fades away slowly due to the decomposition of CrO5 to Cr3+ ions and oxygen. In less acidic solution K2Cr2O7 and H2O2 give salt which is violet coloured and diamagnetic due to the formation of [CrO(O2)(OH)]–. In alkaline medium with 30% H2O2, a red-brown K3CrO8 (diperoxo) is formed. It is tetra peroxospecies [Cr(O2)4]3– and thus the Cr is in +V oxidation state. In ammonical solution a dark red-brown compound, (NH3)3CrO4 - diperoxo compound with Cr(IV) is formed.

(iv)

Potassium dichromate reacts with hydrochloric acid and evolves chlorine gas. K2Cr2O7 + 14HCl  2KCl + 2CrCl3 + 7H2O + 3Cl2

(v)

It acts as a powerful oxidising agent in acidic medium (dilute H2SO4) + –  2Cr+3 + 7H2O. (Eº = 1.33 V) Cr2O 2 7 + 14H + 6e

The oxidation state of Cr changes from + 6 to +3. (a) Iodine is liberated from potassium iodide : + –  2Cr3+ + 7H2O Cr2O 2 7 + 14H + 6e 2–  2 + 2e– × [3] ––––––––––––––––––––––––––––––––––––– +  2Cr3+ + 32 + 7H2O Cr2O 2 7 + 14H + 6¯ (b) Ferrous salts are oxidised to ferric salts : +  6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ + Cr2O 2 7 + 14H

(c) Sulphites are oxidised to sulphates : 2 2 +  Cr2O 2 3SO 4 + 2Cr3+ + 4H2O 7 + 3SO 3 + 8H

(d) H2S is oxidised to sulphur : +  Cr2O 2 2Cr3+ + 7H2O + 3S 7 + 3H2S + 8H

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CHEMISTRY



(e) SO2 is oxidised to H2SO4 : Cr2O72– + 3SO2 + 2H+  2Cr3+ + 3SO42– + H2O ; Chrome alum is obtained when acidified K2Cr2O7 solution is saturated with SO2.  70 ºC K2Cr2O7 + H2SO4 + SO2 + H2O T  K2SO4 . Cr2(SO4)3 . 24H2O

(f) It oxidises ethylalcohol to acetaldehyde and acetaldehyde to acetic acid O] O] C2H5OH [ CH3CHO [ CH3COOH ethyl alcohol acetaldehyde acetic acid

(g) It also oxidises nitrites to nitrates, arsenites to arsenates, HBr to Br2. HI to I2, etc.   (h) K2Cr2O7 + 2C (charcoal)  Cr2O3 + K2CO3 + CO 

(vi)

(vii)

Chromyl chloride test : 4Cl– + Cr2O72– + 6H+  2CrO2Cl2  (deep red) + 3H2O CrO2Cl2 + 4OH–  CrO42– (yellow) + 2Cl– + 2H2O CrO42– (yellow) + Pb2+  PbCrO4  (yellow) Cr2O72– (concentrated solution) + 2Ag+  Ag2Cr2O7  (reddish brown) 2– + Ag2Cr2O7  + H2O boil  Ag2CrO4  + CrO4 + 2H .

(viii)

Cr2O72– + Ba2+ + H2O

2BaCrO4  + 2H+

As strong acid is produced, the precipitation is only partial. But if NaOH or CH3COONa is added, precipitate becomes quantitative. Uses :

(i)

It is used : as a volumetric reagent in the estimation of reducing agents such as oxalic acid, ferrous ions, iodide ions, etc. It is used as a primary standard.

(ii)

for the preparation of several chromium compounds such as chrome alum, chrome yellow, chrome red, zinc yellow, etc.

(iii)

in dyeing, chrome tanning, calico printing, photography etc.

(iv)

as a cleansing agent for glass ware in the form of chromic acid.

Example-19

Solution

An inorganic compound (A) has garnet red prismatic crystals. (A) is moderately soluble in water and dissolves in cold concentrated H2SO4 to yield red crystals (B). In presence of dilute H2SO4 it converts a pungent gas (C) into a yellow turbidity (D) and converts a suffocating gas (E) into a green solution (F). The gas (C) and (E) also combine to produce the yellow turbidity (D). With K and starch in presence of dilute. H2SO4 (A) yields blue colour. (A) and concentrated H2SO4 mixture is used as a cleansing agent for glassware in the laboratory. Identify (A) and explain the reactions. As compound (A) has garnet red prismatic crystals which with cold conc. H2SO4 gives red crystals and a suffocating gas (SO2) turns its solution in water in to green coloured solution, therefore compound (A) may be K2Cr2O7 . K2Cr2O7 + 4H2SO4 + 3H2S  KHSO4 + 2 CrO3 (red crystals) + H2O (A)

(B)

K2Cr2O7 + 4H2SO4 + 3H2S  K2SO4 + Cr2 (SO4)3 + 7H2O + 3S (Yellow) (C)

(D)

K2Cr2O7 + H2SO4 + 3SO2  Cr2 (SO4)3 (Green solution) + K2SO4 + H2O (E)

(F)

K2Cr2O7 + 7H2SO4 + 6K  4K2SO4 + Cr2 (SO4)3 + 7H2O + 32 2 + Starch  Blue colour

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CHEMISTRY MISCELLANEOUS SOLVED PROBLEMS (MSPs) 1.

Among the following statements choose the true or false statement(s). (a) K2Cr2O7 on heating with charcoal gives metallic potassium and Cr2O3. (b)

On heating in current of H2 the crystalline KMnO4 is converted into KOH and Mn3O4.

Ans.

(c) (a)

Hydrated ferric chloride on treatment with 2, 2–dimethoxypropane gives anhydrous ferric chloride. False (b) False (c) True

Sol.

(a)

  K2Cr2O7 + 2C (charcoal)  Cr2O3 + K2CO3 + CO  .

(b)

  2KMnO4 + 5H2  2 KOH + 2MnO + 4H2O.

(c)

OCH3 | FeCl3 . 6H2O + 6CH3 — C — CH3  FeCl3 (anhydrous) + 12CH3OH + 6CH3COCH3 . | OCH3

2.

Ans.

A compound (A) is used in paints instead of salts of lead. Compound (A) is obtained when a white compound (B) is strongly heated. Compound (B) is insoluble in water but dissolves in sodium hydroxide forming a solution of compound (C). The compound (A) on heating with coke gives a metal (D) and a gas (E) which burns with blue flame. (B) also dissolves in ammonium sulphate solution mixed with ammonium hydroxide. Solution of compound (A) in dilute HCl gives a bluish white / white precipitate (F) with excess of K4[Fe(CN)6]. Identify (A) to (F) and explain the reactions. (A) ZnO, (B) Zn(OH)2, (C) Na2 ZnO2 , (D) Zn, (E) CO, (F) K2Zn3 [Fe(CN)6]2

Sol.

  Zn(OH)2 (B)  ZnO (A) + H2O.

Zn(OH)2 (B)  + 2OH–  [Zn(OH)4]2– (C) (soluble complex).   ZnO (A) + C  Zn (D) + CO (E).

Zn(OH)2 (B) + 4NH3  [Zn(NH3)4]2+ (soluble complex) + 2OH–. ZnO + 2HCl  ZnCl2 + H2O. 3ZnCl2 + 2K4[Fe(CN)6]  K2Zn3[Fe(CN)6]2  (bluish white/white) (F) + 6KCl. 3.

Ans. Sol.

An unknown inorganic compound (X) gave the following reactions: (i) The compound (X) on heating gave a residue, oxygen and oxide of nitrogen. (ii) An aqueous solution of compound (X) on addition to tap water gave a turbidity which did not dissolve in HNO3. (iii) The turbidity dissolves in NH4OH. Identify the compound (X) and give equations for the reactions (i), (ii) & (iii). X = AgNO3   2AgNO3 (X)  2 Ag + 2NO2 + O2 . AgNO3 (aq.) + Cl—  AgCl  (white) + NO3— . AgCl + 2NH3  [Ag(NH3)2]+ (soluble complex).

4.

Ans. Sol.

Amongst [TiF6]2– , [CoF6]3– , Cu2 Cl2 and [NiCl4]2– [Atomic number ; Ti = 22, Co = 27, Cu = 29, Ni = 28] the colourless species are : (A) [TiF6]2– and [Cu2Cl2] (B) Cu2Cl2 and [NiCl4]2– (C) [TiF6]2– and [CoF6]3– (D) [CoF6]3– and [NiCl4]2– (A) In [TiF6] 2– the titanium is in +4 oxidation state having the electronic configuration [Ar]18 3d0 4s0. Similarly in Cu2Cl2 the copper is in +1 oxidation state having the electronic configuration [Ar]18 3d10 4s0. As they do not have any unpaired electrons for d-d transition, they are therefore colourless. In [NiCl4]2– the nickel is in +2 oxidation state and electronic configuration is [Ar]18 3d8 4s0. As it has two unpaired electrons, so the complex is coloured. In [CoF6]3–, the cobalt is in +3 oxidation state having electron configuration [Ar] 3d6 4s0. As it has four unpaired electrons, so the complex is coloured.

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CHEMISTRY 5.

Sol.

On the basis of trends in the properties of the 3d-series elements, suggests possible M2+ aqua ions for use as reducing agents, and write a balanced chemical equation for the reaction of one of these ions with O2 in acidic solution. Because oxidation state +2 is most stable for the later elements of 3d-series elements, strong reducing agents include ions of the metals on the left of the series: such ions include V2+ (aq) and Cr2+ (aq) The Fe2+ (aq) ion is only weakly reducing. The Co2+ (aq), Ni2+ (aq), and Cu2+ (aq) ions are not oxidized in water. 2+  0.77 – 0.44 Fe3+    Fe   Fe

The chemical equation for the oxidation is then 4 Fe2+ (aq) + O2(g) + 4H+ (aq)  4Fe3+ (aq) + 2H2O (). 6.

Match the reactions given in column-I with the characteristic(s) of the reaction products given in column-II. Column-I Column-II (A)

Zn  TiCl4 

(p)

One of the products is bright orange coloured but diamagnetic.

(B)

573 K FeCl3  

(q)

One of the products is green coloured and paramagnetic.

(C) (D)

KMnO4   (r) K2Cr2O7 + H2SO4 (cold & conc.) (s) 750 K

One of the products is violet and paramagnetic. One of the products exists as dimer.

Ans.

[A – r] ; [B – s ; [C – q] ; [D – p].

Sol.

Zn  TiCl3, violet (one unpaired electron so d-d transition is possible). (A) TiCl4 

573 K (B) 2FeCl3  

gas dimer..

750 K (C) 2KMnO4   K2MnO4 green (one unpaired electron so d-d transition is possible) + MnO2 + O2 .

(D) K2Cr2O7 + 2H2SO4  2CrO3 bright orange (diamagnetic) + 2KHSO4 + H2O. 7.

Sol.

8.

Sol.

Which of the following is true for the species having 3d4 configuration ? (A) Cr2+ is reducing in nature. (B) Mn3+ is oxidising in nature. (C) Both (A) and (B) (D) None of these 2+ Cr is reducing as its configuration changes from d4 to d3, the latter having a half-filled t32g energy level of 3d orbitals in octahedral crystal field spliting. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability. Therefore, (C) option is correct. Which of the following increasing order of oxidising power is correct for the following species ? VO2+ , MnO4– , Cr2O72– (A) VO2+ < Cr2O72– < MnO4–

(B) VO2+ < MnO4– < Cr2O72–

(C) Cr2O72– < VO2+ < MnO4–

(D) Cr2O72– < MnO4– < VO2+

This is attributed to the increasing stability of the lower species to which they are reduced. MnO4– is reduced to Mn2+ which has stable half filled valence shell electron configuration [3d5]. Cr2O72– is reduced to Cr3+ which has half filled t32g energy level of 3d orbitals in octahedral crystal field spliting VO2+ is reduced to V3+ which has electronic configuration [Ar]18 3d2 4s0. So the order of increasing stability of the reduced species is Mn2+ > Cr3+ > V3+ and, therefore, the increasing order of oxidising power is VO2+ < Cr2O72– < MnO4– . Therefore, (A) option is correct.

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CHEMISTRY 9.

Sol.

Which of the following statement(s) is/are correct ? (A) Transition metals and many of their compounds show paramagnetic behaviour. (B) The enthalpies of atomisation of the transition metals are high (C) The transition metals generally form coloured compounds (D) Transition metals and their many compounds act as good catalyst. (A) As metal ions generally contain one or more unpaired electrons in them & hence their complexes are generally paramagnetic (B) Because of having larger number of unpaired electrons in their atoms, they have stronger inter atomic interaction and hence stronger bonding between the atoms. (C) According to CFT, in presence of ligands the colour of the compound is due to the d-d transition of the electrons. (D) This activity is ascribed to their ability to adopt multiple oxidation state and to form complexes. Therefore, (A,B,C,D) options are correct.

10.

When CO2 is passed into aqueous : (A) Na2CrO4 solution, its yellow colour changes to orange. (B) K2MnO4 solution, it disproportionates to KMnO4 and MnO2 (C) Na2Cr2O7 solution, its orange colour changes to green (D) KMnO4 solution, its pink colour changes to green.

Sol.

 Na2Cr2O7 (orange colour) (A) Na2CrO4 

H

H

 MnO4– + MnO2 , in neutral or acidic medium (B) MnO42–  (C) False - In acidic medium no colour change takes place. –

OH (D) MnO4– + e–   MnO42– ; in strong alkaline medium pink colour of KMnO4 changes to green.

Therefore, (A,B) options are correct. 11.

Sol.

12.

Sol. 13.

Sol.

Which of the following statement(s) is (are) not correct with reference to ferrous and ferric ions (A) Fe3+ gives brown colour with potassium ferricyanide (B) Fe2+ gives blue precipitate with potassium ferricyanide (C) Fe3+ gives red colour with potassium sulphocyanide (D) Fe2+ givesbrowncolour withpotassium sulphocyanide Fe3+ produces red colouration with KSCN but Fe2+ does not give brown colour with KSCN. Therefore, (D) option is correct. STATEMENT-1 : Ammonical silver nitrate converts glucose to gluconic acid and metallic silver is precipitated. STATEMENT-2 : Glucose acts as a weak reducing agent. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (A) Ag2O + C6H12O6  2Ag + C6H12O7. STATEMENT-1 : The number of unpaired electrons in the following gaseous ions : Mn3+, Cr3+, V3+ and Ti3+ are 4, 3, 2 and 1 respectively. STATEMENT-2 : Cr3+ is most stable in aqueous solution among these ions. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (B) Mn3+ = [Ar]18 3d4 , Cr+3 = [Ar]18 3d3 , V3+ = [Ar]18 3d2 , Ti3+ = [Ar]18 3d1 Cr3+ is most stable in aqueous solution because it has half filled t32g energy level of 3d orbitals in octahedral crystal field spliting and according to crystal field theory (CFT) it has highest value of CFSE i.e. 1.2 O.

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CHEMISTRY 14.

S1 : Interstitial compounds have high melting points, higher than those of pure metals. S2 : Permanganate titrations in presence of hydrochloric acid are unsatisfactory. S3 : KMnO4 does not act as an oxidising agent in strong alkaline medium. S4 : KMnO4 on heating in a current of H2 gives MnO. (A) T T F T (B) T F F T (C) T F T T

Sol.

(D) F F T F

S1 : Due to strong interatomic forces. S2 : Some of the hydrochloric acid is oxidised to chlorine and thus we get less volume of KMnO4 than the actual one. OH–

S3 : MnO4– + e–  MnO42– S4 : 2KMnO4 + 5H2  2KOH + 2MnO + H2O. Therefore, (A) option is correct. 15.

Match the reactions in Column I with nature of the reactions/type of the products in Column II. Column I Column II (A) AgNO3(aq) + I2 (excess) + H2O  (p) Disproportionation (B) K2MnO4(aq) + CO2(g)  (q) Comproportionation  (C) Na2Cr2O7 + C 

(r) Redox

(D) CuCl2(aq) + Cu(s) 

(s) One of the products is insoluble in water

Ans.

(A  p, r, s) ; (B  p, r, s) ; (C  r, s) : (D  q, r, s)

Sol.

(A) 5AgNO3(aq) + 3 I 2 (excess) + 3H2O  HIO3 + 5AgI + 5HNO3

V

0

–1

So it is redox and disproportionation reaction. AgI insoluble in water.  VI

 VII

IV

(B) 3K 2MnO 4 (aq) + 2CO2(g)  2KMnO4 + MnO 2 + 2K2CO3 So it is redox and disproportionation reaction. MnO2 insoluble in water. 0

 VI

III

IV

II

 (C) 2C + Na 2Cr2O7  Cr2O3 + Na 2CO3 + CO 

So it is redox reaction. Cr2O3 (green pigment) is insoluble in water. II

0

I

(D) CuCl2 (aq) + Cu (s)  Cu2Cl2 (s) So it is redox and comproportionation reaction. Cu2Cl2 is insoluble in water.

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