Mastering Organic Chemistry and INORGANIC
cbse 12th syllabus...
Mastering Organic Chemistry (Class XII)
Focussing on NCERT questions: The NCERT Part – II comprises of organic chemistry and contributes to 28 marks in the examination. An analysis of previous years’ question papers depicts that many questions are asked as it is from the NCERT textbooks.
The table given below shows the question numbers of the questions in the NCERT book which have been asked in the examination. Note that, many questions are being asked repeatedly. Units
Haloalkanes and Haloarenes
Alcohols, Phenols and Ethers
Aldehydes, Ketones and Carboxylic Acids
Chemistry in Everyday Life
2012 2011 2010 Example 10.2 Example (vi) Table 10.1 10.7
2008 2007 2006 Ex 10.2 Example (i)10.9 (i), 10.6, 10.17 10.14 (vii) Exercise 11.5, 11.20 (ii, 11.3 (ii), 11.18 (ii), 11.15 iv), 11.18 (i, 11.3 (v) 11.21 (ii, 11.22 11.31 (iv) iii) ii, vi) Example 12.5 12.13 (i, (i, iv),Exercise Fig 12.2, 12.12 (ii), iv), 12.16 12.16 (ii), Table 12.10,Exercise 12.11, 12.13 12.17 (i, x, 12.16 (ii), (ii, iv), 12.19, 12.17 12.1, 12.12, (iii, vii), xi), 12.16 12.13 (vii) 12.17 (i, (iv) 12.13 (i) Exercise 12.13 12.12 (i, ii) (ii, iii) ii) (i, vi) Exercise 13.7 Table 13.1, (i and iii), Ex 13.4, 13.3 (I, iii, 13.4 (ii) 13.4 (i,ii, iii13.12 13.2 (v) Exercise 13.11 13.7 (iii, iv) v) a, iv) (ii, v) Exercise 14.12 14.12, 14.12 (i), 14.12 (i, 14.15, 14.17, 14.22 14.12 (ii) 14.13 table 14.3 iii), 14.10 table14.3 15.12, 15.2 15.11 15.12 15.17 15.12, 15.15 15.16 Exercise 16.10 16.15, 16.12, 16.15 16.21 (i) 16.17, 16.21 16.15 16.22 (i)
So, mark these questions in the NCERT book and go through all the NCERT solutions thoroughly. For this you can also refer to the NCERT solutions section on meritnation.com
IUPAC Nomenclature questions: A conclusion from previous years’ papers is that, 90% of the questions related to nomenclature of organic compounds usually come from the tables of common and IUPAC names or in-text examples present in NCERT book. Below is a summarized chart of all the tables and in-text examples meant for nomenclature.
Haloalkanes and Haloarenes
Table 10.1, example 10.1 and example 10.2
Alcohols, Phenols and Ethers Aldehydes, ketones and carboxylic acids Amines
Table 11.1, 11.2, example 11.1 and in-text question 11.3 Table 12.1 and intext questions 12.1, Table 12.3 and in-text question 12.6 Table 13.1
Note: Do not leave even a single compound from the tables and in-text examples.
Name reactions: Important name reactions which have been asked previously are Sandmeyer reaction, Williamson synthesis, Riemer-Tiemann reaction, Kolbe’s reaction, Aldol Condensation, Cannizzaro Reaction, Clemmensen Reduction reaction, Hoffmann Bromamide reaction, Coupling reaction. So,we advise you to revise these reactions.
Distinction test: The distinction tests are usually asked between:
Aliphatic and aromatic compounds Compounds having two different functional groups Compounds having same functional group but different arrangement of atoms (e.g., 1°, 2°, 3°)
Steps for attempting these questions Step - I: See how many marks are allotted to the question. Remember, 1 mark is for 1 test. Step – II: Write the structural formulae of both the compounds. Step – III: See where the two structures differ in. Step – IV: Recall the reactions which you have studied. Step – V: Apply those reactions in the compounds keeping in mind the skeletal structure they differ in.
Conversions: Conversion based questions are surely going to come in exams. Remember there can be multiple steps to reach the final product but the shortest and feasible steps have to be written in the answer-sheet.
Please note CBSE has yet not asked any conversion which consist of more than 3-steps. Steps for attempting these questions Step - I: Read the question very carefully. Step - II: Write the starting compound on the left hand side and the final compound on the right hand side. Step – III: See where do the two structure differ in. (They mostly differ either in functional groups, number of carbon atoms or both) Step – IV: Recall the reactions which you have studied. Step – V: Apply those reactions in initial compound so as to reach to the final compound. Expected areas from where questions can come in 2012
Expected areas from where questions can be asked in 2013 examination IUPAC Nomenclature, SN1 and SN2 reaction (Question: Which Haloalkanes and compound undergoes faster reaction?), DDT, Iodoform Haloarenes IUPAC Nomenclature, Conversions, Alcohols - Boiling points and solubilities, Phenols-Acidic properties, Difference in the Alcohols, Phenols and boiling points of ethers and alcohols, Williamson reaction, Ethers Reimer-Tiemann reaction IUPAC Nomenclature, Nucleophilic addition on carbonyl carbon, Aldol Condensation, Cannizzaro reaction, Clemmensen reduction Aldehydes, Ketones and reaction, Acidity of carboxylic acids (Question: Arrange the CarboxylicAcids following compounds in increasing/ decreasing order of their acidic trends.), Distinction tests, Conversions IUPAC Nomenclature, Basicities of amines (Question: Arrange the following amines in increasing/ decreasing order of their basic strengths.), Hoffmann Bromamide reaction, Coupling Amines reaction, Aniline -Insoluble in water and does not undergo Friedel Crafts reaction Glucose open chain and cyclic structure (Question: Why was the open chain structure of glucose unable to explain its properties?), Proteins-Primary, secondary, tertiary and quarternery structures, Biomolecules Vitamins - Sources and deficiency disease - table 4.3, nucleosides and nucleotides Elastomer, thermoplastic and thermosetting polymer, Rubber, Polymers Nylon 6, nylon 6,6, teflon, bakelite, Buna-N Detergents, Food preservatives, Enzymes, Antifertility drugs, Chemistry in Everyday Life Analgesics, Artificial sweetening agents, biodegradable and nonbiodegradable detergents Unit
To help all you students in this last stage of preparation, we are providing Conversion Schemes , Distinction Tests, Name Reaction List and Name Reaction in Detail. Follow these simple but smart ways and give your 100 % in the examination. Best of Luck! Team Meritnation 10 comments March 9th, 2013
Coordination Compounds (Chemistry Class XII) Some important terms:
Coordination Compounds − Complex compounds in which the transition metal atoms are bound to a number of anions or neutral molecules
Coordination Entity- Constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. Example: [CoCl3(NH3)3] is a coordination entity Central Atom or Ion- The atom or ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement around it in a coordination entity Ligands- Ions or molecules bound to the central metal atom or ion in a coordination entity Coordination Number- Number of ligand-donor atoms bonded directly to the metal Coordination Sphere- The central atom or ion and the ligands attached to it are enclosed in square brackets, which are collectively known as the coordination spheres. Coordination Polyhedron- The spatial arrangement of the ligand atoms which are directly attached to the central atom or ion Oxidation Number of Central Atom- The charge an atom would carry if all the ligands are removed along with the electron pairs that are shared with the central atom Homoleptic complexes: Complexes in which the metal is bound to only one kind of donor group. Example: [Co(NH3)6]3+ Heteroleptic complexes: Complexes in which the metal is bound to more than one kind of donor groups. For example: [Co(NH3)4Cl2]+
Theories related to coordination compounds:
Werner’s theory: In coordination compounds, there are two types of linkages (valences) − primary and secondary. The primary valences are ionisable, and are satisfied by negative ions. The secondary valences are non-ionisable, and are satisfied by negative ions or neutral molecules. The secondary valence is equal to the coordination number of a metal, and remains fixed for a metal. Different coordination numbers have characteristic spatial arrangement of ions or groups bound by the secondary linkages. Valence bond theory: The metal atom or ion under the influence of ligands can use its (n−1)d, ns, np or ns, np, nd orbitals for hybridisation, to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar, and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. Crystal-field theory: An electrostatic model which considers the metal−ligand bond to be ionic, and arises purely from the electrostatic interaction between the metal ion and the ligands. Ligands are treated as point charges in the case of anions, or dipoles in the case of neutral molecules. The five d-orbitals in an isolated gaseous metal atom/ ion are degenerate (i.e., have the same energy). Due to the negative fields of the ligands (either anions or the negative ends of dipolar molecules), the degeneracy of the d-orbitals is lifted, resulting in the splitting of the d-orbitals.
Coordination compounds: Features Formula writing
Coordination compounds Central atom is listed first. The ligands are then listed in the alphabetical order. Polydentate ligands are also listed in the alphabetical order. The formula of the entire coordination entity is enclosed in square
The cation is named first in both positively and negatively charged coordination entities. The ligands are named in alphabetical order before the name of the central atom/ion. Names of the anionic ligands end in −o and those of neutral and cation ligands are the same. To indicate the number of the individual ligands, the prefixes mono−, di−, tri−, etc., are used. If these prefixes are present in the names of ligands, then the terms −bis, −tris, −tetrakis, etc., are used. Oxidation state of the metal is indicated by a Roman numeral in parentheses. If the complex ion is cation, then the metal is named as the element. If the complex ion is anion, then the metal is named with ‘−ate’ ending. The neutral complex molecule is named as the complex cation.
brackets. Ligand abbreviations and formulas for polyatomic ligands are enclosed in parentheses. For the charged coordination entity, the charge is indicated outside the square brackets, as a right superscript, with the number before the sign. The charge of the cation(s) is balanced by the charge of the anion(s).
1. Geometrical- Due to different possible geometrical arrangement of ligands 2. Optical- Due to chirality.
1. Linkage- Only with ambidentate ligand 2. Coordination- Interchange of ligands between cationic and anionic entities of different metal ions present in the complex
Ionization- The counter ion in the complex salt is itself a potential ligand and can displace a ligand, which can then become the counter ion Solvate- Water is involved as a solvent Complexes with unpaired electron(s) in the orbitals are paramagnetic. Complexes with no unpaired electron(s) in the orbitals (i.e., all the electrons are paired) are diamagnetic.
The colour of the coordination compounds is attributed to d−d transition of
electrons. In the absence of ligand, crystal-field splitting does not occur; hence, the substance is colourless. The stability of a complex in a solution refers to the degree of association between the two species involved in the state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant.
Questions that were asked previously: Q. Name the following coordination entities and describe their structure: (2012 Set 3) (i) [Fe(CN)6]4(ii) [Cr(NH3)4Cl2]+ (iii) [Ni(CN)4]2(Atomic numbers: Fe-26, Cr-24, Ni-28) Q. Write the name, stereochemistry and magnetic behavior of the following: (2011 Set 1) (Atomic number: Mn-25, Co-27, Ni-28) (i) K4[Mn(CN)6] (ii) [Co(NH3)5Cl]Cl2 (iii) K2[Ni(CN)4] Q. Name the following coordination compounds according to the IUPAC system of nomenclature: (2010 Set 3) (i) [Co(NH3)4(H2O)Cl]Cl2 (ii) [CrCl2(en)2] Best of luck, Team Meritnation! 3 comments March 9th, 2013
d- and f-Block Elements (Chemistry Class XII) d-Block elements:
General outer electronic configuration: (n -1)d1-10 ns1-2
Generally known as: transition elements/ transition metals Properties
No. 1 Melting and boiling point
First transition series are lower than Occurrence of stronger metallic those of the heavier transition bonding (M-M bonding) in heavier elements. metals 2 Atomic and ionic The atomic sizes of the elements of Increase in nuclear charge and the first transition series are smaller number of electrons. sizes than those of the corresponding heavier elements. The atomic sizes of the elements in Lanthanoid contraction the third transition series are virtually the same as those of the corresponding members in the second transition series. 3 Ionisation The first ionisation enthalpies of the Poor shielding effect of 4f electrons third transition series are higher in the third transition series Enthalpies than those of the first and second transition series. 4 Oxidation states They show variable oxidation Participation of ns and (n-1)d states. electrons in bonding 5 Chemical Many of the transition metals are Presence of empty d-orbitals (as electropositive while few are noble. they can accept electrons) reactivity 6 Magnetic Some metals are paramagnetic Magnetic moment increases with while some are diamagnetic. It increase in number of electrons properties depends on number of electrons. 7 Formation of All metals form coloured d-d transition compounds. Coloured Ions 8 Formation of Transition metals form a large Comparatively smaller size of metal number of complex compounds ions, high ionic charges and Complex availability of d-orbitals for bond Compounds formation 9 Catalytic Transition metals and their They can lend electrons or compounds are known for their withdraw electrons from the Properties catalytic activity reagent, depending on the nature of the reaction 10 Alloy Formation Alloys are readily formed by these Similar radii metals 11 Formation of They form interstitial compounds Hydrogen occupies interstitial sites with hydrogen. in the metal lattice without Interstitial changing the type of the lattice Compounds f-Block elements:
General outer electronic configuration: 4f1-146s2 Generally known as: inner transition elements Properties Atomic and ionic radii Oxidation states
Trends Reason There is a gradual decrease in Lanthanoid contraction atomic and ionic radii of Lanthanoids. The lanthanoids exhibit mainly +3 oxidation state.
Actinoid: These are the radioactive elements. Properties Atomic and ionic radii Oxidation states
Trends There is a gradual decrease in atomic and ionic radii of actinoids . There is a gradual decrease in atomic and ionic radii of actinoids. Exhibit mainly +3 oxidation state. Ionisation enthalpy Lower ionisation enthalpies Magnetic property Paramagnetic
Reason Due to actinoid contraction 5f, 6d, and 7s subshells are of comparable energies. 5f electrons are less effectively shielded than 4f electrons. Presence of unpaired electrons.
Lanthanoid Contraction: The steady decrease in the atomic and ionic radii of the transition metals as the atomic number increases is known as lanthanoid contraction. This is because of filling of 4f orbitals before the 5d orbitals. This contraction in size is quite regular. Actinoid Contraction: The gradual decrease in the ionic radii with the increase in atomic no. is called actinoid contraction. It is due to the imperfect shielding of one 5f electron by the other in the same subshell. Topics/ Questions that were asked previously
Variable oxidation states of transition elements (2007, 2008) Atomic size of transition metals (2012)
Q. There is hardly any increase in atomic size with increasing atomic number in a series of transition metals? (2012 Set 3)
Formation of coloured ions by transition metal ions (2007, 2010) Lanthanoid contraction (2007, 2008, 2009) Oxidation states of actinoids (2009, 2011, 2012 Set 3)
Q. Actinides exhibit a much larger number of oxidation states than the lanthanoids? (2012 Set 3)
Q. With the same d-orbital configuration (d4), Cr2+ is a reducing agent while Mn3+ is an oxidizing agent? (2012 Set 3) NCERT questions which have been asked previously Page 211 220 220 232 234 -
Question Number Example 8.1 Example 8.7 Intext question 8.6 Intext question 8.10 Exercise-8.7 8.11 (ii) 8.11 (iii) 8.11 (iv) 8.21 (i)
Year 2010 2010 2010 2009, 2008 2007 2008 2007, 2010 2008 2009, 2012
Best of luck, Team Meritnation! 5 comments March 9th, 2013
p-Block Elements (Class XII Chemistry) Inert pair effect: Inertness of s subshell electrons (i.e. ns2) towards bond formation. As we move down the group, the lower oxidation state gets stabilised. Q. Which of the two is more stable? SbCl5 or SbCl3 Q. Why is Bi (V) a stronger oxidant than Sb (V)?
(2009 set 1)
Q. The stability of + 5 oxidation state decreases down the group in group 15 of the periodic table. (2010 Set 3) Q. Tendency to form pentahalides decreases down the group in group 15 of the periodic table (2010 Set 3) Disproportionation: Element in a particular oxidation state undergoes self-oxidation and selfreduction. Q. Can HNO3 and H3PO4 undergo disproportionation reaction? Q. Does hydrolysis of XeF6 lead to disproportionation reaction?
Non-metallic hydrides-Bond dissociation enthalpy and basicity of hydrides decreases down the group, Reducing character of hydrides increases down the group, Smaller size and inavailability of d-orbitals affect basic strength Q. Ammonia is a stronger base than phosphine. (2008 Set 1) Q. The acidic strength decreases in the order HCl > H2S > PH3 (2010 Set 3) Q. Which is the stronger acid in aqueous solution? HF or HCl Q. Ammonia has higher boiling point than phosphine. Explain Q. Bond angle in phosphine is lesser than ammonia. Explain Q. H2O is liquid but H2S is gas. Explain Q. Oxygen is a gas but sulphur is a solid? ( 2012 Set 3) Q. All the bonds in SF4 are not equivivalent. (2012 Set 3) Colour of halogen compounds-Halogen absorbs radiations in visible region which excites electrons to higher energy region, so the amount of energy required for excitation is different for each halogen. Q. The halogens are coloured, why? (2012 Set 3) Non-metallic halides-Hexa halides of elements are stable owing to sterically protected six halide atoms, Covalent character-Polarising power: Charge/ radius ionic character is opposite of covalent character Q. NF3 is an exothermic compound but NCl3 is endothermic compound. (2012 Set 3) Q. Why are pentahalides more covalent than trihalides? Q. SF6 is kinetically an inert substance. (2011 Set 1) Q. Solid phosphorus pentachloride behaves as an ionic compound. (2010 Set 3) Q. SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed. (2009 Set 1) Catenation: This self-linking property is due to higher bond strength Q. Sulphur has greater tendency for catenation than oxygen. (2011 Set 3, 2012 Set 3) Q. Catenation tendency is weaker in nitrogen than phosphorus. Explain (2012 Set 3)
Q. The tendency of catenation decreases down the group. Explain Interhalogen compounds: Two or more different halogen atoms reacting with each other to form compounds. Q. Interhalogen compounds are strong oxidising agents. (2007, Set 1) Q. In general interhalogen compounds are more recative than halogens. Why? Q. Why ICl is more reactive than I2? (NCERT, Intext 7.31, 2012 Set 3) Q. ClF3 molecule has a T-shaped structure and not a trigonal planar one. (2010, Set 3) Structures of Oxoacids: phosphorus (2007), sulphur (2007), Structures of Fluoride: sulphur (2008), xenon (2008, 2009), bromine (2009) Interhalogen compounds (2007) Basicity of group 15 elements (2008, 2009) 1. Structures of PCl5, H2SO3, H2SO4, H2S2O8, H2S2O7, HOCl, HClO2, HClO3, HClO4, N2O5, XeOF4 2. Complete the following reaction: o P4 + NaOH + H2O → o P4 + SO2Cl2 → o POCl3 + H2O→ o KMnO4 + HCl → o NaOH + Cl2 → o Cu + HNO3 (dil) → o XeF4 + O2F2 → 3. Conceptual questions:
NF3 is an exothermic compound whereas NCl3 is not. NH3 is a stronger base than PH3. Why? Arrange the following in order of the indicated property: o F2, Cl2, Br2, I2 in increasing order of bond dissociation enthalpy o HF, HCl, HBr, HI in increasing order of acidic strength o NH3, PH3, AsH3, SbH3, BiH3 in increasing order of basic strength What are interhalogen compounds?