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 Marks 2012 2011 2010 2009 2008 2007 2006
Haloalkanes and Haloarenes 4
Example 10.2
(vi) Table 10.1 Example 10.7 Example 10.6, 10.17 Ex 10.2 (i)10.9 (i), 10.14 (vii) - - Alcohols, Phenols and Ethers 4 Exercise 11.5, 11.15 11.18 (ii), 11.31 (iv) 11.20 (ii, iv), 11.18 (i, iii) 11.3 (v) 11.3 (ii), 11.21 (ii, ii, vi) 11.22 - Aldehydes, Ketones and Carboxylic Acids 6 Example 12.5 (i, iv),Exercise 12.10,Exercise 12.12, Exercise 12.13 (i, vi) 12.13 (i, iv), 12.16 (ii, iv), 12.17 (i, ii) Fig 12.2, 12.11, 12.13 (iii, vii), 12.12 (i, ii) 12.16 (ii), 12.19, 12.17 (iv) 12.12 (ii), 12.17 (i, x, xi), 12.16 (ii, iii) Table 12.1, 12.13 (i) 12.16 (ii), 12.13 (vii) Amines 4 Exercise 13.7 (i and iii), Exercise 13.11 (ii, v) 13.4 (ii) Table 13.1, 13.4 (i,ii, iii-a, iv) Ex 13.4, 13.7 (iii, iv) 13.3 (I, iii, v) 13.12 13.2 (v) Biomolecules 4 Exercise 14.12 14.12, 14.13 14.17, 14.22 14.12 (i), table 14.3 14.12 (i, iii), 14.10 14.12 (ii) 14.15, table14.3 Polymers 3 - 15.2 15.11 15.12 15.17 15.12, 15.16 15.12, 15.15 Chemistry in Everyday Life 3 Exercise 16.10 16.15 16.21 (i) 16.15, 16.17, 16.21 (i) 16.12, 16.22 16.15 -
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.
Alcohols, Phenols and Ethers Table 11.1, 11.2, example 11.1 and in-text question 11.3
Aldehydes, ketones and carboxylic acids Table 12.1 and intext questions 12.1, Table 12.3 and in-text question 12.6
Amines 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.
Unit Expected areas from where questions can be asked in 2013 examination
Haloalkanes and Haloarenes
IUPAC Nomenclature, SN1 and SN2 reaction (Question: Which
compound undergoes faster reaction?), DDT, Iodoform
Alcohols, Phenols and Ethers
IUPAC Nomenclature, Conversions, Alcohols - Boiling points and solubilities, Phenols-Acidic properties, Difference in the boiling points of ethers and alcohols, Williamson reaction, Reimer-Tiemann reaction
Aldehydes, Ketones and CarboxylicAcids
IUPAC Nomenclature, Nucleophilic addition on carbonyl carbon, Aldol Condensation, Cannizzaro reaction, Clemmensen reduction reaction, Acidity of carboxylic acids (Question: Arrange the
following compounds in increasing/ decreasing order of their acidic trends.), Distinction tests, Conversions
Amines
IUPAC Nomenclature, Basicities of amines (Question: Arrange
the following amines in increasing/ decreasing order of their basic strengths.), Hoffmann Bromamide reaction, Coupling
reaction, Aniline -Insoluble in water and does not undergo Friedel Crafts reaction
Biomolecules
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, Vitamins - Sources and deficiency disease - table 4.3, nucleosides and nucleotides
Polymers Elastomer, thermoplastic and thermosetting polymer, Rubber,
Nylon 6, nylon 6,6, teflon, bakelite, Buna-N
Chemistry in Everyday Life
Detergents, Food preservatives, Enzymes, Antifertility drugs, Analgesics, Artificial sweetening agents, biodegradable and non-biodegradable detergents
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
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 Coordination compounds
Formula writing 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
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). Nomenclature 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. Isomerism Stereoisomerism
1. Geometrical- Due to different possible geometrical arrangement of ligands
2. Optical- Due to chirality. Structural isomerism
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
Magnetic properties 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.
electrons. In the absence of ligand, crystal-field splitting does not occur; hence, the substance is colourless.
Stability 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:
Generally known as: transition elements/ transition metals
S. No.
Properties Trends Reason
1 Melting and boiling point
First transition series are lower than those of the heavier transition elements.
Occurrence of stronger metallic bonding (M-M bonding) in heavier metals
2 Atomic and ionic sizes
The atomic sizes of the elements of the first transition series are smaller than those of the corresponding heavier elements.
Increase in nuclear charge and number of electrons.
The atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series.
Lanthanoid contraction
3 Ionisation Enthalpies
The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series.
Poor shielding effect of 4f electrons in the third transition series
4 Oxidation states They show variable oxidation
states.
Participation of ns and (n-1)d electrons in bonding
5 Chemical
reactivity
Many of the transition metals are electropositive while few are noble.
Presence of empty d-orbitals (as they can accept electrons) 6 Magnetic
properties
Some metals are paramagnetic while some are diamagnetic. It depends on number of electrons.
Magnetic moment increases with increase in number of electrons 7 Formation of
Coloured Ions
All metals form coloured compounds.
d-d transition
8 Formation of Complex Compounds
Transition metals form a large number of complex compounds
Comparatively smaller size of metal ions, high ionic charges and
availability of d-orbitals for bond formation
9 Catalytic Properties
Transition metals and their compounds are known for their catalytic activity
They can lend electrons or withdraw electrons from the
reagent, depending on the nature of the reaction
10 Alloy Formation Alloys are readily formed by these metals
Similar radii 11 Formation of
Interstitial Compounds
They form interstitial compounds with hydrogen.
Hydrogen occupies interstitial sites in the metal lattice without
changing the type of the lattice
General outer electronic configuration: 4f1-146s2 Generally known as: inner transition elements
Properties Trends Reason
Atomic and ionic radii
There is a gradual decrease in atomic and ionic radii of Lanthanoids.
Lanthanoid contraction
Oxidation states The lanthanoids exhibit mainly +3 oxidation state.
Actinoid: These are the radioactive elements.
Properties Trends Reason
Atomic and ionic radii
There is a gradual decrease in atomic and ionic radii of actinoids .
Due to actinoid contraction Oxidation states There is a gradual decrease in atomic
and ionic radii of actinoids. Exhibit mainly +3 oxidation state.
5f, 6d, and 7s subshells are of comparable energies.
Ionisation enthalpy Lower ionisation enthalpies 5f electrons are less effectively shielded than 4f electrons. Magnetic property Paramagnetic 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. 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 Question Number Year
211 Example 8.1 2010 220 Example 8.7 2010 220 Intext question 8.6 2010 232 Intext question 8.10 2009, 2008 234 Exercise-8.7 2007 - 8.11 (ii) 2008 - 8.11 (iii) 2007, 2010 - 8.11 (iv) 2008 - 8.21 (i) 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 oxidation and
self-reduction.
Q. Can HNO3 and H3PO4 undergo 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?