(i) A basic repeating structural unit of a crystalline solid is called a unit cell.
(ii) A crystal is consisted of large number of unit cells.

(i) This is also called as salt isomerism.
(ii) This type of isomers arises when an ambidentate ligand is bonded to the central metal atom/ion through either of its two different donor atoms. In the below mentioned examples, the nitrite ion is bound to the central metal ion Co³⁺ through a nitrogen atom in one complex and through oxygen atom in other complex.
\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{NO}_{2}\right)\right]^{2+}\)

a. 3MnO₄^2- + 4H⁺ ⟶  2MnO₄^- + MnO₂ + 2H₂O
(Manganate ion) (Permanganate ion)  Manganese dioxide
b. C₆H₅CH₃ \(\overset { acidified }{ \underset { KMnO_{ 4 } }{ \longrightarrow } } \) C₆H₅COOH
 Toluene                         Benzoic Acid
c. 2MnO₄^- + 10Fe²⁺+16H⁺ \(\underrightarrow { { 8H }^{ + } } \)  2Mn²⁺+ 10Fe³⁺  + 8H₂O
d. 2KMnO₄ \(\overset { \triangle }{ \underset { Red\ hot }{ \longrightarrow } } \) K₂MnO₄ + MnO₂ + O₂
(Potassium Permanganate)    (Potassium Manganate)
e. Cr₂O₇^2- + 6I^- + 14H⁺ \(\underrightarrow { { (O) }}\) 2Cr³⁺ + 3I₂ + 7H₂O
                 (Iodide ion)                        Iodine
f. Na₂Cr₂O₇ + 2KCl ⟶ K₂Cr₂O₇ + 2NaCl
(Sodium dichromate)    (Potassium dichromate)

Arrhenius equation is,
\(k=Ae^\left ({ \frac { -Ea }{ RT } } \right )\)
Here,
A \(\rightarrow\) Frequency factor
E_a \(\rightarrow\) Activation energy of the reaction
R \(\rightarrow\) Gas constant
T \(\rightarrow\) Absolute temperature (in K)

a) Primary and secondary nitroalkanes, having α-H, also show an equilibrium mixture of two tautomers namely nitro - and aci - form

b) Difference:

Ca[Co(NH₃)CI(Ox)₂] \(\rightleftharpoons\) Ca²⁺ + [Co(NH₃)Cl(Ox)₂]^2-

Hybridisation of Co = sp³d²
No of cis isomer = 2
No of trans isomer = 1
Total isomer = 3
Three isomers are possible for the given complex.

Metal excess defect:
(i) It arises due to the presence of more number of metal ions as compared to anions.
(ii) Examples: NaCl, KCl
(iii) The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the presence of extra cation.

(iii) For example, when NaCI crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal.
(iv) Chloride ions (Cl^-) diffuse to the surface from the lattice point and combines with Na⁺ ion.
(v) The electron lost by the sodium vapour diffuse into the vacancy created by the Cl^- ions.
(vi) Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl can be written as Na_1+xCl.
Metal deficiency defect:

(i) Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.
(ii) For example, In FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice. To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions. In such cases, overall number of Fe²⁺ and Fe³⁺ ions is less than the O^2- ions.

In bcc unit cell, ΔABC
AC² = AB² + BC²
\(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
\(\\ AC=\sqrt { { a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 2a }^{ 2 } } =\sqrt { 2 } a\)

In ΔACG
AG² = AC² + CG²
\(AG=\sqrt { { AC }^{ 2 }+{ CG }^{ 2 } } \)
\(AG=\sqrt { { \left( \sqrt { 2a } \right) }^{ 2 }+{ a }^{ 2 } } \)
\(AG=\sqrt { { 2a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } \)
\(AG=\sqrt { 3a } \)
\(\sqrt { 3 } a=4r\)
\(r=\frac { \sqrt { 3 } }{ 4 } a\)
∴ Volume of the sphere with radius 'r' \(=\frac { 4 }{ 3 } { \pi r }^{ 3 }\)
\(=\frac{4}{3}\pi { \left( \frac { \sqrt { 3 } }{ 4 } a \right) }^{ 3 }\)\(=\frac { \sqrt { 3 } }{ 16 } \pi { a }^{ 3 }\)
Number of spheres belong to a unit cell in BCC arrangement is equal to two and hence the total volume of all spheres.
(i) Packing fraction = \(=\frac{Total \quad volume \quad occupied \quad by \quad spheres \quad in \quad a \quad unit \quad cell}{volume \quad of \quad the \quad unit \quad cell}\times100\)
\(\therefore\)Volume of all spheres \(=2\times \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 16 } \right) =\frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \)
Packing fraction \(=\frac { \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \right) }{ ({ a }^{ 3 }) } \times 100\)
\(=\frac { \sqrt { 3 } \pi }{ 8 } \times 100\)
\(\\ =\sqrt { 3 } \pi \times 12.5\)
= 1.732 x 3.14 x 12.5
= 68%

Most of the elements exhibit, two types of valence namely primary valence and secondary valence and each element tend to satisfy both the valences.
The primary valence is referred the oxidation state of the metal atom.
The secondary valence as the coordination number. For example, according to Werner, the primary and secondary valences of cobalt are 3 and 6 respectively.
The primary valence of a metal ions ae always satisfied by negative ions.
For example in the complex CoCI₃.6NH₃. The primary valence of Co is +3 and is satisfied by 3CI^- ions.
The secondary valence is satisfied by negative ions, neutral molecules, positive ions or the combination of these.
For example, in CoCl₃.6NH₃ complex primary valence of cobalt +3 and it is satisfied by 3 CI^-.
The secondary valence of cobalt is 6 and is satisfied by six neutral ammonia molecules. where as in CoCI₆.NH₃. 
Secondary valence of Co = 5{It is satisfied five neutral molecules and a Cl^- ion}
According to Werner, there are two spheres of attraction around a metal atom/ion in a complex.

The inner /coordination sphere:
The groups present in this sphere are firmly attached to the metal.
The outer sphere / ionisation sphere:
The groups present in this sphere are loosely bound to the central metal ion and hence can be separated into ions upon dissolving the complex in a suitable solvent.
The primary valencies are non-directional. while the secondary valencies are directional.
The geometry of the complex is determined by the special arrangement of the groups which satisfy the secondary valence.

[Ni (H₂O)₆]²⁺

It has two unpaired electrons. So there is d-d transition. Hence it is green coloured.
[Ni(CN)₄]^2-

There is no unpaired electrons. So it is colourless, as there is no d-d transition.

(a) Elementary reaction
Each and Every single step in a reaction mechanism is called an elementary reaction.
Rate = k[A] [B]
(b)

Hoffmann's bromide reaction:
When Amides are treated with bromine in the presence of aqueous or ethanolic solution of KOH, primary amines with one carbon atom less than the parent amides are obtained.
\(\underset { \quad \quad \quad amide\\ R=Alkyl(or)Aryl }{ R-\overset { \underset { || }{ O } }{ C } -{ NH }_{ 2 } } \overset { { Br }_{ 2 }/KOH }{ \longrightarrow } \underset { Primary\quad amine }{ R-{ NH }_{ 2 }+{ K }_{ 2 }{ CO }_{ 3 } } +KBr+{ H }_{ 2 }O\)
(ii) Hoffmann's ammonolysis:
When Alkyl halides (or) benzylhalides are heated with alcoholic ammonia in a sealed tube, mixtures of 1°, 2° and 3° amines and quaternary ammonium salts are obtained.
\( { CH }_{ 3 }-Br\overset { \ddot { N } { H }_{ 3 } }{ \underset { \Delta }{ \longrightarrow } } \underset { { 1 }^{ 0 }-amine }{ { CH }_{ 3 }-\ddot { N } { H }_{ 2 } } \overset { { CH }_{ 3 }-Br }{ \longrightarrow } \underset { { 3 }^{ o }-amine }{ \left( { CH }_{ 3 } \right) _{ 2 }\ddot { N } H } \overset { { CH }_{ 3 }Br }{ \longrightarrow } \underset { 3^{ o }-amine }{ \left( { { CH }_{ 3 } } \right) _{ 3 }\ddot { N } } \overset { CH_{ 3 }Br }{ \longrightarrow } \underset { Quartenary \ ammonium\ bromide }{ \left( { CH }_{ 3 } \right) _{ 4 }\overset { + }{ N } { Br }^{ - } } \) 
This is a nucleophilic substitution, the halide ion of alkyl halide is substituted by the -NH₂ group. The product primary amine so formed can also has a tendency to act as a nucleophile and hence if excess alkyl halide is taken, further nucleophilic substitution takes place leading to the formation of quarternary ammonium salt. However, if the process is carried out with excess ammonia, primary amine is obtained as the major product. The order of reactivity of alkylhalides with amines
RI > RBr > RCl
(iii) Gabriel phthalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline cannot be prepared by this method because the arylhalides do not undergo nucleophilic substitution with the anion formed by phthalimide.

(iv) Schotten - Baumann reaction :
Aniline reacts with benzoylchloride (C₆H₅COCI) in the presence of NaOH to give N - phenyl benzamide. This reaction is known as Schotten - Baumann reaction. The acylation and benzoylation are nucleophilic substitutions.
\(\underset { Aniline }{ { C }_{ 6 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Benzoyl\quad chloride }{ { C }_{ 6 }{ H }_{ 5 }-\overset { \underset { || }{ O } }{ C } -Cl } \overset { Pyridine }{ \longrightarrow } \underset { N-phenyl\quad benzamide }{ { C }_{ 6 }{ H }_{ 5 }-NH-\overset { \underset { || }{ O } }{ C } -{ C }_{ 6 }{ H }_{ 5 } } +HCl\)
(v) Carbylamine reaction :
Aliphatic (or) aromatic primary amines react with: chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test used to identify the primary amines.
\(\underset { Ethylamine }{ { C }_{ 2 }{ H }_{ 5 }-{ NH }_{ 2 } } +\underset { Chloroform }{ { CHCl }_{ 3 }+3KOH } \longrightarrow \underset { Ethylisocyanide }{ { C }_{ 2 }{ H }_{ 5 }-NC } +3KCl+3{ H }_{ 2 }O\)
(vi) Mustard oil reaction:
(a) When primary amines are treated with carbon disulphide (CS₂), N - alkyldithio carbonic acid is formed which on subsequent treatment with HgCI₂, gives an alkyl isothiocyanate.


(b) When aniline is treated with carbon disulphide, or heated together, S-diphenylthio urea is formed, which on boiling with strong HCI, phenyl isothiocyanate (phenyl mustard oil), is formed.

These reactions are known as Hofmann - Mustard oil reaction. This test is used to identify the primary amines.
(vii) Coupling reactions (or) p-hydroxyazobenzene, p-aminoazobenzene, 2-phenyl azo-4-methyl phenol:
Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds. Coupling generally occurs at the para position. If para position is occupied then coupling occurs at the ortho position. Coupling tendency is enhanced if an electron donating group is present at the para - position to -N₂CI- group. This is an electrophilic substitution.

Aryl fluorides and iodides cannot be prepared by direct halogenation and the cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene. For introducing such a halide group. cyano group -OH, NO₂, etc.. benzenediazonium chloride is a very good intermediate Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(viii) Diazotisation :
Aniline reacts with nitrous acid at low temperature (273 - 278 K) to give benzene. Diazonium chloride which is stable for a short time and slowly decomposes even at low temperatures.This reaction is known as diazotization

(ix) Gomberg reaction :
Benzene diazonium chloride reacts with benzene in the presence of sodium hydroxide to give biphenyl. This reaction in known as the Gomberg reaction

1. The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. The coordination number of atoms in a bcc structure is '8'.

(i) Coordination compounds which possess chirality exhibit optical isomerism similar to organic compounds.
(ii) The pair of two optically active isomers which are mirror images of each other are called enantiomers.
(iii) Their solutions rotate the plane of the plane polarised light either clockwise or anticlockwise and the corresponding isomers are called 'd' (dextrorotatory) and 'I' (levorotatory)  forms respectively.
(iv) The octahedral complexes of type \(\left[\mathrm{M}(\mathrm{xx})_{3}\right]^{\mathrm{n} \pm}, \left[\mathrm{M}(\mathrm{xx})_{2} \mathrm{AB}\right]^{\mathrm{n\pm}}\) and \(\left[\mathrm{M}(\mathrm{xx})_{2} \mathrm{~B}_{2}\right]^{\mathrm{n\pm}}\) exhibit optical isomerism.
Examples:
(i) The optical isomers of \(\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}\) 
(ii) The coordination complex \(\left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right]^{+}\) has three isomers, two optically active cis forms and one optically inactive trans form.

(i) Fe³⁺ - electronic configuration - [Ar] 3d⁵
(ii) It has exactly half-filled stable electronic configuration.
(iii) Fe²⁺ - electronic configuration -[Ar]3d⁶
(iv) It has only partially filled d-orbitals.
Hence Fe³⁺ is more stable than Fe²⁺.

Mn³⁺ has large and negative standard electrode potential E⁰ (-1.18 V) than that of Cr²⁺ which has only -0.91 V. If the standard electrode potential of a metal is large and negative, the metal is a powerful reducing agent because it loses electrons easily. Hence Mn³⁺ is strongly oxidizing while Cr²⁺ is strongly reducing.

La(OH)₃ is more basic than Lu(OH)₃. Due to lanthanide contraction, the size of Ln³⁺ ions decreases regularly with increase in atomic number. According to Fajan's rule, decrease in size of Ln³⁺ ions decreases the basic character between Ln³⁺ and OH^- ion in Ln(OH)₃. So La(OH)₃ is more basic than Lu(OH)₃.

Rate law:
(i) Rate law or rate equation is an expression which relates the rate of a reaction with rate constant and the concentration of reactants.
(ii) For xA + yB → products
(iii) The rate law is r = k[A]^m[B]ⁿ
Rate constant:
(i) It is the rate of the reaction when the concentration of the reactants are taken unity.
In above rate law if[A] = [B] = 1, rate constant k = Rate

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentration is called a zero order reaction.
Let us consider the following by hypothetical zero order reaction.
 A⟶ product 
The rate law can be written as,
Rate = k[A]⁰ 
\(\frac { -d\left[ A \right] }{ dt } =k(1)\ \ \ \therefore \left( { \left[ A \right] }^{ 0 }=1 \right) \)
\(\Rightarrow -d\left[ A \right] =kdt\)
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time 't',
\(-\int _{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }{ d\left[ A \right] } =k\int _{ 0 }^{ t }{ dt } \)
\(-{ \left( \left[ A \right] \right) }_{ \left[ { A }_{ 0 } \right] }^{ \left[ A \right] }=k{ \left( t \right) }_{ 0 }^{ t }\)
[A₀] - [A] = kt   
\(k=\frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \)
Straight line equation y = mx + c
ie., \(\left[ A \right] =-kt+\left[ { A }_{ 0 } \right] \)
⇒ y = c + mx
A plot [A] vs time gives a straight line with a slope of -k and y intercept of [A₀]

+5           +6           +7
VO₂⁺ < Cr₂O₇^2- < MnO₄^-
Greater the oxidation state, higher is the oxidising power.

MnO^-₄ + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

+7       +2+3          2+              3+       4+
MnO₄^- + FeC₂O₄ ⟶ Mn²⁺ + Fe³⁺ + 2CO₂
5e^- acception       3e^- release
5 moles of FeC₂O₄ \(\equiv\) 3 moles of KMnO₄
1 mole of FeC₂O₄ \(\equiv\) (3/5) moles of KMnO₄
1 mole of FeC₂O₄ \(\equiv\) 0.6 moles of KMnO₄

In the complex [M(en)₂(Ox)]Cl For the central metal ion M3+
The primary valence is = +3
The secondary valence = 6
sum of primary valence and secondary valence = 3 + 6 = 9

Molecular formula: MSO₄Cl.6H₂O
Formation of white precipitate with Barium chloride indicates that SO^2-₄ ions are outside the coordination sphere, and no precipitate with AgNO₃ solution indicates that the Cl^- ions are inside the coordination sphere. Since the coordination number of M is 6.Cl^- and 5 H₂O are ligands, remaining 1 H₂O molecular and SO^2-₄ are in the outer coordination sphere.

[\( \overset{+}{Fe}\)(H₂O)₅\( \overset{+}{NO}\)]²⁺ SO^2-₄
+1 and +1 respectively

 The electronic configuration t_2g³, e²_g
[ 3 x (-0.4)+ 2(0.6)]Δ₀
[-1.2 + 1.2] Δ₀ = 0

Complexes given in other options (a), (c) and (d) have symmetry elements and hence they are optically inactive.

If number of close packed atoms = N, then
The number of Tetrahedral holes formed = 2N
The number of Octahedral holes formed = N
Therefore, N:2N = 1:2

If the total number of M atoms is n, then the number of tetrahedral voids = 2n
Given that \(\left( \frac { 1 }{ 3 } \right) ^{rd}\) of tetrahedral voids are occupied.
i.,e \(\left( \frac { 1 }{ 3 } \right) \times 2n\) are occupied by N atoms
\(\therefore\)M : N = n : \(\left( \frac { 2 }{ 3 } \right) n\)
= 1 : \(\frac { 2 }{ 3 }\)
Hence M₃N₂ = 3 : 2

for n ≠ 1 t_1/2 = \(\frac{2^{n-1} -1}{(n- 1) k[A_{0}]^{-1}}\)
for n = 0; t_1/2 = \(\frac{1}{2 k[A_{0}]^{-1}}\)
t_1/2 = \(\frac{[A_{0}]}{2 k}\)
t_1/2 α [A₀] ...(1)
Given [A₀] = 0.002 M; t_1/2 = 10 min
[A₀] = 0.04M; t_1/2 = ?
Substitute in (1)
10 min α 0.02 M...(2)
t_1/2 α 0.04M ....(3)
(3)(2)
⇒ t_1/2 / 10 min
= 0.04 M/0.02 M
t_1/2 = 2 x 10 min = 20 min

\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope.

Packing efficiency = 68%
\(\therefore\)empty space percentage = 100 - 68 = 32%

Given:
At low pressure the reaction follows first order therefore,
Rate α [reactant]¹
Rate α (surface area)
At high pressure due to the complete coverage of surface area, the reaction follows zero order.
Rate α [reactant]⁰
Therefore the rate is independent of surface area.

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 100: [A] = 25
\(6.909=\frac { 2.303 }{ t } \log\frac { \left[ {100 } \right] }{ \left[ 25\right] } \)
\(t =\frac { 2.303 }{ 6.909 } \log(4)\)
\(t =\frac { 1 }{ 3 } \log(2^2)\)
\(= \left( \frac { 2 }{ 3 } \right) \log2\)