(rate constant K is equal to collision frequency A if E_a = 0)
In zero order reactions, increase in the concentration of reactant does not alter the rate.
So statement (i) is wrong.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
if E_a = 0 so, statement (ii) is correct, and statement (iii) is wrong
k = Ae⁰
k = A
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
So statement (iv) and (v) are wrong.

\(3 \sqrt{a} = r_{C_{s+}} + 2r_{C_{f-}} + r_{C_{s+}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) a = ​​ r_{C_{s+}} + r_{C_{f-}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \times 400\)= inter ionic distance

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

As we move from La to Lu, their metallic behaviour because almost similar to that of aluminium.

In \(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
In 2 = \(\frac { { E }_{ a } }{ R } \) \(\left[ \frac { 10 }{ 300\times 310 } \right] \)
E_a = 9300 R In 2
= 53.4 kJ mol^-1

(i) Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
(ii) Water molecules moves freely in and out of these pores hence they are highly porous.

Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF =x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

HNO₃ reacts with basic oxides to form salts and water
ZnO + 2HNO₃\(\longrightarrow \) Zn(NO₃)₂ + H₂O
3FeO + 10HNO₃ \(\longrightarrow \) 3Fe(NO₃)₃ + NO + 5H₂O

(i) In diborane two BH₂ units are linked by two bridged hydrogens.
(ii) It has eight B-H bonds.
(iii) Diborane has only 12 valance electrons.
(iv) The four terminal B-H- bonds is "2c - 2e" bond (two centre - two electron bond.)
(v) Two three centred B - H - B bonds two electrons each. "(3c - 2e)"
(vi) In diborane, the boron is "sp³" hybridised
(vii) Three of the four "sp³" hydridised orbitals contains single electron and the fourth orbital is empty.

(i) The Chalcogens belong to group (16).
(ii) The group consists of elements: Oxygen, Sulphur, Selenium, Tellurium and Polonium.
(iii) These are ore forming elements as most of the ores are oxides and sulphides.
(iv) Chalcos meaning 'ore formers'.

Catenation is an ability of an element to form chain of atoms.
The conditions for catenation.
(a) The valency of element is greater than or equal to two.
(b) Element should have an ability to bond with itself
(c) The self bond must be as strong as Its bond with other elements
(d) Kinetic inertness of catenated compound towards other molecules.
(e) Carbon possesses all the above properties and forms a wide range of compounds with itself and with other elements such as H, O, N, S and halogens.

1. This process is called as Hall-Heroult process.
Cathode: In this method, electrolysis is carried out in an iron tank lined with carbon which acts as the cathode.
Anode: The carbon blocks immersed in the electrolyte acts as a anode.
Eletrolyte: A 20% solution of alumina, obtained from the bauxite ore is mixed with molten Cryolite and is taken in the electrolysis chamber.
2. About 10% calcium chloride is also added to the solution.
3. Here Calcium chloride helps to lower the melting point of the mixture.
Temperature: The fused mixture is maintained at a temperature of above 1270 K.
4. The chemical reactions involved in this process as follows
(a) Ionisaiton of alumina: \({ A }l_{ 2 }{ O }_{ 3 }\longrightarrow { 2Al }^{ 3+ }+{ 3O }^{ 2- }\)
(b) Reaction at cathode: \(2{ Al }^{ 3+ }_{(melt)}+{ 6e }^{ - }\longrightarrow { Al }_{ (l) }\)
(c) Reaction at anode: \(6{ O }^{2-}_{(melt)}\longrightarrow { 3O }_{ 2 }+{ 12e }^{ - }\)
5. Since carbon acts as anode the following reaction also takes place
(a) \({ C }_{ (s) }+{ O }^{ 2- }_{(melt)}\longrightarrow CO+{ 2e }^{ - }\)
(b) \({ C }_{ (s) }+{ 2O }^{ 2- }_{(melt)}\longrightarrow { CO }_{ 2 }+{ 4e }^{ - }\)
6. Due to the above two reactions, anodes are slowly consumed during the electrolysis.
7. The pure aluminium is formed at the cathode. The net electrolysis reaction can be written as
\({ 4Al }^{ 3+ }_{(melt)}+{ 6O }^{ 2- }_{(melt)}+{ 3C }_{ (s) }\longrightarrow { 4Al }_{ (l) }+{ 3CO }_{ 2(g) }\)

(i) Collision theory is based on the kinetic theory of gases. According to this theory, a chemical reaction occurs as a result of collisions between the reacting molecules.
(ii) Let us understand this theory by considering the following reaction.
A₂(g) + B₂(g) ⟶ 2AB(g)
(iii) If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second.
(iv) Rate ∝ number of molecules colliding per litre per second (collision rate).
(v) The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collison rate ∝ [A₂][B₂]
Collision rate = Z [A₂][B₂]
(vi) Where, Z is a constant
(vii) A fraction of effective collisions (f) is given by the following expression
\(f={ e }^{ \frac { { -E }_{ a } }{ RT } }\)
(xiv) This fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collides with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.
(viii) The diagram illustrates the importance of proper alignment of molecules which leads to reaction.

(ix) The fraction of effective collisions (f) having proper orientation is given by the steric factor p.
⇒ Rate = p x f x collision rate
\(\Rightarrow Rate=p\times { e }^{ \frac { -Ea }{ RT } }\times Z\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(1)\)
As per the rate law,
Rate = \(k=\left[ { A }_{ 2 } \right] \left[ { B }_{ 2 } \right] \quad ...(2)\)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is
\(k=pZ{ e }^{ \frac { -Ea }{ RT } }\)

(i) It does not explain the colour of the complex.
(ii) It considers only the spin only magnetic moments and does not consider the other components of magnetic moments.
(iii) It does not provide a quantitative explanation as to why certain complexes are inner orbital complexes and the others are outer orbital complexes for the same metal. For example, [Fe(CN)₆]^4- is diamagnetic (low spin) whereas [FeF₆]^4- is paramagnetic (high spin).

Bonding in metal carbonyls:
(i) In metal carbonyls, the bond between metal atom and the carbonyl ligand consists of two components.
(ii) The first component is an electron pair donation from the carbon atom of carbonyl ligand into a vacant d-orbital of central metal atom.
(iii) This electron pair donation forms \(M\overset { \sigma \ bond }{ \longleftarrow } \text {CO sigma bond}\)
(iv) This sigma bond formation increases the electron density in metal d orbitals and makes the metal electron rich.
(v) In order to compensate for this increased electron density, a filled metal d-orbital interacts with the empty \({ \pi }^{ \bigstar }\) orbital on the carbonyl ligand and transfers the added electron density back to the ligand.
(vi) This second component is called \(\pi\)-back bonding.
(vii) Thus in metal carbonyls, electron density moves from ligand to metal through sigma bonding and from metal to ligand through pi bonding, this synergic effect accounts for strong M⇽CO bond in metal carbonyls.
(viii) This phenomenon is shown diagrammatically as follows.

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%

1. The crude metal is refined by electrolysis. It is carried out in an electrolytic cell
Anode : Impure metal to be refined with dilute acid.
Cathode : Thin strips of pure metal
Electrolyte : Aqueous solution of the salts of the metal with dilute acid.
2. The metal dissolves from the anode, pass into the solution.
3. At the same amount of metal ions from the solution will be deposited at the cathode.
4. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Example: Electrolytic refining of silver.
Cathode: Pure silver
Anode: lmpure silver rods
Electrolyte: Acidified aqueous solution of silver nitrate
5. When a current is passed through the electrodes the following reactions will take place
(a) Reaction at anode: \({ Ag }_{ (s) }\longrightarrow { Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\)
(b) Reaction at cathode: \({ Ag }^{ + }_{ (aq) }+{ 1e }^{ - }\longrightarrow { Ag }_{ (s) }\)
6. During electrolysis, at anode silver loses electrons and form silver ions and the silver ions migrate towards the cathode and get discharged and deposited on the cathode.
7. Copper, Zinc etc can also be refined by this process.

\(M=\frac { d\times { a }^{ 3 }\times NA }{ g } \)
d = Density of the material
a = Length of the edge of the cell.
N_A = Avogadro number
Z = No. of atoms
\(M=\frac { 10.5{ gcm }^{ -3 }{ (4.07\times { 10 }^{ -6 }cm) }^{ 3 }\times \left( 6.023\times { 10 }^{ 23 }{ mol }^{ -1 } \right) }{ 4 } \)
Atomic mass of silver M = 107.08 g mol^-1.

(i) CaC₂
2 + 2x = 0
2x = -2
x = -1
(ii) H₂CO₃
2 + x + (-6) = 0
x = +4
(iii) HCN
1 + x + (-3) = 0
x = +3 - 1 = +2
(iv) CO
x + (-2) = 0
x= +2

Let \(\left[A_{0}\right]=100 \%\), t = 50 minutes
Then [A]=100 - 40 = 60 %
(1) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{\mathrm{o}}\right]}{[\mathrm{A}]}\)
\(=\frac{2.303}{50} \log \left(\frac{100}{60}\right) \)
\(=\frac{2.303}{50} \log 1.667  \)
\(=\frac{2.303}{50} \times 0.2219 \)
\(\mathrm{k}=0.010216 \mathrm{~min}^{-1}  \)
\(\mathrm{k}=1 \times 10^{-2} \mathrm{~min}^{-1}\)
(2)  \( t =\frac{2.303}{0.010216} \log \left(\frac{100}{20}\right) \)
\(t =\frac{2.303}{{0.010216}}\times 0.6990\)
= 225.43 \(\times \) 0.6990
t = 157.58 min.
The time at which the reaction will be 80% complete is 157.58 min.

(i) The chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.
(ii) The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Example:
Let us compare the following two reactions that you carried out in volumetric analysis.
1) Redox reaction between ferrous Ammonium Sulphate (FAS) and KMnO₄.
2) Redox reaction between oxalic acid and KMnO₄.
(i) The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺. In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60^oC.
(ii) The physical state of the reactant also plays an important role to influence the rate of reactions.
(iii) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.
Ex : Na_(s) + I₂(vap)   [Faster]
Na_(s) + I₂(s)   [Slower]
KI_(aq) + Pb(NO3)_2(aq) → PbI₂ (yellow) [Faster]
KI_(s) + Pb(NO3)_2(s) → PbI₂ (yellow)  [Slower]

(i) The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value.
(ii)   The rate constant for a first order reaction is given by
\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(at\quad t={ t }_{ \frac { 1 }{ 2 } };\left[ A \right] =\frac { \left[ { A }_{ 0 } \right] }{ 2 } \)
\(k=\frac { 2.303 }{ t_{ 1/2 } } log\frac { \left[ { A }_{ 0 } \right] }{ \frac { \left[ { A }_{ 0 } \right] }{ 2 } } \)
\(k=\frac { 2.303 }{ { t }_{ 1/2 } } log2\)
\(k=\frac { 2.303\times 0.3010 }{ { t }_{ 1/2 } } =\frac { 0.6932 }{ { t }_{ 1/2 } } \)
\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
This equation has no concentration term So, the half life of a first order reaction is independent of initial concentration.

(i) The actual position of Lanthanides in the periodic table is at group number 3 and period number 6. However, in the sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these fourteen elements following lanthanum show similar chemical properties.
(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties. Hence they are placed separately bottom of the modern periodic table.

(i) EDTA is used in the volumetric determination of a wide variety of metal ions in solution.
Eg. Zn²⁺, Pb²⁺, Ca²⁺, CO²⁺, Ni²⁺,Cu^2+, etc.
(ii) By careful adjustment of the pH and using suitable indication, mixtures of metals can v be analyzed.
Eg. Bi³⁺in the presence of Pb²⁺
(iii) Hardness of water due to the presence of Ca²⁺ and Mg²⁺ ions is estimated by complexometric titrations using EDTA.

[Co (en)₂ Cl₂]⁺ This is an octahedral complex

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'.

Fusion of urea with B(OH)_3' in an atmosphere of ammonia at 800 - 1200 K gives boron nitride.
B(OH)₃ + NH₃ \(\overset { \Delta }{ \longrightarrow } \) BN+ 3H₂O