Slope = k/2.303

Transition elements show variable oxidation state due to their ability of utilising single or multiple 4s and 3d electrons.
(i) Mn (manganese) shows the maximum number of oxidation states. 
(ii) Scandium shows only +3 oxidation state.

Order and molecularity:The important differences between order and molecularity of a reaction are given below:

\(4FeCr_2O_4 + 8Na_2CO_3 + 7O_2\longrightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2\)
\(2Na_2CrO_4 + 2H^+\longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O\)
\(Na_2Cr_2O_7 + 2KCI \longrightarrow K_2Cr_2O_7 + 2NaCI\)
\(\underset{dihromate\ ion\ medium}{\underset{(orange)}{Cr_2O_7^{2-} }+\underset{Basic} {2OH^-} }\longrightarrow \underset{chromate\ ion\\ yellow}{2CrO_4^{2-}} + H_2O\)

(c) : 75% of reaction is completed in two half-lived  i.e., 2 \(\times\) t ₁₂ = 32 min or t _1/2 = 16 min

[Ti(H20)6]³⁺ is colored while [Sc(H20)6]³⁺ is colorless. This is because Ti³⁺ has one electron in the d orbital (3dl) which can absorb energy corresponding to yellow wavelength and undergo a d-d transition (from t2g to eg). But Sc³⁺ is 3JO, i.e., it has no electron in 3d subshell.

Electronic configuration of Mn²⁺ is 3d 5 which is half-filled and hence stable. Therefore, 3rd ionization enthalpy is very high, i.e.,a 3rd electron cannot be lost easily. In the case of Fe²⁺, an electronic configuration is 3d₆. Hence, it can lose one electron easily to give the stable configuration 3d₅.

\((i) \ \frac { dx }{ dt } =k{ \left[ A \right] }{ \left[ B \right] }^{ 2 }.\)
\((ii) \ Rate=k{ \ ab }^{ 2 }\)
If [B] is tripled, Rate = \(ka{ \left( 3b \right) }^{ 2 }=9k \ { ab }^{ 2 }=9 \ times \)
(iii) If both [A]and[B] are doubled, Rate = \(k(2a){ \left( 2b \right) }^{ 2 }=8k \ { ab }^{ 2 }=8 \ times\)

Actinoid Contraction. The decrease in atomic and ionic size along actinoid series, is called actinoid contraction.
Cause: It is due to poor shielding effect of 5f-electrons due to which effective nuclear charge increases.
Comparison Between Lanthanoids and Actinoids: Similarities:
(i) In bothf-orbitals are progressively filled.
(ii) Both form co loured ions.
(iii) Both are paramagnetic in nature.
(iv) Both show contraction, i.e. lanthanoid contraction and actinoid contraction.
Differences between Lanthanoids and Actinoids:

For a first order reaction
\(k=\frac{2.303}{t}\log\frac{[R_0]}{[R]},\) where \([R]_0\) = initial concentration, [R] = conc. after time t
When half of the reaction is completed, [R] = [R]⁰/2. Representing, the time taken for half of the reaction to be completed, by t_1/2' equation becomes:
\(k=\frac{2.303}{t_{1/2}}\log\frac{[R]_0}{[R]_0/2} \)
\(\Rightarrow\ \ \ \ t_{1/2}=\frac{2.303}{k}\log2\)
\(\Rightarrow\ \ \ \ t_{1/2}=\frac{2.303}{k}\times 0.3010\)
\(\Rightarrow\ \ \ \ t_{1/2}=\frac{0.693}{k}\)
The above equation shows that half life first order reaction is independent of the initial concentration of the reactant.
(ii) For a first order reaction
\(\Rightarrow\ t=\frac{2.303}{k}\log\frac{a}{a-x}\)
\(\Rightarrow\ \ \ \ t_{99\cdot /\cdot }=\frac{2.303}{k}\log\frac{100}{1}\)
\( =\frac{2.303}{k}\log 100\)
\( =\frac{2.303\times 2}{k}=\frac{4.606}{k}\)
and \(\Rightarrow\ \ t_{90\cdot /\cdot }=\frac{2.303}{k}\log\frac{100}{10}\)
\(=\frac{2.303}{k}\log{10}=\frac{2.303}{k}\)
\(\frac{t_{99\cdot /\cdot }}{t_{90\cdot /\cdot } }=2\)
\(t_{99\cdot /\cdot }=2\times t_{90\cdot /\cdot }\)

(a) (a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.
Consider the hydrolysis of an ester
\( \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
\(\text { Rate }=k\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]\left[\mathrm{H}_2 \mathrm{O}\right] \)
Hydrolysis of an ester follows first order because the concentration of water, \(\left[\mathrm{H}_2 \mathrm{O}\right]\) remains nearly constant and the change in the concentration of acid will result in very negligible change in the concentration of \(\mathrm{H}^{+} / \mathrm{OH}^{-}\)ion during the course of the reaction.
Thus, both (A) and (R) are correct and (R) is the correct explanation of (A).

(a) : A bimolecular reaction may involve combination of two molecules or exchange of atoms or groups of atoms between the two reactant molecules.

(i) (c) : The reactions of higher order are very rare because of the less chances of the molecules to come together simultaneously and collide.
(ii) (c) : The total number of reactant molecules participating in a chemical reaction is known as its rnolecularity, hence the molecularity = 6 + 3 + 1 = 10.
(iii) (c) : Molecularity mayor may not be equal to the order of a reaction.
(iv) (d) : Order of reaction  \(=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{6+4+3}{12}=\frac{13}{12}\)