MHT CET 2025 Chemistry Question Paper with Answer and Solution

(B) For a weak monoacidic base $BOH$,the dissociation equilibrium is represented as: $BOH \rightleftharpoons B^{+} + OH^{-}$.
Let the initial concentration be $c$ and the degree of dissociation be $\alpha$.
The equilibrium concentrations are: $[BOH] = c(1-\alpha)$,$[B^{+}] = c\alpha$,and $[OH^{-}] = c\alpha$.
The dissociation constant $K_b$ is given by: $K_b = \frac{[B^{+}][OH^{-}]}{[BOH]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}$.
Since the base is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$.
Thus,$K_b \approx c\alpha^2$,which gives $\alpha = \sqrt{\frac{K_b}{c}}$.
The concentration of $[OH^{-}]$ is $c\alpha = c \cdot \sqrt{\frac{K_b}{c}} = \sqrt{K_b \cdot c}$.

(B) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: Concentration $C = 0.1 \ M$ and degree of dissociation $\alpha = 2 \% = 0.02$.
Substituting the values into the formula:
$K_b = 0.1 \times (0.02)^2$
$K_b = 0.1 \times 0.0004$
$K_b = 4 \times 10^{-5}$.

(A) For a weak monoacidic base,the dissociation constant $K_b$ is given by the formula $K_b = C \alpha^2 / (1 - \alpha)$.
Given concentration $C = 0.001 \ M = 10^{-3} \ M$.
Degree of dissociation $\alpha = 1.5 \% = 0.015 = 1.5 \times 10^{-2}$.
Since $\alpha$ is very small,$(1 - \alpha) \approx 1$.
Therefore,$K_b \approx C \alpha^2$.
$K_b = (10^{-3}) \times (1.5 \times 10^{-2})^2$.
$K_b = 10^{-3} \times 2.25 \times 10^{-4} = 2.25 \times 10^{-7}$.

(D) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \frac{c \alpha^2}{1 - \alpha}$.
Given: Concentration $c = 0.01 \ M$,degree of dissociation $\alpha = 5 \% = 0.05$.
Since $\alpha$ is very small,we can approximate $1 - \alpha \approx 1$.
Thus,$K_b \approx c \alpha^2$.
Substituting the values: $K_b = (0.01) \times (0.05)^2$.
$K_b = 10^{-2} \times (2.5 \times 10^{-3}) = 2.5 \times 10^{-5}$.

(D) For a monoacidic base $BOH$,the $pH = 10$.
Since $pH + pOH = 14$,we have $pOH = 14 - 10 = 4$.
Therefore,$[OH^-] = 10^{-pOH} = 10^{-4} \ M$.
The concentration of the base $C = 0.01 \ M = 10^{-2} \ M$.
For a weak base,$[OH^-] = C \times \alpha$,where $\alpha$ is the degree of dissociation.
$10^{-4} = 10^{-2} \times \alpha$.
$\alpha = \frac{10^{-4}}{10^{-2}} = 10^{-2} = 0.01$.
Percentage dissociation $= \alpha \times 100 = 0.01 \times 100 = 1 \%$.

(C) For a monoacidic weak base $BOH$,the concentration $C = 0.01 \ M$ (centimolar).
The degree of dissociation $\alpha = 10 \% = 0.1$.
The concentration of $OH^-$ ions is given by $[OH^-] = C \times \alpha = 0.01 \times 0.1 = 0.001 \ M = 10^{-3} \ M$.
Now,$pOH = -\log[OH^-] = -\log(10^{-3}) = 3$.
Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pH = 14 - pOH = 14 - 3 = 11$.
Therefore,the correct option is $C$.

(C) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C \alpha^2$.
Given: $C = 0.01 \ M = 10^{-2} \ M$ and $K_a = 10^{-4}$.
Substituting the values: $10^{-4} = 10^{-2} \times \alpha^2$.
$\alpha^2 = \frac{10^{-4}}{10^{-2}} = 10^{-2}$.
$\alpha = \sqrt{10^{-2}} = 0.1$.
Percent dissociation = $\alpha \times 100 = 0.1 \times 100 = 10 \%$.

(C) For a dibasic acid $H_2A$,the dissociation is $H_2A \rightleftharpoons 2H^+ + A^{2-}$.
Given concentration $C = M/100 = 0.01 \ M$.
Degree of dissociation $\alpha = 2 \% = 0.02$.
Since the acid is dibasic,the concentration of $H^+$ ions is $[H^+] = 2 \times C \times \alpha$.
$[H^+] = 2 \times 0.01 \times 0.02 = 0.0004 \ M = 4 \times 10^{-4} \ M$.
$pH = -\log[H^+] = -\log(4 \times 10^{-4}) = 4 - \log(4) = 4 - 0.602 = 3.398$.
Thus,the $pH$ is approximately $3.397$.

(A) The relationship between $pH$ and the concentration of hydronium ions is given by the formula: $pH = -\log[H_3O^{+}]$.
Given $pH = 2.8$,we have $2.8 = -\log[H_3O^{+}]$,which implies $\log[H_3O^{+}] = -2.8$.
To find $[H_3O^{+}]$,we take the antilog of $-2.8$: $[H_3O^{+}] = 10^{-2.8}$.
$[H_3O^{+}] = 10^{0.2} \times 10^{-3}$.
Since $10^{0.2} \approx 1.585$,the concentration is $[H_3O^{+}] = 1.585 \times 10^{-3} \ mol \ dm^{-3}$.

(D) For a monobasic acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$.
Given concentration $C = 0.08 \ mol \ dm^{-3}$ and $pH = 2$.
We know that $[H^+] = 10^{-pH} = 10^{-2} = 0.01 \ mol \ dm^{-3}$.
The degree of dissociation $\alpha$ is given by $\alpha = \frac{[H^+]}{C} = \frac{0.01}{0.08} = 0.125$.
The ionisation constant $K_a$ is given by the formula $K_a = \frac{C\alpha^2}{1-\alpha}$.
Substituting the values: $K_a = \frac{0.08 \times (0.125)^2}{1 - 0.125} = \frac{0.08 \times 0.015625}{0.875} = \frac{0.00125}{0.875} \approx 1.428 \times 10^{-3}$.
Re-evaluating the approximation for $K_a \approx C\alpha^2$ (if $\alpha$ is small) or using the exact formula,the closest provided option is $D$.

(C) Salt hydrolysis occurs when a salt is formed from a weak acid and a weak base,a weak acid and a strong base,or a strong acid and a weak base.
Salts formed from a strong acid and a strong base do not undergo hydrolysis in water because their constituent ions do not react with water to change the $pH$ of the solution.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$.
Therefore,$KNO_3$ does not undergo hydrolysis in water.

(A) salt that turns red litmus blue in an aqueous solution must be basic in nature. This occurs due to the hydrolysis of a salt formed from a strong base and a weak acid.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
In water,$KCN$ undergoes hydrolysis as follows:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$
Since $OH^-$ ions are produced,the solution becomes basic,which turns red litmus blue.
$NaNO_3$,$NaCl$,and $KCl$ are salts of strong acids and strong bases,resulting in neutral aqueous solutions.

(B) salt formed from a strong base and a weak acid undergoes anionic hydrolysis to form a basic solution.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,the carbonate ion $(CO_3^{2-})$ reacts with water:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
The production of $OH^-$ ions makes the solution basic.
$NH_4NO_3$ and $NH_4Cl$ are salts of a weak base and a strong acid,forming acidic solutions.
$Na_2SO_4$ is a salt of a strong base and a strong acid,forming a neutral solution.

(B) salt formed from a strong base and a weak acid undergoes anionic hydrolysis to produce a basic solution.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,the carbonate ion $(CO_3^{2-})$ reacts with water:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
The production of $OH^-$ ions makes the solution basic.
$NH_4NO_3$ is a salt of a weak base and a strong acid (acidic).
$NaNO_3$ is a salt of a strong base and a strong acid (neutral).
$CuSO_4$ is a salt of a weak base and a strong acid (acidic).

(D) salt that turns blue litmus red in an aqueous solution must be acidic in nature.
This occurs when a salt is formed from a strong acid and a weak base.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$,making it basic.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,making it basic.
$NaNO_3$ is a salt of a strong base $(NaOH)$ and a strong acid $(HNO_3)$,making it neutral.
$CuCl_2$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(HCl)$.
Due to the hydrolysis of $Cu^{2+}$ ions,the solution becomes acidic,turning blue litmus red.

(C) salt formed from a strong base and a weak acid undergoes anionic hydrolysis to form a basic solution.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
When dissolved in water,the $CN^-$ ion reacts with water:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$.
The production of $OH^-$ ions makes the solution basic.
$NaNO_3$ is a salt of a strong acid and a strong base (neutral).
$CH_3COONH_4$ is a salt of a weak acid and a weak base (nearly neutral).
$NH_4F$ is a salt of a weak base and a weak acid (nearly neutral).

(A) salt that turns blue litmus red in an aqueous solution must be acidic in nature.
$CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$.
In aqueous solution,$Cu^{2+}$ ions undergo hydrolysis: $Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$.
The release of $H^+$ ions makes the solution acidic,which turns blue litmus red.
$Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$,making it basic.
$Na_2SO_4$ and $NaNO_3$ are salts of strong acids and strong bases,making them neutral.

(A) salt that turns red litmus blue in an aqueous solution must be basic in nature.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
In water,it undergoes anionic hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
The production of $OH^-$ ions makes the solution basic,which turns red litmus blue.
$CuCl_2$ and $NH_4Cl$ are salts of strong acids and weak bases,making them acidic.
$KCl$ is a salt of a strong acid and a strong base,making it neutral.

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$.
Given: $pK_{a} = 4.680$,$[Salt] = 0.02 \ M$,$[Acid] = 0.01 \ M$.
Substituting the values: $pH = 4.680 + \log \frac{0.02}{0.01}$.
$pH = 4.680 + \log(2)$.
Since $\log(2) \approx 0.301$,we get $pH = 4.680 + 0.301 = 4.981$.

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$
Given:
$pK_a = 4.7447$
$[\text{Salt}] = 0.05 \ M$
$[\text{Acid}] = 0.01 \ M$
Substituting the values:
$pH = 4.7447 + \log \left( \frac{0.05}{0.01} \right)$
$pH = 4.7447 + \log(5)$
Since $\log(5) \approx 0.6990$:
$pH = 4.7447 + 0.6990 = 5.4437$
Rounding to two decimal places,the $pH$ is $5.44$.

(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
Given:
$pH = 5.5$
$pK_{a} = 4.5$
$[Acid] = 0.1 \ M$
Substituting the values:
$5.5 = 4.5 + \log \frac{[Salt]}{0.1}$
$5.5 - 4.5 = \log \frac{[Salt]}{0.1}$
$1 = \log \frac{[Salt]}{0.1}$
Taking antilog on both sides:
$10^{1} = \frac{[Salt]}{0.1}$
$[Salt] = 10 \times 0.1 = 1.0 \ M$

(C) For a sparingly soluble salt of the type $AX$,the dissociation is given by: $AX(s) \rightleftharpoons A^+(aq) + X^-(aq)$.
Let the solubility of the salt be $S \ mol \ dm^{-3}$.
Then,$[A^+] = S$ and $[X^-] = S$.
The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [A^+][X^-] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13}$.
Therefore,$S^2 = 4.9 \times 10^{-13} = 49 \times 10^{-14}$.
Taking the square root on both sides: $S = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.

(D) For a saturated solution of $Ba(OH)_2$,the dissociation is $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$.
Given $pH = 12$,then $pOH = 14 - 12 = 2$.
Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \ M$.
From the stoichiometry,$[Ba^{2+}] = \frac{1}{2} [OH^-] = \frac{1}{2} \times 10^{-2} = 0.5 \times 10^{-2} \ M$.
The solubility product constant $K_{sp}$ is given by $K_{sp} = [Ba^{2+}][OH^-]^2$.
Substituting the values: $K_{sp} = (0.5 \times 10^{-2}) \times (10^{-2})^2 = 0.5 \times 10^{-2} \times 10^{-4} = 0.5 \times 10^{-6} = 5 \times 10^{-7}$.
Thus,the correct option is $D$.

(A) The dissociation of $AgBr$ is given by: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.
Let the solubility be $s = 7.1 \times 10^{-7} \ mol \ dm^{-3}$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ag^+][Br^-] = s \times s = s^2$.
Substituting the value of $s$: $K_{sp} = (7.1 \times 10^{-7})^2$.
$K_{sp} = 50.41 \times 10^{-14} = 5.041 \times 10^{-13}$.

(B) The dissociation of $Ca_{3}(PO_{4})_{2}$ is represented as:
$Ca_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_{4}^{3-}(aq)$
If the solubility is $S \ mol \ dm^{-3}$,then the concentration of ions at equilibrium is:
$[Ca^{2+}] = 3S$
$[PO_{4}^{3-}] = 2S$
The solubility product expression is:
$K_{sp} = [Ca^{2+}]^{3} [PO_{4}^{3-}]^{2}$
Substituting the values:
$K_{sp} = (3S)^{3} (2S)^{2}$
$K_{sp} = (27S^{3}) (4S^{2})$
$K_{sp} = 108S^{5}$

(B) The dissociation of the salt $BA_2$ is given by: $BA_2 (s) \rightleftharpoons B^{2+} (aq) + 2A^- (aq)$.
Let the solubility of the salt be $s = 4 \times 10^{-4} \ mol \ dm^{-3}$.
Then,$[B^{2+}] = s$ and $[A^-] = 2s$.
The solubility product constant $K_{sp}$ is defined as: $K_{sp} = [B^{2+}][A^-]^2$.
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.
Now,substitute $s = 4 \times 10^{-4}$:
$K_{sp} = 4 \times (4 \times 10^{-4})^3 = 4 \times (64 \times 10^{-12}) = 256 \times 10^{-12} = 2.56 \times 10^{-10}$.

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.
Let the solubility of $PbI_2$ be $s \ mol \ dm^{-3}$.
Then,$[Pb^{2+}] = s$ and $[I^-] = 2s$.
The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.
Substituting the values: $1.08 \times 10^{-7} = (s)(2s)^2 = 4s^3$.
Solving for $s$: $s^3 = \frac{1.08 \times 10^{-7}}{4} = 0.27 \times 10^{-7} = 27 \times 10^{-9}$.
Taking the cube root: $s = \sqrt[3]{27 \times 10^{-9}} = 3 \times 10^{-3} \ mol \ dm^{-3}$.

(B) The dissociation of calcium carbonate is given by: $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$.
Let the solubility be $s = 6.4 \times 10^{-5} \ mol \ dm^{-3}$.
The solubility product constant is given by: $K_{sp} = [Ca^{2+}][CO_3^{2-}] = s \times s = s^2$.
Substituting the value of $s$: $K_{sp} = (6.4 \times 10^{-5})^2$.
$K_{sp} = 40.96 \times 10^{-10} = 4.096 \times 10^{-9}$.

(B) For a sparingly soluble salt of the type $AX_2$,the dissociation equilibrium is given by:
$AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$
If $s$ is the solubility of the salt in $mol \ dm^{-3}$,then the concentration of $A^{2+}$ is $s$ and the concentration of $X^-$ is $2s$.
The solubility product expression is:
$K_{sp} = [A^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$
Given $s = 1 \times 10^{-4} \ mol \ dm^{-3}$.
Substituting the value of $s$:
$K_{sp} = 4 \times (1 \times 10^{-4})^3 = 4 \times 10^{-12}$
Therefore,the correct option is $B$.

(B) For a sparingly soluble salt like $NiS$,the dissociation equilibrium is given by:
$NiS(s) \rightleftharpoons Ni^{2+}(aq) + S^{2-}(aq)$
Let the solubility of $NiS$ be $s \ mol \ dm^{-3}$.
Then,$[Ni^{2+}] = s$ and $[S^{2-}] = s$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [Ni^{2+}][S^{2-}] = s \times s = s^2$
Given $K_{sp} = 4.9 \times 10^{-5}$.
$s^2 = 4.9 \times 10^{-5} = 49 \times 10^{-6}$
$s = \sqrt{49 \times 10^{-6}} = 7.0 \times 10^{-3} \ mol \ dm^{-3}$
Therefore,the correct option is $B$.

(A) $1$. Convert solubility from $g \ dm^{-3}$ to $mol \ dm^{-3}$ (molarity,$S$):
$S = \frac{1.12 \times 10^{-4} \ g \ dm^{-3}}{112 \ g \ mol^{-1}} = 1.0 \times 10^{-6} \ mol \ dm^{-3}$.
$2$. For a binary salt $(AB)$,the dissociation is $AB \rightleftharpoons A^+ + B^-$.
$3$. The solubility product expression is $K_{sp} = [A^+][B^-] = S \times S = S^2$.
$4$. Substitute the value of $S$: $K_{sp} = (1.0 \times 10^{-6})^2 = 1.0 \times 10^{-12} \ mol^2 \ dm^{-6}$.

(A) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.
Given $K_{sp} = 8.0 \times 10^{-28}$ and $[S^{2-}] = 1 \times 10^{-11} \ mol \ dm^{-3}$.
Substituting the values into the equation: $8.0 \times 10^{-28} = [Pb^{2+}] \times (1 \times 10^{-11})$.
Therefore,$[Pb^{2+}] = \frac{8.0 \times 10^{-28}}{1 \times 10^{-11}} = 8.0 \times 10^{-17} \ mol \ dm^{-3}$.

(C) The dissociation of the salt $AB_2$ is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$.
Let the solubility of $AB_2$ be $s \ mol \ dm^{-3}$.
Then,$[A^{2+}] = s$ and $[B^-] = 2s$.
The solubility product expression is $K_{sp} = [A^{2+}][B^-]^2$.
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 2.56 \times 10^{-10}$.
So,$4s^3 = 2.56 \times 10^{-10}$.
$s^3 = 0.64 \times 10^{-10} = 64 \times 10^{-12}$.
$s = \sqrt[3]{64 \times 10^{-12}} = 4 \times 10^{-4} \ mol \ dm^{-3}$.

(C) For a salt of the type $AB$,the solubility product $K_{sp}$ is related to solubility $S$ by the equation: $K_{sp} = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13}$.
Therefore,$S = \sqrt{K_{sp}} = \sqrt{4.9 \times 10^{-13}}$.
$S = \sqrt{49 \times 10^{-14}} = 7 \times 10^{-7} \ mol \ dm^{-3}$.
Thus,the correct option is $C$.

(C) The dissociation of salt $B_2A$ is given by: $B_2A(s) \rightleftharpoons 2B^+(aq) + A^{2-}(aq)$.
Let the solubility of $B_2A$ be $s \ mol \ dm^{-3}$.
Then,$[B^+] = 2s$ and $[A^{2-}] = s$.
The solubility product expression is: $K_{sp} = [B^+]^2 [A^{2-}] = (2s)^2 (s) = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$.
So,$4s^3 = 3.2 \times 10^{-11}$.
$s^3 = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
$s = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ mol \ dm^{-3}$.

(A) For a sparingly soluble salt of the type $BA$,the dissociation equilibrium is given by: $BA(s) \rightleftharpoons B^+(aq) + A^-(aq)$.
Let the solubility of the salt be $s \ mol \ dm^{-3}$.
Then,$[B^+] = s$ and $[A^-] = s$.
The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [B^+][A^-] = s \times s = s^2$.
Given $K_{sp} = 2.7 \times 10^{-10}$.
Therefore,$s^2 = 2.7 \times 10^{-10}$.
$s = \sqrt{2.7 \times 10^{-10}} = \sqrt{27 \times 10^{-11}} = \sqrt{2.7} \times 10^{-5} \approx 1.643 \times 10^{-5} \ mol \ dm^{-3}$.
Thus,the ionic concentration of each ion is $1.643 \times 10^{-5} \ mol \ dm^{-3}$.

(A) The reaction $CO + H_2O \rightleftharpoons CO_2 + H_2$ is known as the water-gas shift reaction.
In industrial processes,this reaction is carried out at approximately $500^{\circ} C$ using iron chromite $(Fe_2O_3-Cr_2O_3)$ as a catalyst to increase the yield of hydrogen.

(D) $SO_2$,$CO_2$,and $N_2O_5$ are acidic oxides because they react with water to form acids ($H_2SO_3$,$H_2CO_3$,and $HNO_3$ respectively).
$N_2O$ (nitrous oxide) is a neutral oxide.
Therefore,$N_2O$ is not acidic in nature.

(C) The chemical formula for thiosulfuric acid is $H_2S_2O_3$.
In its structure,the central sulfur atom is bonded to another sulfur atom via a double bond $(S=S)$.
The central sulfur atom is also bonded to two hydroxyl groups $(-OH)$ via single bonds and one oxygen atom via a double bond $(S=O)$.
Thus,the bonds formed by the central sulfur with oxygen atoms are two $S=O$ double bonds and two $S-OH$ single bonds.
However,considering the standard representation of $H_2S_2O_3$ as $O=S(OH)_2=S$,the central sulfur atom forms two double bonds with oxygen/sulfur and two single bonds with oxygen atoms.
Therefore,the correct description is two double bonds and two single bonds.

(C) First,assign oxidation numbers to the elements in the reaction: $I_2^0 + K^{+1}Cl^{+5}O_3^{-2} \rightarrow I^{+1}Cl^{-1} + K^{+1}I^{+5}O_3^{-2}$.
In this reaction,$I_2$ (oxidation number $0$) is oxidized to $I^{+1}$ (in $ICl$) and $I^{+5}$ (in $KIO_3$).
Since $I_2$ is being oxidized,it acts as a reducing agent.
$Cl$ in $KClO_3$ changes from $+5$ to $-1$ (in $ICl$).
The change in oxidation number of $Cl$ is $|-1 - 5| = 6$. Thus,the oxidation number of $Cl$ decreases by $6$.
Therefore,the statement 'Oxidation number of $Cl$ decreases by $6$' is correct.

(C) An oxidising agent is a species that gains electrons and gets reduced. The strength of an oxidising agent is determined by its standard reduction potential $(E^{\circ}_{red})$. Higher positive values of $E^{\circ}_{red}$ indicate a stronger tendency to gain electrons,making the species a stronger oxidising agent.
Among the given options:
$1$. $Li$ is a strong reducing agent (it loses electrons easily).
$2$. $Li^{+}$ has a very negative reduction potential and is a very weak oxidising agent.
$3$. $F_2$ has the highest standard reduction potential $(+2.87 \ V)$,making it the strongest oxidising agent among the options.
$4$. $F^{-}$ is the reduced form of fluorine and cannot be further reduced,so it acts as a reducing agent.
Therefore,$F_2$ is the strongest oxidising agent.

(C) In the reactant $NH_4NO_2$,the nitrogen atoms are present in two different oxidation states:
$1$. In the ammonium ion $(NH_4^+)$,the oxidation state of $N$ is $x + 4(+1) = +1$,so $x = -3$.
$2$. In the nitrite ion $(NO_2^-)$,the oxidation state of $N$ is $x + 2(-2) = -1$,so $x = +3$.
In the product $N_2$,the oxidation state of $N$ is $0$.
Since the nitrogen atom in $NH_4^+$ $(-3)$ increases its oxidation state to $0$ in $N_2$,it undergoes oxidation.
Since the nitrogen atom in $NO_2^-$ $(+3)$ decreases its oxidation state to $0$ in $N_2$,it undergoes reduction.
Therefore,nitrogen is both oxidised and reduced in this reaction.

(B) reducing agent is a species that donates electrons and gets oxidized itself.
For a species to act as a reducing agent,it must be able to lose electrons.
$Li$ is a strong reducing agent because it easily loses an electron to form $Li^+$.
$F_2$ is a strong oxidizing agent as it easily gains electrons.
$F^-$ has a complete octet and a very high electronegativity,making it extremely difficult to remove an electron from it.
$Li^+$ has a stable noble gas configuration $(1s^2)$ and cannot lose further electrons easily to act as a reducing agent.
However,comparing $Li^+$ and $F^-$,$Li^+$ is a cation and cannot be oxidized further,making it the weakest reducing agent among the given options.

(B) In the tetrathionate ion $(S_4O_6^{2-})$,the structure consists of a chain of four sulfur atoms.
For the terminal sulfur atoms (numbered $1$ and $4$): Each is bonded to three oxygen atoms (two double-bonded and one single-bonded) and one sulfur atom. The oxidation state is calculated as: $x + 2(-2) + 1(-2) + 1(-1) = -1$ (considering the charge contribution). More simply,using electronegativity: $S$ bonded to $O$ gains $+2$ for each double bond and $+1$ for the single bond,totaling $+5$.
For the central sulfur atoms (numbered $2$ and $3$): Each is bonded to two other sulfur atoms. Since the atoms are identical,the bond between them contributes $0$ to the oxidation state. Thus,the oxidation state of the central sulfur atoms is $0$.
Therefore,the oxidation states for sulfur atoms $1, 2, 3, 4$ are $+5, 0, 0, +5$ respectively.

(D) redox reaction is one in which there is a change in the oxidation state of the elements involved.
In option $D$,the reaction is $Zn + 2 AgCN \longrightarrow 2 Ag + Zn(CN)_2$.
The oxidation state of $Zn$ changes from $0$ (in elemental form) to $+2$ (in $Zn(CN)_2$),which is oxidation.
The oxidation state of $Ag$ changes from $+1$ (in $AgCN$) to $0$ (in elemental form),which is reduction.
Since both oxidation and reduction occur,this is a redox reaction.
Options $A$,$B$,and $C$ are double displacement reactions where no change in oxidation state occurs.

(D) The oxidation state of a metal in a compound is considered fractional if it represents the average oxidation state of multiple atoms of the same element in different oxidation environments within the same molecule.
$Fe_3O_4$ is a mixed oxide of $FeO$ and $Fe_2O_3$,where $Fe$ exists in $+2$ and $+3$ states. The average oxidation state is $+8/3$.
$Mn_3O_4$ is a mixed oxide of $MnO$ and $Mn_2O_3$,where $Mn$ exists in $+2$ and $+3$ states. The average oxidation state is $+8/3$.
$Pb_3O_4$ is a mixed oxide of $2PbO$ and $PbO_2$,where $Pb$ exists in $+2$ and $+4$ states. The average oxidation state is $+8/3$.
In $Na_2S_4O_6$ (sodium tetrathionate),the sulfur atoms are not metals. The question asks for a metal. However,if we evaluate the oxidation states of the metals provided in the options,all $Fe$,$Mn$,and $Pb$ in these specific mixed oxides show fractional average oxidation states. Given the options,$Na_2S_4O_6$ is the only compound where the central element $(S)$ is a non-metal,and the metal $(Na)$ is in a fixed $+1$ oxidation state.

(C) For a compound to be electrically neutral,the sum of the oxidation numbers of all atoms must be zero.
Let the formula be $X_a(Y_bZ_c)_d$.
Checking option $C$: $X_3(YZ_4)_3$.
Oxidation state of $X = +3$,$Y = +5$,$Z = -2$.
Total charge $= 3(+3) + 3(+5 + 4(-2)) = 9 + 3(5 - 8) = 9 + 3(-3) = 9 - 9 = 0$.
Since the sum of oxidation numbers is zero,$X_3(YZ_4)_3$ is a possible formula.

(D) Stock notation represents the oxidation state of a metal in a compound using Roman numerals in parentheses.
In $Hg_2Cl_2$,the oxidation state of $Hg$ is $+1$,so it is $Hg(I)Cl$. In $HgCl_2$,the oxidation state of $Hg$ is $+2$,so it is $Hg(II)Cl_2$.
Option $A$ and $B$ are correct because $Hg(I)$ is the reduced form of $Hg(II)$.
Option $C$ is correct as stock notation explicitly states the oxidation state.
Option $D$ is incorrect because 'Aurous' refers to $Au(I)$,whereas $Au(III)Cl_3$ is 'Auric chloride'. Therefore,$Au(III)Cl_3$ is not Aurous chloride.

(C) In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $1 + x + 4(-2) = 0$,so $x = +7$.
In $Mn_2O_3$,the oxidation state of $Mn$ is calculated as: $2x + 3(-2) = 0$,so $2x = +6$,which means $x = +3$.
The change in oxidation state per $Mn$ atom is $|7 - 3| = 4$.
Since there are $2$ $Mn$ atoms in $Mn_2O_3$,the total number of electrons transferred is $2 \times 4 = 8$.
However,the question asks for the number of electrons transferred per $Mn$ atom involved in the reduction process from $KMnO_4$ to $Mn_2O_3$.
Considering the stoichiometry $2KMnO_4 \rightarrow Mn_2O_3$,the total change is $2 \times (7 - 3) = 8$ electrons.
Given the options provided,the change per $Mn$ atom is $4$.

(B) In $KMnO_4$,let the oxidation number of $Mn$ be $x$. The sum of oxidation numbers is $1 + x + 4(-2) = 0$,so $x - 7 = 0$,which gives $x = +7$.
In $MnO_2$,let the oxidation number of $Mn$ be $y$. The sum of oxidation numbers is $y + 2(-2) = 0$,so $y - 4 = 0$,which gives $y = +4$.
The difference in oxidation number is $|7 - 4| = 3$.

(B) The given rate law is $r = k[A][B]$.
Let the initial rate be $r_1 = k[A][B]$.
For option $A$: $r_2 = k(2[A])(2[B]) = 4k[A][B] = 4r_1$.
For option $B$: $r_2 = k(2[A])[B] = 2k[A][B] = 2r_1$.
For option $C$: $r_2 = k(0.5[A])(2[B]) = k[A][B] = r_1$.
For option $D$: $r_2 = k[A](0.5[B]) = 0.5k[A][B] = 0.5r_1$.
Thus,the rate of reaction doubles when the concentration of $A$ is doubled and the concentration of $B$ is kept constant.

(D) The order of a reaction is defined as the sum of the powers of the concentration terms in the rate law expression.
Given the rate law: $r = k[A]^{1.5}[B]^{2.5}$.
Order of reaction = $1.5 + 2.5 = 4$.
Therefore,the correct option is $D$.

(A) The initial rate is given by $r_1 = k[A]^a[B]^b$.
When the concentration of $A$ is doubled $([A]' = 2[A])$ and $B$ is halved $([B]' = \frac{1}{2}[B])$,the new rate $r_2$ is:
$r_2 = k(2[A])^a(\frac{1}{2}[B])^b$
$r_2 = k \cdot 2^a \cdot [A]^a \cdot (2^{-1})^b \cdot [B]^b$
$r_2 = 2^a \cdot 2^{-b} \cdot k[A]^a[B]^b$
$r_2 = 2^{a-b} \cdot r_1$
Therefore,the ratio $\frac{r_2}{r_1} = 2^{a-b}$.

(A) The rate law is given by the expression $R = k[NO_2]^2[CO]^0$.
Comparing this with the general rate law $R = k[A]^x[B]^y$,we can see that the exponent of the concentration of $CO$ is $0$.
Therefore,the order of the reaction with respect to $CO$ is $0$.

(A) The given rate law is $r = k[A]^2[B]$.
Given values are $k = 6.25 \ M^{-2} \ s^{-1}$,$[A] = 1 \ M$,and $[B] = 0.2 \ M$.
Substituting these values into the rate equation:
$r = 6.25 \times (1)^2 \times (0.2)$
$r = 6.25 \times 1 \times 0.2$
$r = 1.25 \ M \ s^{-1}$.
Therefore,the correct option is $A$.

(B) The unit of the rate constant for a reaction of order $n$ is given by $(\text{concentration})^{1-n} \times (\text{time})^{-1}$.
For a first-order reaction $(n=1)$,the unit is $(\text{time})^{-1}$.
Given the unit of the rate constant is $hour^{-1}$,which matches the unit for a first-order reaction.
Therefore,the order of the reaction is $1$.

(B) The rate law for the reaction $H_{2(g)} + I_{2(g)} \longrightarrow 2 HI_{(g)}$ is given by $Rate = k[H_2][I_2]$.
Since the sum of the powers of the concentration terms in the rate law is $1 + 1 = 2$,this reaction is a second-order reaction.
Option $B$ is the correct answer.

(B) Given rate of reaction $r = 3.6 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.
Rate law is $r = k[A][B]^2$.
Given concentrations are $[A] = 0.2 \ M$ and $[B] = 0.1 \ M$.
Substituting the values in the rate law:
$3.6 \times 10^{-2} = k \times (0.2) \times (0.1)^2$
$3.6 \times 10^{-2} = k \times (0.2) \times (0.01)$
$3.6 \times 10^{-2} = k \times (0.002)$
$k = \frac{3.6 \times 10^{-2}}{2 \times 10^{-3}}$
$k = 1.8 \times 10^1 = 18 \ mol^{-2} \ dm^6 \ s^{-1}$.
Thus,the correct option is $B$.

(C) The given reaction is $2 NO_{2(g)} \longrightarrow 2 NO_{(g)} + O_{2(g)}$.
This reaction is a well-known example of a second-order reaction at temperatures below $500 \ K$.
The rate law for this reaction is expressed as: $\text{Rate} = k[NO_2]^2$.
Since the exponent of the concentration term $[NO_2]$ is $2$,the order of the reaction is $2$.

(A) For a complex reaction,the rate of the overall reaction is determined by the slowest step,which is known as the rate-determining step.
In the given mechanism,the first step is the slow step:
$NO_2Cl_{(g)} \longrightarrow NO_{2(g)} + Cl_{(g)}$ (slow)
Therefore,the rate of the reaction depends only on the concentration of the reactant involved in this slow step.
The rate law is given by: $r = k[NO_2Cl]$.

(C) The unit of the rate constant $(k)$ for a reaction of order $n$ is given by $(mol \ L^{-1})^{1-n} \,s^{-1}$.
For a first-order reaction $(n=1)$, the unit is $(mol \ L^{-1})^{1-1} \,s^{-1} = s^{-1}$.
Since the given unit of the rate constant is $s^{-1}$, the reaction is a first-order reaction.

(D) The order of a reaction is defined as the sum of the powers of the concentration terms of the reactants in the rate law expression.
It is an experimental quantity and is not necessarily related to the stoichiometric coefficients of the balanced chemical equation.
Crucially,the order of a reaction can be zero,a fraction,or a whole number.
Therefore,the statement that it is always a whole number is incorrect.

(C) For a first-order reaction,the half-life $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Here,$k$ is the rate constant of the reaction.
As seen in the formula,$t_{1/2}$ is independent of the initial concentration of the reactant.
Therefore,if the initial concentration is doubled,the half-life remains unchanged.

(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given:
Initial concentration $[A]_0 = 20 \ mmol \ dm^{-3}$
Final concentration $[A]_t = 8 \ mmol \ dm^{-3}$
Time $t = 40 \ minute$
Substituting the values:
$k = \frac{2.303}{40} \log \frac{20}{8}$
$k = \frac{2.303}{40} \log(2.5)$
Since $\log(2.5) \approx 0.3979$:
$k = \frac{2.303 \times 0.3979}{40} \approx 0.0229 \ minute^{-1}$
Rounding off,we get $k \approx 0.023 \ minute^{-1}$.

(B) For a first order reaction,the rate constant $k$ is related to half-life $t_{1/2}$ by $k = \frac{0.693}{t_{1/2}}$.
Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \ R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
Substituting $k = \frac{0.693}{t_{1/2}}$,we get $\log \frac{t_{1/2, 1}}{t_{1/2, 2}} = \frac{E_a}{2.303 \ R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$.
Given $T_1 = 400 \ K$,$t_{1/2, 1} = 900 \ \text{min}$,$T_2 = 300 \ K$,and $\frac{E_a}{2.303 \ R} = 1305.6 \ K$.
$\log \frac{900}{t_{1/2, 2}} = 1305.6 \left( \frac{1}{400} - \frac{1}{300} \right) = 1305.6 \left( \frac{3 - 4}{1200} \right) = 1305.6 \left( -\frac{1}{1200} \right) = -1.088$.
$\frac{900}{t_{1/2, 2}} = 10^{-1.088} \approx 0.08166$.
$t_{1/2, 2} = \frac{900}{0.08166} \approx 11021 \ \text{min}$.
The closest option is $11025.0 \ \text{min}$.

(C) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 3 \ minute$,so $k = \frac{0.693}{3} = 0.231 \ minute^{-1}$.
The time $t$ required for a first order reaction is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.
If the concentration is reduced by $90 \%$,then $[A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0$.
Substituting the values: $t = \frac{2.303}{0.231} \log \frac{[A]_0}{0.10[A]_0} = \frac{2.303}{0.231} \log(10)$.
Since $\log(10) = 1$,$t = \frac{2.303}{0.231} \approx 9.97 \ minute$.

(A) For a first order reaction,the half-life $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given the rate constant $k = 1.386 \times 10^{-3} \ s^{-1}$.
Substituting the value of $k$ into the formula:
$t_{1/2} = \frac{0.693}{1.386 \times 10^{-3} \ s^{-1}}$.
$t_{1/2} = \frac{0.693}{1.386} \times 10^{3} \ s$.
$t_{1/2} = 0.5 \times 1000 \ s = 500 \ s$.

(D) For a zero-order reaction,the rate law is given by $r = K[A]^0 = K$.
Integrating the rate equation,we get $[A]_t = [A]_0 - Kt$.
At half-life $(t = t_{1/2})$,the concentration of the reactant $[A]_t = \frac{[A]_0}{2} = \frac{a}{2}$.
Substituting these values into the integrated rate equation: $\frac{a}{2} = a - K t_{1/2}$.
Rearranging for $t_{1/2}$: $K t_{1/2} = a - \frac{a}{2} = \frac{a}{2}$.
Therefore,$t_{1/2} = \frac{a}{2K}$.

(A) For a first order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.
Given $t_{1/2} = 20 \ min$,so $k = \frac{0.693}{20} = 0.03465 \ min^{-1}$.
The time $t$ required for the concentration to become $(1/10)^{th}$ of the initial concentration is given by the formula $t = \frac{2.303}{k} \log(\frac{[A]_0}{[A]_t})$.
Here,$[A]_t = \frac{[A]_0}{10}$,so $\frac{[A]_0}{[A]_t} = 10$.
Substituting the values: $t = \frac{2.303}{0.03465} \times \log(10)$.
Since $\log(10) = 1$,$t = \frac{2.303}{0.03465} \approx 66.46 \ min$.

(A) For a first-order reaction,the rate constant $k$ is related to the half-life $t_{1/2}$ by the formula $k = \frac{0.693}{t_{1/2}}$.
Given for reaction $A$: $t_{1/2, A} = 75 \ min$.
Given for reaction $B$: $t_{1/2, B} = 2.5 \ h = 2.5 \times 60 \ min = 150 \ min$.
The rate constant for $A$ is $k_A = \frac{0.693}{75}$.
The rate constant for $B$ is $k_B = \frac{0.693}{150}$.
The ratio of rate constants is $\frac{k_A}{k_B} = \frac{0.693 / 75}{0.693 / 150} = \frac{150}{75} = 2.0$.
Thus,the ratio is $2.0$.

(D) The decomposition of hydrogen peroxide $(2H_2O_{2_{(aq)}} \rightarrow 2H_2O_{(l)} + O_{2_{(g)}})$ is a well-known example of a first-order reaction.
In this reaction,the rate of decomposition depends on the concentration of $H_2O_2$ raised to the power of $1$,i.e.,$\text{Rate} = k[H_2O_2]^1$.

(D) For a first order reaction,the integrated rate equation is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given: $k = 1.15 \times 10^{-3} \ s^{-1}$,$[A]_0 = 5 \ g$,$[A]_t = 3 \ g$
Substituting the values: $1.15 \times 10^{-3} = \frac{2.303}{t} \log \frac{5}{3}$
$t = \frac{2.303}{1.15 \times 10^{-3}} \log(1.666)$
$t = \frac{2.303}{1.15 \times 10^{-3}} \times 0.2218$
$t \approx 444 \ s$
Therefore,the correct option is $D$.

(D) For a first order reaction,the half-life $(t_{1/2})$ is the time taken for $50 \%$ completion. Given $t_{1/2} = 16 \ minutes$.
In $32 \ minutes$,the number of half-lives elapsed is $n = \frac{32}{16} = 2$.
The fraction of reactant remaining after $n$ half-lives is given by $(\frac{1}{2})^n$.
Remaining fraction = $(\frac{1}{2})^2 = \frac{1}{4} = 0.25$ or $25 \%$.
The percentage of reactant reacted = $100 \% - 25 \% = 75 \%$.

(A) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Here,$[A]_0 = 100$ and $[A]_t = 100 - 99 = 1$.
Given $k = 23.03 \ min^{-1}$.
Substituting the values: $23.03 = \frac{2.303}{t} \log \frac{100}{1}$.
$23.03 = \frac{2.303}{t} \log(10^2)$.
$23.03 = \frac{2.303 \times 2}{t}$.
$t = \frac{4.606}{23.03} = 0.2 \ min$.

(C) For a first-order reaction, the integrated rate equation is $\ln [A]_t = -kt + \ln [A]_0$.
Converting this to base $10$ logarithm: $\log [A]_t = -\frac{k}{2.303}t + \log [A]_0$.
The slope of the graph $\log [A]_t$ versus '$t$' is given by $m = -\frac{k}{2.303}$.
Given slope $m = -2.5 \times 10^{-3} \,s^{-1}$.
Therefore, $-\frac{k}{2.303} = -2.5 \times 10^{-3} \,s^{-1}$.
$k = 2.5 \times 10^{-3} \times 2.303 \,s^{-1} = 5.7575 \times 10^{-3} \,s^{-1}$.
Thus, the rate constant is $5.757 \times 10^{-3} \,s^{-1}$.

(A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given that the reaction is $20 \%$ completed,the remaining concentration $[A]_t$ is $80 \%$ of the initial concentration $[A]_0$.
So,$[A]_t = 0.80 [A]_0$ and $t = 10 \ min$.
Substituting these values: $k = \frac{2.303}{10} \log \frac{1}{0.8} = \frac{2.303}{10} \log(1.25)$.
Since $\log(1.25) \approx 0.0969$,we get $k = \frac{2.303 \times 0.0969}{10} \approx 0.0223 \ min^{-1}$.

(C) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.
Given,$[A]_0 = 0.4 \ M$,$[A]_t = 0.1 \ M$,and $t = x \ \text{hours}$.
Substituting these values: $k = \frac{2.303}{x} \log \frac{0.4}{0.1} = \frac{2.303}{x} \log 4 = \frac{2.303}{x} \times 2 \log 2$.
Since $\log 2 \approx 0.3010$,$k = \frac{2.303 \times 2 \times 0.3010}{x} = \frac{1.386}{x}$.
The half-life $t_{1/2}$ is given by: $t_{1/2} = \frac{0.693}{k}$.
Substituting $k$: $t_{1/2} = \frac{0.693}{1.386 / x} = \frac{0.693 \times x}{1.386} = \frac{x}{2} \ \text{hours}$.

(D) reaction intermediate is a substance that is produced in one step of a reaction mechanism and consumed in a subsequent step.
In the given mechanism:
Step $i$: $2 ClO^{-} \rightarrow ClO_2^{-} + Cl^{-}$
Step $ii$: $ClO_2^{-} + ClO^{-} \rightarrow ClO_3^{-} + Cl^{-}$
The species $ClO_2^{-}$ is produced in step $i$ and consumed in step $ii$.
Therefore,$ClO_2^{-}$ is the reaction intermediate.

(B) The Arrhenius equation is given by: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} [\frac{T_2 - T_1}{T_1 T_2}]$.
Given: $k_1 = 0.026 \ s^{-1}$ at $T_1 = 290 \ K$,$k_2 = 0.58 \ s^{-1}$ at $T_2 = 300 \ K$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $\ln(\frac{0.58}{0.026}) = \frac{E_a}{8.314} [\frac{300 - 290}{300 \times 290}]$.
$\ln(22.307) = \frac{E_a}{8.314} [\frac{10}{87000}]$.
$3.1048 = \frac{E_a}{8.314} \times 1.1494 \times 10^{-4}$.
$E_a = \frac{3.1048 \times 8.314}{1.1494 \times 10^{-4}} \approx 224600 \ J \ mol^{-1} = 224.6 \ kJ \ mol^{-1}$.
The closest option is $224.55 \ kJ \ mol^{-1}$.

(C) Given: $T_1 = 27^{\circ} C = 300 \ K$,$T_2 = 37^{\circ} C = 310 \ K$.
The rate constant doubles,so $k_2 = 2k_1$.
Using the Arrhenius equation: $\log(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} \times (\frac{T_2 - T_1}{T_1 T_2})$.
Substituting the values: $\log(2) = \frac{E_a}{2.303 \times 8.314} \times (\frac{310 - 300}{300 \times 310})$.
$0.3010 = \frac{E_a}{19.147} \times (\frac{10}{93000})$.
$E_a = \frac{0.3010 \times 19.147 \times 93000}{10} \approx 53598 \ J \ mol^{-1} \approx 53.6 \ kJ \ mol^{-1}$.

(D) The Arrhenius equation is given by $k = A \cdot e^{-E_a / RT}$.
Taking the logarithm on both sides,we get $\log k = \log A - \frac{E_a}{2.303 RT}$.
Given values are $A = 1.6 \times 10^{13} \ s^{-1}$ and $\frac{E_a}{2.303 RT} = 21$.
Substituting these values into the equation: $\log k = \log(1.6 \times 10^{13}) - 21$.
$\log k = \log(1.6) + \log(10^{13}) - 21$.
$\log k = 0.204 + 13 - 21$.
$\log k = 13.204 - 21 = -7.796$.
$k = \text{antilog}(-7.796) = \text{antilog}(-8 + 0.204) = 1.6 \times 10^{-8} \ s^{-1}$.

(A) Aspirin is chemically known as $2$-acetoxybenzoic acid.
It is prepared by the acetylation of salicylic acid.
It contains an ester linkage $(-COO-)$ and a carboxylic acid group $(-COOH)$,not an amide linkage.
Therefore,the statement that it contains an amide linkage is false.
Additionally,aspirin is known to cause stomach irritation as a side effect,although it is generally safer than salicylic acid.

(B) Antiseptics are chemical substances that prevent the growth of microorganisms or kill them and are applied to living tissues such as wounds,cuts,ulcers,and diseased skin surfaces.
$Thymol$ is a well-known antiseptic compound.
$Salvarsan$ is an antibiotic (specifically an anti-syphilitic drug).
$Sulphanilamide$ is a sulpha drug (antibacterial).
$Chloramphenicol$ is a broad-spectrum antibiotic.
Therefore,the correct option is $B$.

(D) Gallic acid is a naturally occurring phenolic acid found in various plants. Among the given options,$Indian \ gooseberry$ (also known as $Amla$) is a rich source of gallic acid.

(B) Mustard gas is a chemical warfare agent known as $2,2'$-dichlorodiethyl sulfide.
Its chemical formula is $ClCH_2CH_2SCH_2CH_2Cl$.

(C) Curcumin is a bright yellow chemical produced by some plants. It is the principal curcuminoid of turmeric $(Curcuma \ longa)$.
It is well-known for its potent $Antioxidant$ and $Anti-inflammatory$ properties.
Therefore,among the given options,$Antioxidant$ is the correct property exhibited by curcumin.

(B) Silver nanoparticles $(AgNPs)$ are widely known for their potent antimicrobial properties. They are frequently used to coat filter materials in water purification systems because they effectively kill bacteria and prevent biofilm formation,acting as an efficient bacterial disinfectant.

(A) Eugenol is a chemical compound that is the primary component of clove oil. It is responsible for the characteristic aroma and flavor of cloves. Therefore,the correct source of Eugenol is $A$. Clove.

(D) Clove contains eugenol,which is a well-known compound used for its analgesic (pain-relieving) and antimicrobial properties. It is commonly used in dentistry to relieve toothache.

(D) The structure of $Paracetamol$ is $N-(4-hydroxyphenyl)acetamide$.
It contains an amide functional group $(-NH-CO-)$ in its structure.
$Aspirin$ and $Methyl \ salicylate$ are esters.
$Curcumin$ contains phenolic and enolic groups but not an amide linkage.

(A) $Sodium \ lauryl \ sulphate$ $(CH_3(CH_2)_{11}SO_4^-Na^+)$ is a classic example of an anionic detergent.
In anionic detergents,the anionic part of the molecule is involved in the cleansing action.
These detergents are commonly used in toothpaste and household cleaning products.
Therefore,the correct statement is that it is an anionic detergent used in toothpaste.

(B) The ligand $en$ stands for ethylenediamine,which is a bidentate ligand.
In the given options,$Bis(ethylenediamine)dithiocyanatoplatinum(IV)$ contains the ligand $en$ as indicated by the name 'ethylenediamine'.
The formula for this complex is $[Pt(en)_2(SCN)_2]^{2+}$.
Therefore,the correct option is $B$.

(B) The chemical formula for the Tetracyanonickelate$(II)$ ion is $[Ni(CN)_4]^{2-}$.
In this complex,the ligand is the cyanide ion $(CN^-)$.
Each cyanide ion acts as a monodentate ligand,meaning it donates one electron pair through the carbon atom.
Since there are $4$ cyanide ligands,the total number of donor atoms is $4 \times 1 = 4$.

(B) The chemical formula of cisplatin is $[Pt(NH_{3})_{2}Cl_{2}]$.
In this coordination complex,the central metal ion is platinum $(Pt^{2+})$.
The ligands attached to the central metal ion are two ammonia $(NH_{3})$ molecules and two chloride $(Cl^{-})$ ions.
Therefore,the correct option is $B$.

(A) According to the spectrochemical series,ligands are arranged in increasing order of their field strength.
Among the given options,$CN^{-}$ is a strong field ligand because it is capable of causing pairing of electrons in the $d$-orbitals of the central metal ion.
$I^{-}$ is a weak field ligand,while $SCN^{-}$ and $C_2O_4^{2-}$ are intermediate or weak field ligands compared to $CN^{-}$.

(A) Cisplatin is a coordination complex with the chemical formula $[Pt(NH_3)_2Cl_2]$.
In this complex,the central metal atom is $Pt$ (Platinum).
It is bonded to two $NH_3$ ligands and two $Cl^-$ ligands.
Since each ligand is monodentate,the total number of coordinate bonds formed by the central metal atom is $2 + 2 = 4$.
Therefore,the coordination number of $Pt$ in cisplatin is $4$.

(D) The chemical formula for pentaamminecarbonatocobalt$(III)$ chloride is $[Co(NH_3)_5(CO_3)]Cl$.
In this coordination compound,the chloride ion $(Cl^-)$ is present outside the coordination sphere as a counter ion.
When this compound reacts with excess silver nitrate $(AgNO_3)$,the chloride ion outside the coordination sphere reacts to form silver chloride $(AgCl)$ precipitate:
$[Co(NH_3)_5(CO_3)]Cl + AgNO_3 \rightarrow [Co(NH_3)_5(CO_3)]NO_3 + AgCl(s)$.
Since there is only $1$ mole of ionizable $Cl^-$ per mole of the complex,$1$ mole of $AgCl$ will be precipitated.

(C) neutral ligand is a ligand that carries no net electrical charge.
Among the given options:
$1$. Nitrato $(NO_3^-)$ is an anionic ligand with a charge of $-1$.
$2$. Cyano $(CN^-)$ is an anionic ligand with a charge of $-1$.
$3$. Aqua $(H_2O)$ is a neutral ligand with a net charge of $0$.
$4$. Iodo $(I^-)$ is an anionic ligand with a charge of $-1$.
Therefore,the correct option is $C$.

(A) neutral ligand is a ligand that carries no net electrical charge.
$1$. Carbonyl $(CO)$ is a neutral ligand.
$2$. Sulphato $(SO_4^{2-})$ carries a charge of $-2$.
$3$. Oxalato $(C_2O_4^{2-})$ carries a charge of $-2$.
$4$. Bromo $(Br^-)$ carries a charge of $-1$.
Therefore,the correct option is $A$.

(B) An octahedral complex is formed when the central metal ion is surrounded by $6$ ligands in an octahedral geometry. Therefore,the coordination number of the central metal ion in an octahedral complex is $6$.