(a) (i) Cu , because the electronic configuration of Cu is $3 d^{10} 4 s^1$. So, second electron needs to be removed from completely filled $d$-orbital which is very difficult.

(ii) Zinc, because of electronic configuration of $\mathrm{Zn}=3 d^{10} 4 s^2$ and $\mathrm{Zn}^{2+}=3 d^{10}$ which is fully filled and hence is very stable. Removal of third electron requires very high energy.

(iii) Zinc, because of it has completely filled 3d subshell and no unpaired electron is available for metallic bonding.

(b) (i) Carbonyl $\mathrm{M}(\mathrm{CO})_5$ is $\mathrm{Fe}(\mathrm{CO})_5$

According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic number of nearest inert gas EAN is calculated as

$$\begin{aligned} \text { EAN } & =\text { number of electrons in metal }+2 \times(\mathrm{CO}) \\ & =\text { atomic number of nearest inert gas } \end{aligned}$$

In $M(\mathrm{CO})_5=x+2 \times(5)=36$ ( Kr is the nearest inert gas)

$x=26$ (atomic number of metal)

So, the metal is Fe (iron).

(ii) $\mathrm{MO}_3 \mathrm{~F}$ is $\mathrm{MnO}_3 \mathrm{~F}$.

In MO $\mathrm{F}_3$

Let us assume that oxidation state of $M$ is $x$

$$x+3 \times(-2)+(-1)=0$$

or, $x=+7$ i.e., $M$ is in +7 oxidation state of +7 . Hence, the given compound is $\mathrm{MnO}_3 \mathrm{F}$.

When small atoms like $\mathrm{H}, \mathrm{C}$ and $\mathrm{N}$ get trapped inside the crystal lattice of transition metals.

(a) Such compounds are called interstitial compounds.

(b) Their characteristic properties are;

(i) They have high melting points, higher than those of pure metals.

(ii) They are very hard.

(iii) They retain metallic conductivity.

(iv) They are chemically inert.

(a) Reaction between iodide and persulphate ions is

$2 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \xrightarrow{\mathrm{Fe}(\mathrm{III})} \mathrm{I}_2+2 \mathrm{SO}_4^{2-}$

$$\begin{aligned} &\text { Role of } \mathrm{Fe} \text { (III) ions }\\ &\begin{aligned} 2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} & \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2 \\ 2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} & \longrightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-} \end{aligned} \end{aligned}$$

(b) (i) Vanadium $(\mathrm{V})$ oxide used in contact process for oxidation of $\mathrm{SO}_2$ to $\mathrm{SO}_3$.

(ii) Finely divided iron in Haber's process in conversion of $\mathrm{N}_2$ and $\mathrm{H}_2$ to $\mathrm{NH}_3$.

(iii) $\mathrm{MnO}_2$ in preparation of oxygen from $\mathrm{KClO}_3$.

Since, compound $(\mathrm{C})$ on treating with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and NaCl gives $\mathrm{Cl}_2$ gas, so it is manganese dioxide $\left(\mathrm{MnO}_2\right)$. It is obtained alongwith $\mathrm{MnO}_4^{2-}$ when $\mathrm{KMnO}_4$ (violet) is heated.

Thus,

$$\begin{array}{ll} (A)=\mathrm{KMnO}_4 & (B)=\mathrm{K}_2 \mathrm{MnO}_4 \\ (C)=\mathrm{MnO}_2 & (D)=\mathrm{MnCl}_2 \end{array}$$

$\underset{[A]}{\mathrm{KMnO}_4} \xrightarrow{\Delta} \underset{[B]}{\mathrm{K}_2 \mathrm{MnO}_4}+\underset{[C]}{\mathrm{MnO}_2}+\mathrm{O}_2$

$2 \mathrm{MnO}_2+4 \mathrm{KOH}+\mathrm{O}_2 \longrightarrow 2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}$

$$\mathrm{MnO}_2+4 \mathrm{NaCl}+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{[D]}{\mathrm{MnCl}_2}+4 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$$