When an oxide of manganese $(\mathrm{A})$ is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved.
It is the method of preparation of potassium permanganate (purple).
Thus,
$$\begin{array}{ll} (A)=\mathrm{MnO}_2 & (B)=\mathrm{K}_2 \mathrm{MnO}_4 \\ (C)=\mathrm{KMnO}_4 & (D)=\mathrm{KIO}_3 \end{array}$$
$$ \underset{[A]}{2 \mathrm{MnO}_2}+4 \mathrm{KOH}+\mathrm{O}_2 \longrightarrow \underset{[B]}{2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}}$$
$3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \underset{[C]}{\mathrm{2MnO_4^-}} + \mathrm{MnO_2+2H_2O}$
$\mathrm{2MnO_4^- + H_2O+KI}\longrightarrow \underset{[A]}{2 \mathrm{MnO}_2} + \mathrm{2OH}^- + \underset{[D]}{\mathrm{KIO}_3}$
On the basis of lanthanoid contraction, explain the following:
(i) Nature of bonding in $\mathrm{La}_2 \mathrm{O}_3$ and $\mathrm{Lu}_2 \mathrm{O}_3$.
(ii) Trends in the stability of oxo salts of lanthanoids from La to Lu.
(iii) Stability of the complexes of lanthanoids.
(iv) Radii of 4 d and 5 d block elements.
(v) Trends in acidic character of lanthanoid oxides.
(i) As the size decreases covalent character increases. Therefore, $\mathrm{La}_2 \mathrm{O}_3$ is more ionic and $\mathrm{Lu}_2 \mathrm{O}_3$ is more covalent.
(ii) As the size decreases from La to Lu, stability of oxo salts also decreases.
(iii) Stability of complexes increases as the size of lanthanoids decreases.
(iv) Radii of $4 d$ and $5 d$-block elements will be almost same.
(v) Acidic character of oxides increases from La to Lu.
(a) Answer the following questions
(i) Which element of the first transition series has highest second ionisation enthalpy?
(ii) Which element of the first transition series has highest third ionisation enthalpy?
(iii) Which element of the first transition series has lowest enthalpy of atomisation?
(b) Identify the metal and justify your answer.
(i) Carbonyl $\mathrm{M}(\mathrm{CO})_5$
(ii) $\mathrm{MO}_3 \mathrm{F}$
(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 $\mathrm{MnO}_3 \mathrm{F}$.
Mention the type of compounds formed when small atoms like $\mathrm{H}, \mathrm{C}$ and $\mathrm{N}$ get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.
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) Transition metals can act as catalysts because these can change their oxidation state. How does Fe (III) catalyse the reaction between iodide and persulphate ions?
(b) Mention any three processes where transition metals act as catalysts.
(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 $\mathrm{KClO}_3$.