Assertion (A) The highest oxidation state of osmium is +8 .
Reason (R) Osmium is a 5 d -block element.
Identify A to E and also explain the reaction involved.
The substances from $A$ to $E$ are
$$A=\mathrm{Cu} ; B=\mathrm{Cu}\left(\mathrm{NO}_3\right)_2 ; C=\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} ; D=\mathrm{CO}_2 ; E=\mathrm{CaCO}_3$$
(i) $\mathrm{CuCO}_3 \xrightarrow{\Delta} \mathrm{CuO}+\mathrm{CO}_2] \times 2$
(ii) $2 \mathrm{CuO}+\mathrm{CuS} \longrightarrow \underset{[A]}{3 \mathrm{Cu}}+\mathrm{SO}_2$
(iii) $\underset{[A]}{\mathrm{Cu}}+4 \mathrm{HNO}_3$ (conc.) $\longrightarrow \underset{[B]}{\mathrm{Cu}\left(\mathrm{NO}_3\right)_2}+2 \mathrm{NO}+2 \mathrm{H}_2 \mathrm{O}$
(iv) $\underset{[B]}{\mathrm{Cu}^{2+}}+\mathrm{NH}_3 \longrightarrow \underset{\substack{\text { [C] (Blue solution)}}}{\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]}$
(v) $$Ca{(OH)_2} + \mathop {C{O_2}}\limits_{[D]} \buildrel {} \over \longrightarrow \mathop {CaC{O_3}}\limits_{[E]\,(Milky)} + {H_2}O$$
(vi) $\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2$
When a chromite ore $(A)$ is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound $(C)$ is treated with KCl , orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is an orange compound. It is formed when $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ reacts with KCl . In acidic medium, yellow coloured $\mathrm{CrO}_4^{2-}$ (chromate ion) changes into dichromate.
The given process is the preparation method of potassium dichromate from chromite ore.
$$A=\mathrm{FeCr}_2 \mathrm{O}_4 ; B=\mathrm{Na}_2 \mathrm{CrO}_4 ; C=\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 ; D=\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 .$$
(i) $$\mathop {4FeC{r_2}{O_4}}\limits_{[A]} + 8N{a_2}C{O_3} + 7{O_2}\buildrel {} \over \longrightarrow \mathop {8N{a_2}Cr{O_4}}\limits_{[B]} + 2F{e_2}{O_3} + 8C{O_2}$$
(ii) $2 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{H}^{+} \longrightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O}$
(iii) $$\mathop {N{a_2}C{r_2}{O_7}}\limits_{[C]} + 2KCl\buildrel {} \over \longrightarrow \mathop {{K_2}C{r_2}{O_7}}\limits_{[D]} + 2NaCl$$
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.