Assertion (A) Cu cannot liberate hydrogen from acids.
Reason (R) Because it has positive electrode potential.
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,
$$\begin{array}{ll} (A)=\mathrm{MnO}_2 & (B)=\mathrm{K}_2 \mathrm{MnO}_4 \\ (C)=\mathrm{KMnO}_4 & (D)=\mathrm{KIO}_3 \end{array}$$
$$ \underset{[A]}{2 \mathrm{MnO}_2}+4 \mathrm{KOH}+\mathrm{O}_2 \longrightarrow \underset{[B]}{2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}}$$
$3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow \underset{[C]}{\mathrm{2MnO_4^-}} + \mathrm{MnO_2+2H_2O}$
$\mathrm{2MnO_4^- + H_2O+KI}\longrightarrow \underset{[A]}{2 \mathrm{MnO}_2} + \mathrm{2OH}^- + \underset{[D]}{\mathrm{KIO}_3}$