Transition elements show high melting points. Why?
Transition elements show high melting point, due to involvement of greater number of electrons in the interatomic bonding from $(n-1) d$-orbitals in addition to $n s$ electrons in forming metallic bond. Thus, large number of electrons participate forming large number of metallic bond.
When $\mathrm{Cu}^{2+}$ ion is treated with KI , a white precipitate is formed. Explain the reaction with the help of chemical equation.
When $\mathrm{Cu}^{2+}$ ion is treated with KI , it produces $\mathrm{Cu}_2 \mathrm{I}_2$ white precipitate in the final product.
$$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow \underset{\text { (White ppt.) }}{\mathrm{Cu}_2 \mathrm{I}_2}+\mathrm{I}_2$$
(In this reaction, $\mathrm{CuI}_2$ is formed which being unstable, dissociates into $\mathrm{Cu}_2 \mathrm{I}_2$ and $\mathrm{I}_2$ ).
Out of $\mathrm{Cu}_2 \mathrm{Cl}_2$ and $\mathrm{CuCl}_2$, which is more stable and why?
Among $\mathrm{Cu}_2 \mathrm{Cl}_2$ and $\mathrm{CuCl}_2, \mathrm{CuCl}_2$ is more stable. Stability of complex can be explained on the basis of stability of oxidation state of copper. Stability of $\mathrm{Cu}^{2+}(a q)$ rather than $\mathrm{Cu}^{+}(a q)$ is due to much more negative value of $\left[\Delta_{\text {hyd }} H^{\mathrm{S}}\right.$ of $\left.\mathrm{Cu}^{2+}(\mathrm{aq})\right]$ than $\mathrm{Cu}^{+}$which more than compensates for the second ionisation enthalpy of Cu .
When a brown compound of manganese $(\mathrm{A})$ is treated with HCl it gives a gas (B). The gas taken in excess, reacts with $\mathrm{NH}_3$ to give an explosive compound (C). Identify compounds A, B and C.
$\mathrm{MnO}_2$ is the brown compound of Mn which reacts with HCl to give $\mathrm{Cl}_2$ gas. This gas forms an explosive compound $\mathrm{NCl}_3$ when treated with $\mathrm{NH}_3$. Thus, $A=\mathrm{MnO}_2 ; B=\mathrm{Cl}_2 ; C=\mathrm{NCl}_3$ and reactions are as follows
(i) $$\underset{[A]}{\mathrm{MnO}_2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_2+\underset{[B]}{\mathrm{Cl}_2}+2 \mathrm{H}_2 \mathrm{O} $$
(ii) $\mathrm{NH}_3+\underset{\text { (Excess) }}{3 \mathrm{Cl}_2} \longrightarrow \underset{[\mathrm{C}]}{\mathrm{NCl}_3}+3 \mathrm{HCl}$
Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Oxygen can form multiple bonds with metals, while fluorine can't form multiple bond with metals. Hence, oxygen has more ability to stabilize higher oxidation state rather than fluorine.