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.
Although $\mathrm{Cr}^{3+}$ and $\mathrm{Co}^{2+}$ ions have same number of unpaired electrons but the magnetic moment of $\mathrm{Cr}^{3+}$ is 3.87 BM and that of $\mathrm{Co}^{2+}$ is 4.87 BM, Why?
Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in $\mathrm{Cr}^{3+}$ ion. However, appreciable orbital contribution takes place in $\mathrm{Co}^{2+}$ ion.
Ionisation enthalpies of $\mathrm{Ce}, \mathrm{Pr}$ and Nd are higher than $\mathrm{Th}, \mathrm{Pa}$ and U . Why?
$\mathrm{Ce}, \mathrm{Pr}$ and Nd are lanthanoids and have incomplete $4 f$ shell while $\mathrm{Th}, \mathrm{Pa}$ and U are actinoids and have $5 f$ shell incomplete. In the beginning, when $5 f$-orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5 -electrons will therefore, be more effectively shielded from the nuclear charge than $4 f$ electrons of the corresponding lanthanoids.
Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids.