Question

Which is best oxidising agent?

• A. $G{e}^{4+}$
• B. $P{b}^{4+}$
• C. $S{n}^{4+}$
• D. $S{n}^{2+}$

Answer 1

Answer: (B)

$P{b}^{4+}$

The inertness of $s$ subshell electrons towards the bond formation is called inert pair effect or it can be said as the inactiveness of electrons present in the outermost shell (i.e. $n{s}^2$) to get unpaired and involve in bond formation is called inert pair effect.

For example:

1) In 13th group, thallium can exhibit +1 and+3 oxidation states but it is stable in +1 oxidation state only due to inert pair effect

2) In the 14th group, lead shows both +2 and +4 oxidation states but it is stable in +2 oxidation state due to inert pair effect.

Due to this inert pair effect,

Stability of $P{b}^{2+}>S{n}^{2+}>G{e}^{2+}$ lower oxidation state stability increases down the group
$P{b}^{4+}<S{n}^{4+}<G{e}^{4+}$ higher oxidation state stability decreases down the group.
$P{b}^{2+}>P{b}^{4+}$
$S{n}^{4+}>S{n}^{2+}$
$G{e}^{4+}>G{e}^{2+}$

Thus, $P{b}^{4+}+2e^{-}\to P{b}^{2+}$
Thus, $P{b}^{4+}$ salts are better oxiding agent being easily reduced to $P{b}^{2+}$.

Hence, the correct option is $\text{B}$