Question

The correct code for stability of oxidation states of Sn and Pb is :
$\left[1\right]P{b}^{2+}>P{b}^4,S{n}^{2+}>S{n}^{4+}$
$\left[2\right]P{b}^{2+}<P{b}^4,S{n}^{2+}<S{n}^{4+}$
$\left[3\right]P{b}^{2+}>P{b}^4,S{n}^{2+}<S{n}^{4+}$
$\left[4\right]P{b}^{2+}<P{b}^4,S{n}^{2+}>S{n}^{4+}$
$\left[5\right]S{n}^{2+}<P{b}^{2+},S{n}^{4+}>P{b}^{4+}$
$\left[6\right]S{n}^{2+}<P{b}^{2+},S{n}^{4+}<P{b}^{4+}$

• A. $\left[5\right]$ & $\left[6\right]$
• B. $\left[1\right]$,$\left[3\right]$,$\left[5\right]$ & $\left[6\right]$
• C. $\left[3\right]$ & $\left[5\right]$
• D. $\left[2\right]$ & $\left[4\right]$

Answer 1

The correct option is C $\left[3\right]$ & $\left[5\right]$

The correct code for stability of oxidation states of Sn and Pb is $\left[3\right]P{b}^{2+}>P{b}^4,S{n}^{2+}<S{n}^{4+}$ and $\left[5\right]S{n}^{2+}<P{b}^{2+},S{n}^{4+}>P{b}^{4+}$

On moving down the carbon family, the stability of $+4$ oxidation state decreases and the stability of $+2$ oxidation state increases.

$G{e}^{2+}$ compounds are unstable and readily oxidised to $G{e}^{4+}$ compounds.
$S{n}^{2+}$ compounds are reducing agents.
$P{b}^{4+}$ compounds act as oxidising agents and are easily reduced to $P{b}^{2+}$ compounds.