Question

When acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution is added to ${\mathrm{Sn}}^{2+}$ salt, then ${\mathrm{Sn}}^{2+}$ changes to:

• A. $\mathrm{Sn}$
• B. ${\mathrm{Sn}}^{+}$
• C. ${\mathrm{Sn}}^{3+}$
• D. ${\mathrm{Sn}}^{4+}$

Answer 1

The correct option is D

${\mathrm{Sn}}^{4+}$

The explanation for the correct option:

(D) ${\mathrm{Sn}}^{4+}$

• The electronic configuration of $\mathrm{Sn}$ is $\left[\mathrm{Kr}\right]4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^2$.
• According to this configuration, $\mathrm{Sn}$ can exist in $+2$ and $+4$ oxidation state by either losing two or four electrons.
• As acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution is an oxidising agent, therefore, it causes the oxidation of ${\mathrm{Sn}}^{2+}$ions to the highest oxidation state.
• Thus, ${\mathrm{Sn}}^{2+}$ is oxidised to ${\mathrm{Sn}}^{4+}$ and $\mathrm{Cr}$ in $+6$ oxidation state is reduced to $+3$ oxidation state.

$$\begin{gathered} 3{\mathrm{Sn}}^{2+}+{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+14{\mathrm{H}}^{+}\to 3{\mathrm{Sn}}^{4+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Tin}\left(\mathrm{II}\right)\mathrm{dichromate}\mathrm{ion}\mathrm{Stannic}\mathrm{Chromium}\left(\mathrm{III}\right) \\ \mathrm{cation} \end{gathered}$$

The explanation for the incorrect options:

(A) $\mathrm{Sn}$

• Oxidation refers to an increase in the oxidation state.
• Thus, the oxidising agent cannot change ${\mathrm{Sn}}^{2+}$to $\mathrm{Sn}$.

(B) ${\mathrm{Sn}}^{+}$

• Oxidation refers to an increase in the oxidation state.
• Thus, the oxidising agent cannot change ${\mathrm{Sn}}^{2+}$to ${\mathrm{Sn}}^{+}$.

(C) ${\mathrm{Sn}}^{3+}$

• Oxidation refers to an increase in the oxidation state.
• Thus, the oxidising agent cannot change ${\mathrm{Sn}}^{2+}$to ${\mathrm{Sn}}^{3+}$ as it is not the highest oxidation state.

Hence, option (D) is correct. When acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution is added to ${\mathrm{Sn}}^{2+}$ salt, then ${\mathrm{Sn}}^{2+}$ changes to ${\mathrm{Sn}}^{4+}$.