Question

Find the odd one out with respect to the valence electronic configuration of the central(underlined) atom.

• A. P${\mathrm{Cl}}_5$
• B. Sn${\mathrm{Cl}}_4$
• C. Cl${\mathrm{F}}_3$
• D. S${\mathrm{F}}_4$

Answer 1

The correct option is B

Sn${\mathrm{Cl}}_4$

Explanation of the correct option:

• The electronic configuration of the central atom is as follows for each of them
• $\mathrm{P}-\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^3$
• $\mathrm{Sn}-\left[\mathrm{Kr}\right]4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^2$
• $\mathrm{Cl}-\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^5$
• $\mathrm{S}-\left[\mathrm{Ne}\right]3{\mathrm{s}}^{2}3{\mathrm{p}}^4$

So, among the given electronic configuration we can see that only the valence shell of Tin ($\mathrm{Sn}$) belongs to ‘n=5’, whereas for the others the valence shell is ‘n=3’.

Thus, the odd one out is ${\mathrm{SnCl}}_4$ among the given options.

Hence, the correct option is (B).