Q44 / 170 · Manipal (MET) 2015 · Chemistry
The isoelectronic pair is
- Correct answerA$\mathrm{Cl}_2 \mathrm{O}, \mathrm{ICl}_2^{-}$
- B$\mathrm{Cl}_2^{-}, \mathrm{ClO}_2$
- C$\mathrm{IF}_2^{+}, \mathrm{I}_3^{-}$
- D$\mathrm{ClO}_2^{-}, \mathrm{Cl}_2^{+}$
Solution
$\begin{aligned} & \mathrm{Cl}_2 \mathrm{O}=2 \times 17+16=42 \text { electrons } \\ & \mathrm{ICl}_2^{-}=53+2 \times 17+1=88 \text { electrons } \\ & \mathrm{Cl}_2^{-}=2 \times 17+1=35 \text { electrons }\end{aligned}$
$\begin{aligned}
& \mathrm{IF}_2^{+}=53+2 \times 9-1=70 \text { electrons } \\
& \mathrm{ClO}_2=17+2 \times 8=33 \text { electrons } \\
& \mathrm{ClO}_2=17+2 \times 8+1=33 \text { electrons } \\
& \mathrm{ClO}_2^{-}=17+2 \times 8+1=34 \text { electrons } \\
& \mathrm{Cl}_2^{+}=17+2 \times 9-1=34 \text { electrons }
\end{aligned}$
$\mathrm{ClO}_2^{-}$and $\mathrm{ClF}_2^{+}$contain 34 electrons each.
Hence they are isoelectronic.
Hence, the correct answer is option: 'A'.