Question
Of the following sets which one does NOT contain isoelectronic species?
A.
$$BO_3^{3 - },CO_3^{2 - },NO_3^ - $$
B.
$$SO_3^{2 - },CO_3^{2 - },NO_3^ - $$
C.
$$C{N^ - },{N_2},C_2^{2 - }$$
D.
$$PO_4^{3 - },SO_4^{2 - },CIO_4^ - $$
Answer :
$$SO_3^{2 - },CO_3^{2 - },NO_3^ - $$
Solution :
Calculating number of electrons
\[\begin{gathered}
1.\,\,\,\left. \begin{gathered}
BO_3^{3 - } \to 5 + 8 \times 3 + 3 = 32 \hfill \\
CO_3^{2 - } \to 6 + 8 \times 3 + 2 = 32 \hfill \\
NO_3^ - \to 7 + 8 \times 3 + 1 = 32 \hfill \\
\end{gathered} \right\}\,\,{\text{iso - electronic }}\,{\text{species}} \hfill \\
\hfill \\
2.\,\,\,\,\left. \begin{gathered}
SO_3^{2 - } \to 16 + 8 \times 3 + 2 = 42 \hfill \\
CO_3^{2 - } \to 32 \hfill \\
NO_3^ - \to 32 \hfill \\
\end{gathered} \right\}\,\,{\text{not iso - electronic}}\,{\text{ species}} \hfill \\
\hfill \\
{\text{3}}{\text{.}}\,\,\,\,\left. \begin{gathered}
C{N^{2 - }} \to 6 + 7 + 1 = 14 \hfill \\
{N_2} \to 7 \times 2 = 14 \hfill \\
C_2^{2 - } \to 6 \times 2 + 2 = 14 \hfill \\
\end{gathered} \right\}\,\,{\text{iso - electronic}}\,{\text{ species}} \hfill \\
\hfill \\
4.\,\,\,\,\left. \begin{gathered}
PO_4^{3 - } \to 15 + 8 \times 4 + 3 = 50 \hfill \\
SO_4^{2 - } \to 16 \times 4 + 8 + 2 = 50 \hfill \\
CLO_4^ - \to 17 + 8 \times 4 + 1 = 50 \hfill \\
\end{gathered} \right\}\,\,{\text{iso - electronic }}\,{\text{species}} \hfill \\
\end{gathered} \]
Hence the species in option (B) are not isoelectronic.