Question

Which of the following does not contain isoelectronic species?

• A. $P{O}_4^{3-}$, $S{O}_4^{2-}$, $Cl{O}_4^{-}$
• B. $C{N}^{-}$, $N_2$, $C_2^{2-}$
• C. $S{O}_3^{2-}$, $C{O}_3^{2-}$, $N{O}_3^{-}$
• D. $B{O}_3^{3-}$, $C{O}_3^{2-}$, $N{O}_3^{-}$

Answer 1

The correct option is C $S{O}_3^{2-}$, $C{O}_3^{2-}$, $N{O}_3^{-}$

Isoelectronic species are those which contain the same number of electrons. It is obtained by adding $6$ electrons for carbon, $7$ electrons for nitrogen, $8$ electrons for oxygen, $5$ electrons for boron, $16$ electrons for sulphur, $15$ electrons for phosphorus, $17$ electrons for chlorine, $+1$ for negative charge and $-1$ for positive charge.

$S{O}_3^{2-}$, $C{O}_3^{2-}$, $N{O}_3^{-}$ contains different electrons . Hence, they are not isoelctronic.

$S{O}_3^{2-}$ has $42$ electrons, $C{O}_3^{2-}$ has $32$ electrons. $N{O}_3^{-}$ has $32$ electrons.

$C$ and $N$ belongs to the second period whereas $S$ belongs to the third period. Hence, they cannot form isoelectronic species.