The isoelectronic set of ions is:

Question

Quantrex Academy · Accepted Answer

Correct answer: C. N ^ 3 - , O ^ 2 - , F ^ - and N a ^ +

In Isoelectronic species, number of electrons should be same
Number of electrons present in

{\text{Li}}^{\text{+}}\text{= 3} - \text{1 = 2}

Number of electrons present in

{\text{Na}}^{\text{+}}\text{= 11} - \text{1 =}\text{10}

(A)

{F}^{-} = 9 + 1 = 10{e}^{-}

L{i}^{+} = 3 - 1 = 2{e}^{-}

N{a}^{+} = 11 - 1 = 10{e}^{-}

M{g}^{+ 2} = 12 - 2 = 10{e}^{-}

(B)

{N}^{- 3} = 7 + 3 = 10{e}^{-}

L{i}^{+} = 3 - 1 = 2{e}^{-}

M{g}^{+ 2} = 12 - 2 = 10{e}^{-}

N{a}^{+} = 11 - 1 = 10{e}^{-}

(C)

{\text{F}}^{-}\text{= 9+1 = 10}

{\text{O}}^{\text{2} -}\text{= 8+2 = 10}

{\text{N}}^{\text{3} -}\text{= 7+3 = 10}

M{g}^{2 +} = 12 - 2 = 10

(d)

L{i}^{+} = 3 - 1 = 2{e}^{-}

N{a}^{+} = 11 - 1 = 10{e}^{-}

{O}^{- 2} = 8 + 2 = 10{e}^{-}

{F}^{-} = 9 + 1 = 10{e}^{-}

Hence,

{N}^{3 -}, {O}^{2 -}, {F}^{-}, N{a}^{+}

are isoelectronic.

Options