Question

In which pair the two species are isoelectronic
a) $Ne,F^{-}$
b) $Ar,B{e}^{2+}$
c) $K,F^{-}$
d) ${S^2}^{-},He$

Answer 1

Hint: Isoelectronic are those atoms or ions which entails the same number of electrons. Atomic number describes the number of electrons present in the shell of atoms. Besides, atomic number also represents the number of protons present in the atomic nucleus.

Complete answer:
Atomic number describes the number of electrons present in the shell of atoms. Atomic numbers for Neon and fluorine are $10$ and $9$ , respectively. Therefore, $Ne$ has $10$ electrons while $F$ has $9$ electrons. In the formation of anion, a gain of electrons takes place. Therefore, $F^{-}$ has 10 electrons.
Hence, $Ne$ and $F^{-}$ species are isoelectronic and option (a) is correct.
The atomic numbers of $Ar$ and $Be$ are $18$ and $4$ , respectively. Therefore, $Ar$ has $18$ electrons. In the formation of cation, loss of electrons takes place. Therefore, $B{e}^{2+}$ has $2$ . Since $Ar$ and $B{e}^{2+}$ don't have the same number of electrons hence are not isoelectronic.
Therefore option (b) is incorrect.
The atomic numbers of $K$ and $F$ are $19$ and $9$ , respectively. Therefore, $K$ has 19 electrons while $F^{-}$ has 10 electrons. Since, these two species don't have the same number of electrons hence are not isoelectronic.
Therefore option (c) is incorrect.
The atomic number of $S$ and $He$ are $16$ and $2$ , respectively. Therefore, $S^{2-}$ has $18$ electrons while $He$ has $2$ . Since, these two species don't have the same number of electrons hence are not isoelectronic.
Therefore option (d) is incorrect.
From the overall discussion, we conclude that option (a) is correct.

Note:
It is important to note that atomic number describes the number of electrons present in the shell of atoms. $Ne$ and $F^{-}$ species have the same number of electrons. Therefore, option (a) is correct.