Write the electronic configurations of the following ions:

Solution:

(i) (a) $\mathbf{H}^{-}$ion

The electronic configuration of $\mathrm{H}$ atom is $1 \mathrm{~s}^{1}$.

A negative charge on the species indicates the gain of an electron by it

$\therefore$ Electronic configuration of $\mathrm{H}^{-}=1 s^{2}$

(b) $\mathrm{Na}^{+}$ion

The electronic configuration of Na atom is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$.

A positive charge on the species indicates the loss of an electron by it.

$\therefore$ Electronic configuration of $\mathrm{Na}^{+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{0}$ or $1 s^{2} 2 s^{2} 2 p^{6}$

(c) $\mathrm{O}^{2-}$ ion

The electronic configuration of 0 atom is $1 s^{2} 2 s^{2} 2 p^{4}$.

A dinegative charge on the species indicates that two electrons are gained by it

$\therefore$ Electronic configuration of $\mathrm{O}^{2-}$ ion $=1 s^{2} 2 s^{2} \mathrm{p}^{6}$

(d) $\mathrm{F}^{-}$ion

The electronic configuration of $\mathrm{F}$ atom is $1 s^{2} 2 s^{2} 2 p^{5}$.

A negative charge on the species indicates the gain of an electron by it.

$\therefore$ Electron configuration of $\mathrm{F}^{-}$ion $=1 s^{2} 2 s^{2} 2 p^{6}$

(ii) (a) $3 s^{1}$

Completing the electron configuration of the element as

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$

Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

Atomic number of the element = 11

(b) $2 p^{3}$

Completing the electron configuration of the element as

$1 s^{2} 2 s^{2} 2 p^{3}$

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c) $3 p^{5}$

Completing the electron configuration of the element as

$1 s^{2} 2 s^{2} 2 p^{5}$

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(iii) (a) [He] 2s $^{1}$

The electronic configuration of the element is [He] $2 s^{1}=1 s^{2} 2 s^{1}$.

Atomic number of the element = 3

Hence, the element with the electronic configuration [He] $2 s^{1}$ is lithium (Li).

(b) [Ne] $3 s^{2} 3 p^{3}$

The electronic configuration of the element is [Ne] $3 s^{2} 3 p^{3}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$.

Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] $3 s^{2} 3 p^{3}$ is phosphorus (P).

(c) $[\mathrm{Ar}] 4 s^{2} 3 d^{1}$

The electronic configuration of the element is [Ar] $4 s^{2} 3 d^{1}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{1} . \therefore$

Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] $4 s^{2} 3 d^{1}$ is scandium (Sc).