(i) (a) ${\mathrm{H}}^{-}$ ion
The electronic configuration of the 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 ${\mathrm{H}}^{-}$ = $1{\mathrm{s}}^2$

(b) ${\mathrm{Na}}^{+}$ ion
The electronic configuration of the Na atom is $1s^{2}2s^{2}2p^{6}3s^1$.
A positive charge on the species indicates the loss of an electron by it.
$\therefore$ Electronic configuration of $N{a}^{+}=1s^{2}2s^{2}2p^{6}3s^0\text{\hspace{0.17em}or\hspace{0.17em}}1s^{2}2s^{2}2p^6$

(c) ${\mathrm{O}}^{2-}$ ion
The electronic configuration of 0 atom is $1s^{2}2s^{2}2p^4$.
A dinegative charge on the species indicates that two electrons are gained by it.

$\therefore$ Electronic configuration of $O^{2-}\text{\hspace{0.17em}ion\hspace{0.17em}}=1s^{2}2s^2{p}^6$

(d) ${\mathrm{F}}^{-}$ion
The electronic configuration of F atom is $1s^{2}2s^{2}2p^5$
A negative charge on the species indicates the gain of an electron by it.
$\therefore$ Electron configuration of $F^{-}\text{\hspace{0.17em}ion\hspace{0.17em}}=1s^{2}2s^{2}2p^6$

(ii) (a) $3{\mathrm{s}}^1$
Completing the electron configuration of the element as $1{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{p}}^6<mspace\; width="1em"></mspace>3{\mathrm{s}}^1$.
$\therefore$ Number of electrons present in the atom of the element
= 2 + 2 + 6 + 1 = 11
Atomic number of the element = 11

(b) $2{\mathrm{p}}^3$
Completing the electron configuration of the element as $1{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{p}}^3$.
$\therefore$ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7
$\therefore$ The atomic number of the element = 7

(c) $3{\mathrm{p}}^5$
Completing the electron configuration of the element as $1{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{p}}^5$.
$\therefore$ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9
$\therefore$ The atomic number of the element = 9

(iii) (a) [He] $2{\mathrm{s}}^1$
The electronic configuration of the element is [He] $2{\mathrm{s}}^1<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{s}}^1$.
$\therefore$ The atomic number of the element = 3

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

(b) [Ne] $3{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{p}}^3$
The electronic configuration of the element is [Ne] $3{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{p}}^3=<mspace\; width="1em"></mspace>1{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{p}}^6<mspace\; width="1em"></mspace>3{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{p}}^3$.
$\therefore$ The atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] $3{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{p}}^3$ is phosphorus (P).

(c) [Ar] $4{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{d}}^1$
The electronic configuration of the element is [Ar] $4{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{d}}^1=<mspace\; width="1em"></mspace>1{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{s}}^2<mspace\; width="1em"></mspace>2{\mathrm{p}}^6<mspace\; width="1em"></mspace>3{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{p}}^6<mspace\; width="1em"></mspace>4{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{d}}^1$.
The atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] $4{\mathrm{s}}^2<mspace\; width="1em"></mspace>3{\mathrm{d}}^1$ is scandium (Sc).