To identify the anion 'X', we need to determine the total number of electrons from its given molecular orbital (MO) electronic configuration and then match it with the options provided.

Step 1: Count the total number of electrons in the given MO configuration.
The configuration is $\mathrm{KK}{\sigma }^{\ast }2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\pi 2{\mathrm{p}}_{\mathrm{x}}^2\pi 2{\mathrm{p}}_{\mathrm{y}}^2\sigma 2{\mathrm{p}}_{\mathrm{z}}^2{\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}{}^1$.
The 'KK' represents the filled K shells (1s orbitals) of both atoms, which account for 4 electrons (2 from each atom).
Number of electrons from the given configuration: $2(\text{from }{\sigma }^{\ast }2\mathrm{s})+2(\text{from }\sigma 2\mathrm{s})+2(\text{from }\pi 2{\mathrm{p}}_{\mathrm{x}})+2(\text{from }\pi 2{\mathrm{p}}_{\mathrm{y}})+2(\text{from }\sigma 2{\mathrm{p}}_{\mathrm{z}})+1(\text{from }{\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}})=11$ electrons.
Total electrons = $4(\text{from KK})+11=15$ electrons.

Step 2: Determine the number of electrons for each option.
A) ${\mathrm{N}}_2^{-}$: Nitrogen (N) has 7 electrons. So, ${\mathrm{N}}_2$ has $2\times 7=14$ electrons. For ${\mathrm{N}}_2^{-}$, total electrons = $14+1=15$ electrons.
B) ${\mathrm{O}}_2^{-}$: Oxygen (O) has 8 electrons. So, ${\mathrm{O}}_2$ has $2\times 8=16$ electrons. For ${\mathrm{O}}_2^{-}$, total electrons = $16+1=17$ electrons.
C) ${\mathrm{N}}_2^{2-}$: For ${\mathrm{N}}_2^{2-}$, total electrons = $14+2=16$ electrons.
D) ${\mathrm{O}}_2^{2-}$: For ${\mathrm{O}}_2^{2-}$, total electrons = $16+2=18$ electrons.

Step 3: Match the total number of electrons.
The anion 'X' has 15 electrons, which matches the total number of electrons in ${\mathrm{N}}_2^{-}$.

Step 4: Verify the MO configuration for ${\mathrm{N}}_2^{-}$.
For ${\mathrm{N}}_2^{-}$, with 15 electrons, the MO configuration follows the order for molecules with $\le 14$ electrons for the $2p$ orbitals (where $\pi 2p$ are lower in energy than $\sigma 2p$).
The correct order for ${\mathrm{N}}_2^{-}$ (15 electrons) is: $\mathrm{KK}\sigma 2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\pi 2{\mathrm{p}}_{\mathrm{x}}^2\pi 2{\mathrm{p}}_{\mathrm{y}}^2\sigma 2{\mathrm{p}}_{\mathrm{z}}^2{\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}{}^1$.
The given configuration is $\mathrm{KK}{\sigma }^{\ast }2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\pi 2{\mathrm{p}}_{\mathrm{x}}^2\pi 2{\mathrm{p}}_{\mathrm{y}}^2\sigma 2{\mathrm{p}}_{\mathrm{z}}^2{\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}{}^1$. There seems to be a typo in the question's provided configuration, as it lists ${\sigma }^{\ast }2{\mathrm{s}}^2$ twice. Assuming the first ${\sigma }^{\ast }2{\mathrm{s}}^2$ is actually $\sigma 2{\mathrm{s}}^2$, the configuration becomes $\mathrm{KK}\sigma 2{\mathrm{s}}^2{\sigma }^{\ast }2{\mathrm{s}}^2\pi 2{\mathrm{p}}_{\mathrm{x}}^2\pi 2{\mathrm{p}}_{\mathrm{y}}^2\sigma 2{\mathrm{p}}_{\mathrm{z}}^2{\pi }^{\ast }2{\mathrm{p}}_{\mathrm{x}}{}^1$. This configuration correctly accounts for 15 electrons and matches the expected MO configuration for ${\mathrm{N}}_2^{-}$.

Option Analysis:
A) ${\mathrm{N}}_2^{-}$: Total 15 electrons. The MO configuration matches (assuming the typo correction). This is the correct answer.
B) ${\mathrm{O}}_2^{-}$: Total 17 electrons. The MO configuration would have more electrons in the ${\pi }^{\ast }2p$ orbitals.
C) ${\mathrm{N}}_2^{2-}$: Total 16 electrons. The MO configuration would have one more electron in the ${\pi }^{\ast }2p$ orbital compared to the given configuration.
D) ${\mathrm{O}}_2^{2-}$: Total 18 electrons. The MO configuration would have all ${\pi }^{\ast }2p$ orbitals filled.

Correct Answer: (A)