Question

Which of the following are the correct stability order for ${\mathrm{N}}_2$ and its given ions?

• A. ${\mathrm{N}}_2>{{\mathrm{N}}_2}^{+}={{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{2-}$
• B. ${{\mathrm{N}}_2}^{+}>{{\mathrm{N}}_2}^{-}>{\mathrm{N}}_2>{{\mathrm{N}}_2}^{2-}$
• C. ${{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{-}>{\mathrm{N}}_2>{{\mathrm{N}}_2}^{2-}$
• D. ${\mathrm{N}}_2>{{\mathrm{N}}_2}^{+}>{{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{2-}$

Answer 1

The correct option is D

${\mathrm{N}}_2>{{\mathrm{N}}_2}^{+}>{{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{2-}$

Explanation for the correct options:

(D) ${\mathrm{N}}_2>{{\mathrm{N}}_2}^{+}>{{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{2-}$

• Bond stability ∞ bond order.
• $\mathrm{N}$ have $7$ electrons.
• The electronic configuration of nitrogen is $1$s²$2$s²$2$p³.
• In case of ${\mathrm{N}}_2$, $2\mathrm{N}$ are bonded together and they get their stable configuration.
• In ${\mathrm{N}}_2$, $6$ electrons are there in unfilled orbitals and their bond order will be $\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$2=3$}\right.$.
• ${{\mathrm{N}}_2}^{+}$ have donated one electron so their electronic configuration is $1$s²$2$s²$2$p².
• ${{\mathrm{N}}_2}^{+}$ contain the total number of electrons is $5$ and its bond order will be $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=2.5$.
• ${{\mathrm{N}}_2}^{-}$ have accepted one electron so their electronic configuration is $1$s²$2$s²$2$p⁴.
• ${{\mathrm{N}}_2}^{-}$ contain the total number of electrons is $7$ and its bond order will be $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=2.5$.
• ${{\mathrm{N}}_2}^{2-}$ have accepted two-electron so their electronic configuration is $1$s²$2$s²$2$p⁵.
• ${{\mathrm{N}}_2}^{2-}$contain the total number of electrons is $8$ and its bond order will be $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=2$.
• ${{\mathrm{N}}_2}^{+}$ and ${{\mathrm{N}}_2}^{-}$ have the same bond order but later has one electron in antibonding orbital, which reduces its stability as compared to ${{\mathrm{N}}_2}^{+}$.

Explanation for the incorrect options:

(A) ${\mathrm{N}}_2>{{\mathrm{N}}_2}^{+}={{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{2-}$

• The bond order of ${{\mathrm{N}}_2}^{+}$ and ${{\mathrm{N}}_2}^{-}$is the same $=2.5$.
• But ${{\mathrm{N}}_2}^{-}$ has has one electron in the antibonding orbital, which reduces its stability as compared to ${{\mathrm{N}}_2}^{+}$.
• Hence, this option is not correct.

(B) ${{\mathrm{N}}_2}^{+}>{{\mathrm{N}}_2}^{-}>{\mathrm{N}}_2>{{\mathrm{N}}_2}^{2-}$

• The stability of the bond of ${{\mathrm{N}}_2}^{+}$ and ${{\mathrm{N}}_2}^{-}$never be more than ${\mathrm{N}}_2$.
• Because the bond order of ${\mathrm{N}}_2$is more than ${{\mathrm{N}}_2}^{+}$ and ${{\mathrm{N}}_2}^{-}$.

(C) ${{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{-}>{\mathrm{N}}_2>{{\mathrm{N}}_2}^{2-}$

• ${{\mathrm{N}}_2}^{2-}$is less stable than ${{\mathrm{N}}_2}^{-}$,${{\mathrm{N}}_2}^{+}$ and ${\mathrm{N}}_2$.
• So this order is incorrect.

Hence, option (D) is correct., ${\mathrm{N}}_2>{{\mathrm{N}}_2}^{+}>{{\mathrm{N}}_2}^{-}>{{\mathrm{N}}_2}^{2-}$ is the correct stability order for ${\mathrm{N}}_2$ and its given ions.