CHEM_REMOVED - CHEMICAL BONDING Question with Solution | TestHub

Bond order $=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}$

${\mathrm{N}}_{\mathrm{B}}=$ Number of electrons in bonding molecular orbitals.

${\mathrm{N}}_{\mathrm{A}·\mathrm{B}}=$ Number of electrons in anti bonding molecular orbitals.

${\mathrm{C}}_2^{2-}$ has $14$ electrons, it is isoelectronic with ${\mathrm{N}}_2$ where Bond order $=3$,

${\mathrm{N}}_2^{2-}$ has $16$ electrons is isoelectronic with ${\mathrm{O}}_2$ whose bond order $=2$

${\mathrm{O}}_2^{2-}$ has $18$ electrons and is isoelectronic with ${\mathrm{F}}_2$ whose bond order $=1$

So, the correct order of Bond order is

${\mathrm{C}}_2^{2-}>{\mathrm{N}}_2^{2-}>{\mathrm{O}}_2^{2-}$.