Higher the bond order, more is the stability. The bond order of the given species can be calculated as
(i) $\mathrm{N}_2^{-}=$Diatomic molecule

Number of electrons $=(7+7+1)=15 e^{-}$
i.e. B.O. = 2.5
(ii) $\mathrm{C}_2=$ Diatomic molecule

Number of electrons $=(6+6)=12 e^{-}$
i.e.
$$
\text { B.O. }=2.0
$$
(iii) $\mathrm{Ne}_2=$ Number of bonding and antibonding electrons are equal i.e. B.O. $=0$
$\therefore \mathrm{Ne}_2$ does not exist.
(iv) $\mathrm{O}_2^{2-}=$ Diatomic molecule

Number of electrons $=(8+8+2) e^{-}=18 e^{-}$ i.e. B.O. $=1.0$

Hence, correct sequence of B.O. is
Hence, correct sequence of B.O. is
$$
\underset{2.5}{\mathrm{~N}_2^{-}}>\underset{2.0}{\mathrm{C}_2}>\underset{1.0}{\mathrm{O}_2^{2-}}>\underset{0}{\mathrm{Ne}_2 .}
$$

Order of stability is $\mathrm{N}_2^{-}>\mathrm{C}_2>\mathrm{O}_2^{2-}>\mathrm{Ne}_2$.