Question

Stability of the species $L{i}_2,L{i}_2^{-}$and $L{i}_2^{+}$ increases in the order of

• A. $L{i}_2<L{i}_2^{-}<L{i}_2^{+}$
• B. $L{i}_2^{-}<L{i}_2<L{i}_2^{+}$
• C. $L{i}_2<L{i}_2^{+}<L{i}_2^{-}$
• D. $L{i}_2^{-}<L{i}_2^{+}<L{i}_2$

Answer 1

Answer: (D)

$L{i}_2^{-}<L{i}_2^{+}<L{i}_2$
Bond order $\propto$ stability of species
$L{i}_2=\sigma 1s^2<\sigma 1s^2<\sigma 2s^2;B.O=\frac{4-2}{2}=1$
$L{i}_2^{-}=\sigma 1s^2<\sigma 0s^2<\sigma 2s^2<\sigma 2s^1;B.0=\frac{4-3}{2}=0.5$
$L{i}_2^{+}=\sigma 1s^2<\sigma 0s^2<\sigma 2s^1;B.O=\frac{3-2}{2}=0.5$
$L{i}_2^{+}$ and $L{i}_2^{-}$ both have same bond order, but $L{i}_2^{-}$ has more number of electron in anti-bonding molecular orbitals (experience more inter electronic repulsion) and hence least stable among three species.