JEE MAIN - Chemistry (2013 (Offline) - No. 18)
Stability of the species Li2, $$Li_2^−$$ and $$Li_2^+$$ increases in the order of:
Li2 < $$Li_2^+$$ < $$Li_2^-$$
$$Li_2^+$$ < $$Li_2^-$$ < Li2
$$Li_2^-$$ <$$Li_2^+$$ < Li2
$$Li_2^-$$ < Li2 < $$Li_2^+$$
Explanation
Li2 = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}} \,$$
$$ \therefore $$ Bond order = $${1 \over 2}\left( {4 - 2} \right)$$ = 1
$$Li_2^+$$ = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^1}}} \,$$
$$ \therefore $$ Bond order = $${1 \over 2}\left( {3 - 2} \right)$$ = 0.5
$$Li_2^-$$ = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}} \,\,\sigma _{1{s^1}}^ * \,$$
$$ \therefore $$ Bond order = $${1 \over 2}\left( {4 - 3} \right)$$ = 0.5
The bond order of $$Li_2^+$$ and $$Li_2^-$$ is same but $$Li_2^+$$ is more stable than $$Li_2^-$$ because $$Li_2^+$$ is smaller in size and has 2 electrons in antibonding orbitals whereas $$Li_2^-$$ has 3 electrons in antibonding orbitals. Hence $$Li_2^+$$ is more stable than $$Li_2^-$$.
$$ \therefore $$ Bond order = $${1 \over 2}\left( {4 - 2} \right)$$ = 1
$$Li_2^+$$ = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^1}}} \,$$
$$ \therefore $$ Bond order = $${1 \over 2}\left( {3 - 2} \right)$$ = 0.5
$$Li_2^-$$ = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}} \,\,\sigma _{1{s^1}}^ * \,$$
$$ \therefore $$ Bond order = $${1 \over 2}\left( {4 - 3} \right)$$ = 0.5
The bond order of $$Li_2^+$$ and $$Li_2^-$$ is same but $$Li_2^+$$ is more stable than $$Li_2^-$$ because $$Li_2^+$$ is smaller in size and has 2 electrons in antibonding orbitals whereas $$Li_2^-$$ has 3 electrons in antibonding orbitals. Hence $$Li_2^+$$ is more stable than $$Li_2^-$$.
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