According to Molecular Orbital Theory,

Question

Aicharya · Accepted Answer

A] The molecular orbital energy configuration of

C_{2}^{2 -}

is

σ_{1 s}^{2}, σ_{1 s}^{* 2}, σ_{2 s}^{2}, σ_{2 s}^{* 2}, π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2}, σ_{2 p_{z}}^{2}

In the MO of

C_{2}^{2 -}

there is no unpaired electron hence it is diamagnetic
B] Bond order of

O_{2}^{2 +}

is 3 and

O_{2}

is 2 therefore bond length of

O_{2}

is greater than

O_{2}^{2 +}

C] The molecular orbital energy configuration of

N_{2}^{+}

is

σ_{1 s}^{2}, σ_{1 s}^{* 2}, σ_{2 s}^{2}, σ_{2 s}^{* 2}, π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2}, σ_{2 p_{z}}^{1}

Bond order of

N_{2}^{+} = \frac{1}{2} (9 - 4) = 2.5

The molecular orbital energy configuration of

N_{2}^{-}

is

σ_{1 s}^{2}, σ_{1 s}^{* 2}, σ_{2 s}^{2}, σ_{2 s}^{* 2}, π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2}, σ_{2 p_{z}}^{2}, π_{2 p_{x}}^{* 1} = π_{2 p_{y}}^{*}

Bond order of

N_{2}^{-} = \frac{1}{2} (10 - 5) = 2.5

D]

H e_{2}^{+}

has less energy as compare to two isolated He atoms

Solution