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Molecular Orbital Energy Diagrams

Which of the ...

Question

Which of the following statement is incorrect?

During

N_{2}^{+}

formation, one electron is removed from the bonding molecular orbital of

N_{2}

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During

O_{2}^{+}

formation, one electron is removed from the antibonding molecular orbital of

O_{2}

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During

O_{2}^{-}

formation, one electron is added to the bonding molecular orbital of

O_{2}

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During

N_{2}^{-}

formation, one electron is added to the antibonding molecular orbital of

N_{2}^{-}

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Solution

The correct option is C During $O_{2}^{-}$ formation, one electron is added to the bonding molecular orbital of $O_{2}$
$O_{2} : (σ 1 s)^{2} (σ^{} 1 s)^{2} (σ 2 s)^{2} (σ^{} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π^{} 2 p_{x}^{1} = π^{} 2 p_{y}^{1})$
During $O_{2}^{-}$ formation, 1 electron is added to the antibonding molecular orbital of $O_{2}$
$O_{2}^{-} : (σ 1 s)^{2} (σ^{} 1 s)^{2} (σ 2 s)^{2} (σ^{} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π^{} 2 p_{x}^{2} = π^{} 2 p_{y}^{1})$

$N_{2}$ : $(σ 1 s)^{2} (σ^{} 1 s)^{2} (σ 2 s)^{2} (σ^{} 2 s)^{2} (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (σ 2 p_{z})^{2}$
So, during formation of $N_{2}^{+}$ , one electron is removed from the bonding molecular orbital

And during $N_{2}^{-}$ molecular orbital formation, one elctron is added to the antibonding molecular orbital.

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