Give the electron configuration,magnetic property,bond order,and energy diagram for the Nitrogen $(N_{2})$ molecule.
(N/A) For $N_{2}$ $(Z=7)$: The electronic configuration of $N$ is $1s^{2} 2s^{2} 2p^{3}$.
Total electrons in $N_{2} = 14$.
The molecular orbital $(MO)$ configuration for $N_{2}$ is:
$KK(\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\pi 2p_{x})^{2} = (\pi 2p_{y})^{2} (\sigma 2p_{z})^{2}$
Bond order $(BO)$ $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 4) = 3$.
This indicates a triple bond in $N_{2}$.
Magnetic Property: Since all electrons are paired,$N_{2}$ is diamagnetic.
The energy diagram is provided below.
Total electrons in $N_{2} = 14$.
The molecular orbital $(MO)$ configuration for $N_{2}$ is:
$KK(\sigma 2s)^{2} (\sigma^{*} 2s)^{2} (\pi 2p_{x})^{2} = (\pi 2p_{y})^{2} (\sigma 2p_{z})^{2}$
Bond order $(BO)$ $= \frac{1}{2} (N_{b} - N_{a}) = \frac{1}{2} (10 - 4) = 3$.
This indicates a triple bond in $N_{2}$.
Magnetic Property: Since all electrons are paired,$N_{2}$ is diamagnetic.
The energy diagram is provided below.
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$III$) Bond order can assume any value including zero up to four
$IV$) $NO_3^-$ and $BO_3^{3-}$ have the same bond order for the $X-O$ bond (where $X$ is the central atom)
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