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  • Use the molecular orbital energy level diagram to show that N_{2} would be expected to have a triple bond, F_{2} a single bond and Ne_{2} no bond.

Use the molecular orbital energy level diagram to show that N_{2} would be expected to have a triple bond, F_{2} a single bond and Ne_{2} no bond.

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The general sequence of the energy level of molecular orbitals for nitrogen is:

\sigma 1s<\sigma ^{*}1s<\sigma 2s<\sigma ^{*}2s<\pi 2p_{x}=\pi 2p_{y}<\sigma 2p_{z}

The molecular orbitals in the sequence of their energy levels for the given molecules have hereby been described below:

N_{2}\sigma 1s^{2}\; \sigma^{*} 1s^{2}\; \sigma 2s^{2}\sigma^{*} 2s^{2}\pi 2p{_{x}}^{2}=\pi 2p{_{y}}^{2}\sigma 2p{_{z}}^{2}

Now, we know that the,

Bond order for N_{2}=\frac{8-2}{2}=3

Now, as the bond order is 3, it means that N_{2} will have a triple bond.

The molecular orbital of Fluorine has been described below: -

F_{2}= \sigma 1s^{2}, \sigma ^{*} 1s^{2}, \sigma 2s^{2}, \sigma ^{*} 2s^{2}, \sigma \; 2px^{2}, \pi 2px^{2} = \pi 2py^{2}

The bond order of F_{2}=\frac{10-8}{2}=1

Thus, we can imply that when the bond order is 1, the number of bonds must also be 1.

Ne_{2}= \sigma 1s^{2}, \sigma ^{*} 1s^{2}, \sigma 2s^{2}, \sigma ^{*} 2s^{2}, \sigma \; 2px^{2}, \pi 2px^{2} = \pi 2py^{2},\pi ^{*}2p{_{x}}^{2}=\pi ^{*}2p{_{y}}^{2}

Now, we can calculate the bond order of Ne_{2}=\frac{10-10}{2}=0

Thus, Ne_{2} has no bond.

Posted by

infoexpert22

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