The energy of \sigma 2p_{z} molecular orbital is greater than \pi 2p_{x} and \pi 2p_{y} molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species :

N_{2},N_{2}^{+},N_{2}^{-},N{_{2}}^{2+}

Answers (1)

We know that the general sequence of the energy level of the molecular orbital can be written as,

\sigma 1s<\sigma ^{*}1s<\sigma 2s<\sigma ^{*}2s<\pi 2px=\pi 2py<\sigma 2pz

N2\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y<\sigma 2p^{2}z

N2^{+}\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y<\sigma 2p^{1}z

N2^{-}\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y\sigma 2p^{2}z\sigma 2p^{2}x

N{_{2}}^{2+}\; \sigma 1s^{2}<\sigma ^{*}1s^{2}<\sigma 2s^{2}<\sigma ^{*}2s^{2}<\pi 2p^{2}x=\pi 2p^{2}y

We are also aware that, Bond order = ½ [electrons in BMO – electrons in ABMO]

In case of N_{2}=10-\frac{4}{2}=3

Hence, the bond order for N{_{2}}^{+}=9-\frac{4}{2}=2.5

Hence, the bond order for N{_{2}}^{-}=10-\frac{5}{2}=2.5

Hence, the bond order for N{_{2}}^{2+}=8-\frac{4}{2}=2

Therefore, we can conclude that the order of stability is:

N2>N{_{2}}^{-}>N{_{2}}^{+}>N{_{2}}^{2+}

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