Question

Bond order in no+ is greater than in no molecule why

Answer 1

Your NO+ diagram will have one less valence electron and your NO- diagram will have one more valence electron. The stronger bonds will have the higher bond order, greater stability, higher energy and shorter bond. ... This gives it greater energy because the molecule needs a larger energy than if it were packed loosely.

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Answer 2

The bond order of $NO^{+}$ is greater than NO because the number of bonding electrons are more than the non-bonding electrons.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration of $NO^{+}$ and NO will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 7 electrons present in nitrogen and 8 electrons in oxygen.

(a) The number of electrons present in $NO^{+}$ molecule = 7 + 8 - 1 = 14

The molecular orbital configuration of $NO^{+}$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The formula of bonding order = $\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})$

The bonding order of $NO^{+}$ = $\frac{1}{2}\times (10-4)=3$

(b) The number of electrons present in NO molecule = 7 + 8 = 15

The molecular orbital configuration of NO molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],[(\pi_{2p_x}^*)^1=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The bonding order of NO = $\frac{1}{2}\times (10-5)=2.5$

From this we conclude that the bond order of $NO^{+}$ is greater than NO because the number of bonding electrons are more than the non-bonding electrons.

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