Question

how to find the bond order in NO2- and NO3- . pls explain.

Answer 1

In nitrite ion, NO2-, there are two equivalent resonance structures. Each N-O bond order in NO2- is 1.5. Similarly in nitrate ion, NO3-, there are three equivalent resonance structures.

hope this answer helps u

Answer 2

Answer : The bond order of $NO^{2-}$ and $NO^{3-}$ are, 1.5 and 1 respectively.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration of $NO^{2-}$ and  $NO^{3-}$ 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 $NO^{2-}$ molecule = 7 + 8 + 2 = 17

The molecular orbital configuration of $NO^{2-}$ molecule will be,

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

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

The bonding order of $NO^{2-}$ = $\frac{1}{2}\times (10-7)=1.5$

(b) The number of electrons present in $NO^{3-}$ molecule = 7 + 6 + 3 = 18

The molecular orbital configuration of $NO^{3-}$ molecule will be,

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

The bonding order of $NO^{3-}$ = $\frac{1}{2}\times (10-8)=1$

Hence, the bond order of $NO^{2-}$ and $NO^{3-}$ are, 1.5 and 1 respectively.