Question

Which will show highest bond energy o2+ o2 o2- o22-?

Answer 1

Mate here is your answer ....

O2+ will show highest bond energy among other .

hope it helps you.

Answer 2

Answer : $O_2^{+}$ shows highest bond energy.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration 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 8 electrons in oxygen.

(a) The number of electrons present in $O_2$ molecule = 2(8) = 16

The molecular orbital configuration of $O_2$ molecule will be,

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

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

The bonding order of $O_2$ = $\frac{1}{2}\times (10-6)=2$

(b) The number of electrons present in $O_2^-$ molecule = 2(8) + 1 = 17

The molecular orbital configuration of $O_2^-$ 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 bond order = $\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})$

The bonding order of $O_2^-$ = $\frac{1}{2}\times (10-7)=1.5$

(c) The number of electrons present in $O_2^{2-}$ molecule = 2(8) + 2 = 18

The molecular orbital configuration of $O_2^{2-}$ molecule will be,

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

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

The bonding order of $O_2^{2-}$ = $\frac{1}{2}\times (10-8)=1$

(d) The number of electrons present in $O_2^+$ molecule = 2(8) -1 = 15

The molecular orbital configuration of $O_2^+$ molecule will be,

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

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

The bonding order of $O_2^+$ = $\frac{1}{2}\times (10-5)=2.5$

From this we conclude that the bond order from lowest to highest will be,

$O_2^{2-}<O_2^-<O_2<O_2^+$

As we know that, the higher the bond order, the higher will be bond energy. So, the order of bond energy will be:

$O_2^{2-}<O_2^-<O_2<O_2^+$

Hence, $O_2^{+}$ shows highest bond energy.