N_2 molecule has a greater bond dissociation energy than N^+_2 ion whereas O_2 molecule has a lower bond dissociation energy than O^+_2 ion. Explain in terms of molecular orbital theory.

N_2(14):( )^2( *ls)^2( 2S)^2( *2s)^2( 2p_x^2= 2p^2_x)( 2p_z)^2; B.O.= 12(10-4)=3 N^+_2(13):( )^2( *ls)^2( 2s)^2( 2p_x^2= 2p_x^2)( 2p_z)^1 B.O.=1 2 (9-4)= 52 Since N_2 has higher bond order than N_2^+, therefore N_2 has higher bond dissociation energy than N^+_2. O_2(16):( )^2( *ls)^2( 2s)^2( *2s)^2( 2p_z)^2( 2p_x= 2p_y^2)( *2p_x^1= *2p_y^1) B.O.= 12(10-6)= 42=2 O^+_2(15):( )^2( *ls)^2( 2s)^2( *2s)^2( 2p_z)^2 ( 2p_x^2= ^2_y ( *2p^1_x) B.O.=1 2 (10-5)= 52 O^+_2 has higher bond order than O_2, therefore, it has high bond dissociation energy than O_2.

N_2 molecule has a greater bond dissociation energy than N^+_2 io…

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$\text{N}_2$ molecule has a greater bond dissociation energy than $\text{N}^+_2$ ion whereas $\text{O}_2$ molecule has a lower bond dissociation energy than $\text{O}^+_2$ ion. Explain in terms of molecular orbital theory.

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$\text{N}_2(14):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{S})^2(\sigma*\text{2s})^2(\pi2\text{p}_\text{x}^2=\pi2\text{p}^2_\text{x})(\sigma2\text{p}_\text{z})^2;$
$\text{B.O.}=\frac12(10-4)=3$
$\text{N}^+_2(13):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\pi2\text{p}_\text{x}^2=\pi2\text{p}_\text{x}^2)(\sigma2\text{p}_\text{z})^1$
$\text{B.O.}=\frac{1}{2}(9-4)=\frac52$
Since $\text{N}_2$ has higher bond order than $\text{N}_2^+,$ therefore $\text{N}_2$ has higher bond dissociation energy than $\text{N}^+_2.$
$\text{O}_2(16):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\sigma*2\text{s})^2(\sigma2\text{p}_\text{z})^2(\pi2\text{p}_\text{x}=\pi2\text{p}_\text{y}^2)(\pi*2\text{p}_\text{x}^1=\pi*2\text{p}_\text{y}^1)$
$\text{B.O.}=\frac12(10-6)=\frac42=2$
$\text{O}^+_2(15):(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\sigma*\text{2s})^2(\sigma2\text{p}_\text{z})^2{(\pi2\text{p}_\text{x}^2=\pi\text{p}^2_\text{y}}(\pi*2\text{p}^1_\text{x})$
$\text{B.O.}=\frac{1}{2}(10-5)=\frac52$
$\text{O}^+_2$ has higher bond order than $\text{O}_2$, therefore, it has high bond dissociation energy than $\text{O}_2$.

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