$$\,$$ Bond order $$ = {1 \over 2}$$ [N
b $$-$$ N
a]
N
b = No of electrons in bonding molecular orbital
N
a $$=$$ No of electrons in anti bonding molecular orbital
In O atom 8 electrons present, so in O
2, 8 $$ \times $$ 2 = 16 electrons present.
Then in $$O_2^ + $$ no of electrons = 15
in $$O_2^ - $$ no of electrons = 17
in $$O_2^{2 + }$$ no of electrons = 14
$$\therefore\,\,\,\,$$ Molecular orbital configuration of O
2 (16 electrons) is
$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
$$\therefore\,\,\,\,$$N
a = 6
N
b = 10
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ N
b = 10
N
a = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
$$\therefore\,\,\,\,$$ N
b = 10
N
a = 7
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
Molecular orbital configuration of O $$_2^{2 + }$$ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^0}^ * $$
$$\therefore\,\,\,\,$$ N
b = 10
N
a = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 4] = 3
|
$$O_2^-$$ |
$$O_2$$ |
$$O_2^+$$ |
$$O_2^{2+}$$ |
| B.O. : |
1.5 |
2.0 |
2.5 |
3.0 |