JEE MAIN - Chemistry (2004 - No. 63)
The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species?
Bond length in NO+ is greater than in NO
Bond length is unpredictable
Bond length in NO+ in equal to that in NO
Bond length in NO is greater than in NO+
Explanation
Molecular orbital configuration of NO (15 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
Similarly Molecular orbital configuration of NO+ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
Note :
(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
So, bond length in NO > NO+
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 5
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
Similarly Molecular orbital configuration of NO+ (14 electrons) is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
$$\therefore\,\,\,\,$$ Nb = 10
Na = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
Note :
(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
So, bond length in NO > NO+
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