JEE MAIN - Chemistry (2020 - 7th January Evening Slot - No. 12)
The bond order and the magnetic characteristics of CN-
are :
3, paramagnetic
$$2{1 \over 2}$$, paramagnetic
3, diamagnetic
$$2{1 \over 2}$$, diamagnetic
Explanation
CN- has 14 electrons.
Moleculer orbital configuration of CN- is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
Here is no unpaired electron so it is diamagnetic.
Nb = 10
Na = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
Moleculer orbital configuration of CN- is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
Here is no unpaired electron so it is diamagnetic.
Nb = 10
Na = 4
$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
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