JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 12)
Among the following, the molecule expected
to be stabilized by anion formation is :
C2, O2, NO, F2
C2, O2, NO, F2
C2
NO
O2
F2
Explanation
C2 has 12 electrons.
Moleculer orbital configuration of C2 is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}$$
Bond order of C2 = $${{8 - 4} \over 2}$$ = 2
C2- has 13 electrons.
Moleculer orbital configuration of C2- is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}$$ $${\sigma _{2p_z^1}}$$
Bond order of C2- = $${{9 - 4} \over 2}$$ = 2.5
As we know, higher is the bond order more will be the stability of the molecule.
So we can say, stability of C2 increases when it forms C2- ion.
Moleculer orbital configuration of C2 is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}$$
Bond order of C2 = $${{8 - 4} \over 2}$$ = 2
C2- has 13 electrons.
Moleculer orbital configuration of C2- is
$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}$$ $${\sigma _{2p_z^1}}$$
Bond order of C2- = $${{9 - 4} \over 2}$$ = 2.5
As we know, higher is the bond order more will be the stability of the molecule.
So we can say, stability of C2 increases when it forms C2- ion.
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