JEE MAIN - Chemistry (2018 (Offline) - No. 25)
According to molecular orbital theory, which of the following will not be a viable molecule?
$${\rm H}e_2^{2 + }$$
$${\rm H}e_2^{ + }$$
$${\rm H}_2^{- }$$
$${\rm H}_2^{2 - }$$
Explanation
Note :
According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
(a)$$\,\,\,$$ Configuration of $$He_2^{2 + }$$ (2 electrons) is = $${\sigma _{1{s^2}}}$$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$0) = 1
(b)$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5
(c) $$\,\,\,$$ Configuration of $$H_2^ - $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$
$$\therefore$$ Bond order = $${1 \over 2}$$ (2 $$-$$ 1) = 0.5
(d) $$\therefore\,\,\,$$ Configuration of $$He_2^{2 - }$$ is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * $$
$$\therefore\,\,\,$$ Bond order = $$ = {1 \over 2}$$ (2 $$-$$ 2) = 0
$$\therefore\,\,\,$$ $$H_2^{2 - }$$ is not a viable molecule.
According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
(a)$$\,\,\,$$ Configuration of $$He_2^{2 + }$$ (2 electrons) is = $${\sigma _{1{s^2}}}$$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$0) = 1
(b)$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$
$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5
(c) $$\,\,\,$$ Configuration of $$H_2^ - $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$
$$\therefore$$ Bond order = $${1 \over 2}$$ (2 $$-$$ 1) = 0.5
(d) $$\therefore\,\,\,$$ Configuration of $$He_2^{2 - }$$ is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * $$
$$\therefore\,\,\,$$ Bond order = $$ = {1 \over 2}$$ (2 $$-$$ 2) = 0
$$\therefore\,\,\,$$ $$H_2^{2 - }$$ is not a viable molecule.
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