JEE Main 2018 — Chemical Bonding And Molecular Structure Question with Solution

Note :

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

(a) Configuration of $$H{e}_2^{2+}$$ (2 electrons) is = $${\sigma }_{1s^2}$$

$$\therefore$$ Bond order = $$\frac{1}{2}$$ (2 $$-$$0) = 1

(b) Configuration of $$H{e}_2^{+}$$ (3 electrons) is = $${\sigma }_{1s^2}$$ $${\sigma }_{1s^1}^{\ast }$$

$$\therefore$$ Bond order = $$\frac{1}{2}$$ (2 $$-$$1) = 0.5

(c) Configuration of $$H_2^{-}$$ (3 electrons) is = $${\sigma }_{1s^2}$$ $${\sigma }_{1s^1}^{\ast }$$

$$\therefore$$ Bond order = $$\frac{1}{2}$$ (2 $$-$$ 1) = 0.5

(d) $$\therefore$$ Configuration of $$H{e}_2^{2-}$$ is = $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }$$

$$\therefore$$ Bond order = $$=\frac{1}{2}$$ (2 $$-$$ 2) = 0

$$\therefore$$ $$H_2^{2-}$$ is not a viable molecule.