KCET · Chemistry · Chemical Bonding and Molecular Structure

The correct statement with regard to \(\mathrm{H}_{2}^{+}\)and \(\mathrm{H}_{2}^{-}\) is

A both \(\mathrm{H}_{2}^{+}\)and \(\mathrm{H}_{2}^{-}\)are equally stable
B both \(\mathrm{H}_{2}^{+}\)and \(\mathrm{H}_{2}^{-}\)do not exist
C \(\mathrm{H}_{2}^{-}\)is more stable than \(\mathrm{H}_{2}^{+}\)
D \(\mathrm{H}_{2}^{+}\)is more stable than \(\mathrm{H}_{2}^{-}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{H}_{2}^{+}\)is more stable than \(\mathrm{H}_{2}^{-}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{H}_{2}^{+}: \sigma 1 \mathrm{~s}^{1}, \stackrel{*}{\sigma} 1 \mathrm{~s}^{0}\)
Bond order \(=\frac{1}{2}\)
\(\mathrm{H}_{2}^{-}: \sigma 1 \mathrm{~s}^{2}, \stackrel{*}{\sigma} 1 \mathrm{~s}^{1}\)
Bond order \(=\frac{2-1}{2}=\frac{1}{2}\)
The bond order of \(\mathrm{H}_{2}^{+}\)and \(\mathrm{H}_{2}^{-}\)are same but \(\mathrm{H}_{2}^{+}\)is more stable than \(\mathrm{H}_{2}^{-}\). It is due to the presence of one electron in the antibonding molecular orbital in \(\mathrm{H}_{2}^{-}\).

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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