Question
Question: The correct statement with regard to \( H_{2}^{+} \) and \( H_{2}^{-} \) (A) Both \( H_{2}^{+} \)...
The correct statement with regard to H2+ and H2−
(A) Both H2+ and H2− are equally stable
(B) Both H2+ and H2− do not exist
(C) H2− is more stable than H2+
(D) H2+ is more stable than H2−
Solution
Hint : The composition of H2+ and H2− defines the stability order. H2+ does not include any electron in the antibonding molecular orbital. H2− includes electrons in antibonding molecular orbital.The stability can be easily identified from the electron presence in bonding and antibonding molecular orbital.
Complete Step By Step Answer:
The electron presence in anti-bonding orbital result in repulsion and decrease of stability
The stability order follows the molecular orbital theory
The formula for bonding order can be regarded as follows
Bonding order =21 (Number of bonding electrons -Number of antibonding electrons)
The number of antibonding electrons in H2+ is 0
The number of antibonding electrons in H2− is 1
The bonding order of H2+ is =21(1−0)=0.5
The bonding order of H2− is =21(2−1)=0.5
H2+ and H2− molecule have the same bond order
Presence of anti-bonding electron decreases the stability of the molecule
Therefore, H2+ is more stable than H2−
Both H2+ and H2− are equally stable is incorrect statement
Both H2+ and H2− do not exist is incorrect statement
H2− is more stable than H2+ is incorrect statement
The correct statement is H2+ is more stable than H2−
Due to the electron-electron repulsion in H2− , the stability of H2− is low
The configuration of H2+ is 1 electron in 1 s bonding orbital
The configuration of H2− is 1 electron in 1 s bonding orbital and 1 electron in 1 s antibonding orbital
Even though the bond order of both H2+ and H2− are same, the stability is different
Therefore, H2+>H2− .
Note :
The stability order follows the molecular orbital theory. The electron presence in anti-bonding orbital results in repulsion and decrease of stability. Electron-electron repulsion in H2− , the stability of H2− is low.