Question

The following graph is given, between total energy and distance between the two nuclei for species $H_2{}^{+},H_2,H{e}_2{}^{+}\&H{e}_2$, which of the following statements is correct :

• A. $H{e}_2^{+}$is more stable than $H_2{}^{+}$.
• B. Bond dissociation energy of $H_2^{+}$is more than bond dissociation energy of $H{e}_2^{+}$.
• C. Since bond orders of $H{e}_2^{+}$and $H_2^{+}$are equal hence both will have equal bond dissociation energy.
• D. Bond length of $H_2^{+}$is less than bond length of $H_2$.

Answer 1

Answer: (B)

From the graph
$B.E.$ of $H_2>B.E.$ of $H_2^{+}>B.E$ of $H{e}_2^{+}>BE$ of $H{e}_2$
where $BE=$ bond energy or bond dissociation energy and
$B.L.$ of $H_2<B.L.$ of $H_2^{+}<B.L.$ of $H{e}_2^{+}<B.L.$ of $H{e}_2$
where $B.L.$ bond length
so stability order $=H_2>H_2^{+}>H{e}_2^{+}>H{e}_2$