Question

Which of the following relation(s) is/are correct?

• A. $O>O_2(1^{st}$ ionization energy)
• B. $N_2^{+}>N^{+}$ (electron affinity)
• C. $O>Se$ (electron affinity)
• D. $H_2>H(1^{st}$ ionization energy)

Answer 1

Answer: (A), (B), (D)

The correct options are
A $O>O_2(1^{st}$ ionization energy)
B $N_2^{+}>N^{+}$ (electron affinity)
D $H_2>H(1^{st}$ ionization energy)

a) If an electron is removed from an antibonding orbital, the ionization energy of the molecule $X_2$ is lower than that of the corresponding atom X. To get the $1^{st}$ ionization energy we remove an electron from the antibonding orbital of $O_2$, thus $O>O_2$.
b) The electron affinity of $N_2^{+}$ is higher than the $N^{+}$ because here we are adding an electron in the bonding orbital of $N_2$.
c) The electron affinity of O is lower than Se because of a high interelectronic repulsion in oxygen due to its small size.
d) This is true because to get the $1^{st}$ ionization energy we remove an electron from the bonding molecular orbital of $H_2$.