Question

According to molecular orbital theory, which of the following statements is(are) correct?

• A. $H{e}_2^{+}$ has the same energy as two isolated $He$ atoms
• B. $O_2^{2+}$ is expected to have a longer bond length than $O_2$
• C. $C_2^{2-}$ is expeced to be diamagnetic
• D. $N_2^{+}$ and $N_2^{-}$ have the same bond order

Answer 1

Answer: (C), (D)

$N_2^{+}$ and $N_2^{-}$ have the same bond order

For $C_2^{2-}$
Number of electrons $=14$
Molecular orbital electronic configuration; $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y^2=\pi 2p_z^2\sigma 2p_x^2$
Number of bonding electrons = 10
Number of antibonding electrons = 4
Hence, Bond order = 3
Magnetic character = Diamagnetic

For $O_2^{2+}$
Number of electrons $=14$
Molecular orbital electronic configuration; $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y^2=\pi 2p_z^2\sigma 2p_x^2$
Number of bonding electrons = 10
Number of antibonding electrons = 4
Hence, Bond order = 3
Magnetic character = Diamagnetic

For $O_2$
Number of electrons $=16$
Molecular orbital electronic configuration; $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_x^2\pi 2p_y^2=\pi 2p_z^2{\pi }^{\ast }2p_y^2={\pi }^{\ast }2p_z^2$
Number of bonding electrons = 10
Number of antibonding electrons = 6
Hence, Bond order = 2
Magnetic character = Paramagnetic

For $N_2^{+}$
Number of electrons $=13$
Molecular orbital electronic configuration; $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y^2=\pi 2p_z^2\sigma 2p_x^1$
Number of bonding electrons = 9
Number of antibonding electrons = 4
Hence, Bond order = 2.5
Magnetic character = Paramagnetic

For $N_2^{-}$
Number of electrons $=15$
Molecular orbital electronic configuration; $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_x^2\pi 2p_y^2=\pi 2p_z^2{\pi }^{\ast }2p_y^2={\pi }^{\ast }2p_z^1$
Number of bonding electrons = 10
Number of antibonding electrons = 5
Hence, Bond order = 2.5
Magnetic character = Paramagnetic

For $H{e}_2^{+}$
Number of electrons $=3$
Molecular orbital electronic configuration; $\sigma 1s^2{\sigma }^{\ast }1s^1$
Number of bonding electrons = 2
Number of antibonding electrons = 1
Hence, Bond order = 0.5
Magnetic character = Paramagnetic

Thus (a) is correct.
(b) Bond order of $O_2^{2+}>O_2$, thus bond length of $O_2^{2+}<O_2$ Thus, incorrect.
(c) $N_2^{+}$ and $N_2^{-}$ have same bond order.
(d) $H{e}_2^{+}$ with bond order = $0.5$ is more stable thus, less energy than isolated $He$ atoms. Thus, (d) is incorect.