In which of the following options the species changes from paramagnetic to diamagnetic and bond order increases.

Question

Among the species O$_2^+$, N$_2^-$, N$_2^{2-}$ and O$_2^-$ which have same bond order as well as paramagnetic in nature.

Answer 1

To determine which species changes from paramagnetic to diamagnetic and experiences an increase in bond order, let's analyze each option:

$N_2 \to N_{2}^{+}$
- Magnetic Property: $N_2$ is diamagnetic (all electrons paired). $N_{2}^{+}$ still remains diamagnetic because the unpaired electron is removed from the non-bonding orbitals.
- Bond Order: The bond order of $N_2$ is 3, and for $N_{2}^{+}$, the bond order becomes 2.5 (decreases).
$O_2 \to O_{2}^{+}$
- Magnetic Property: $O_2$ is paramagnetic (due to two unpaired electrons in $\pi^*$ orbitals). $O_{2}^{+}$ remains paramagnetic as it still has one unpaired electron.
- Bond Order: The bond order of $O_2$ is 2, and for $O_{2}^{+}$, the bond order increases to 2.5.
$NO \to NO^+$ [Correct Answer]
- Magnetic Property: $NO$ is paramagnetic with one unpaired electron. $NO^+$ is diamagnetic as the removal of an electron pairs up all remaining electrons.
- Bond Order: The bond order of $NO$ is 2.5, and for $NO^+$, it increases to 3.
$O_2 \to O^{+}$
- Magnetic Property: $O_2$ is paramagnetic. $O^{+}$, if considered to exist as a single charged oxygen, is a distinct situation and does not naturally follow typical molecular orbital calculations as it becomes a monoatomic oxygen rather than a molecular species.
- Moreover, the bond order concept is not directly applicable to single atoms in the same way as it is to molecular species.

Thus, the option $NO \to NO^+$ is correct as it transitions from paramagnetic to diamagnetic, and its bond order increases from 2.5 to 3.

Answer 2

To determine which species changes from paramagnetic to diamagnetic and experiences an increase in bond order, let's analyze each option:

$N_2 \to N_{2}^{+}$
- Magnetic Property: $N_2$ is diamagnetic (all electrons paired). $N_{2}^{+}$ still remains diamagnetic because the unpaired electron is removed from the non-bonding orbitals.
- Bond Order: The bond order of $N_2$ is 3, and for $N_{2}^{+}$, the bond order becomes 2.5 (decreases).
$O_2 \to O_{2}^{+}$
- Magnetic Property: $O_2$ is paramagnetic (due to two unpaired electrons in $\pi^*$ orbitals). $O_{2}^{+}$ remains paramagnetic as it still has one unpaired electron.
- Bond Order: The bond order of $O_2$ is 2, and for $O_{2}^{+}$, the bond order increases to 2.5.
$NO \to NO^+$ [Correct Answer]
- Magnetic Property: $NO$ is paramagnetic with one unpaired electron. $NO^+$ is diamagnetic as the removal of an electron pairs up all remaining electrons.
- Bond Order: The bond order of $NO$ is 2.5, and for $NO^+$, it increases to 3.
$O_2 \to O^{+}$
- Magnetic Property: $O_2$ is paramagnetic. $O^{+}$, if considered to exist as a single charged oxygen, is a distinct situation and does not naturally follow typical molecular orbital calculations as it becomes a monoatomic oxygen rather than a molecular species.
- Moreover, the bond order concept is not directly applicable to single atoms in the same way as it is to molecular species.

Thus, the option $NO \to NO^+$ is correct as it transitions from paramagnetic to diamagnetic, and its bond order increases from 2.5 to 3.

The Correct Option is C