Concept: This question involves understanding the molecular orbital theory (MOT) for diatomic molecules, specifically how the removal of an electron affects bond order and magnetic properties (paramagnetism/diamagnetism).

Why (D) is correct:

Step 1: Determine the electronic configuration and properties of O\u2082.
The total number of electrons in O\u2082 is 16. The molecular orbital configuration is:
$(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{\ast }2p_x{)}^1({\pi }^{\ast }2p_y{)}^1$.
Bond order (BO) = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-6)=2$.
Since there are two unpaired electrons in the ${\pi }^{\ast }2p$ orbitals, O\u2082 is paramagnetic.

Step 2: Determine the electronic configuration and properties of O\u2082\u207a.
When O\u2082 is converted to O\u2082\u207a, one electron is removed from the highest occupied molecular orbital (HOMO), which is an antibonding ${\pi }^{\ast }2p$ orbital.
The total number of electrons in O\u2082\u207a is 15. The molecular orbital configuration is:
$(\sigma 1s{)}^2({\sigma }^{\ast }1s{)}^2(\sigma 2s{)}^2({\sigma }^{\ast }2s{)}^2(\sigma 2p_z{)}^2(\pi 2p_x{)}^2(\pi 2p_y{)}^2({\pi }^{\ast }2p_x{)}^1$.
Bond order (BO) = $\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-5)=2.5$.
Since there is one unpaired electron, O\u2082\u207a is paramagnetic.

Step 3: Compare the properties.
- Bond order: O\u2082 (2) to O\u2082\u207a (2.5). The bond order increases.
- Paramagnetic character: O\u2082 has 2 unpaired electrons, O\u2082\u207a has 1 unpaired electron. The number of unpaired electrons decreases, so the paramagnetic character decreases.

Option Analysis:
A) both paramagnetic character and bond order increase. (Incorrect, paramagnetic character decreases)
B) bond order decreases. (Incorrect, bond order increases)
C) paramagnetic character increases. (Incorrect, paramagnetic character decreases)
D) paramagnetic character decreases and the bond order increases. (Correct)

Correct Answer: (D)