In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?

We have to check, for every given process, two things simultaneously: whether the bond order $$\left(BO\right)$$ has increased and whether a paramagnetic molecule (containing at least one unpaired electron) has become diamagnetic (all electrons paired) after the process.

First of all, recall the formula that comes from Molecular Orbital Theory:

$$BO=\dfrac{N_b-N_a}{2}$$

where $$N_b$$ is the total number of electrons present in bonding molecular orbitals and $$N_a$$ is the total number of electrons present in antibonding molecular orbitals.

The magnetic property is judged as follows: if any molecular orbital has an unpaired electron, the species is paramagnetic; if every electron is paired, the species is diamagnetic.

Now we shall apply these ideas one‐by‐one to every option.

Option A : $$O_2\;\rightarrow\;O_2^{+}$$

• For $$O_2$$ the total number of electrons is $$16$$. The valence-shell molecular orbital configuration (for elements after atomic number $$7$$) is

$$\sigma_{2p_z}^2\;\pi_{2p_x}^2=\pi_{2p_y}^2\;\pi_{2p_x}^{*\,1}=\pi_{2p_y}^{*\,1}$$

This gives $$N_b=10,\;N_a=6$$ and hence

$$BO(O_2)=\dfrac{10-6}{2}=2$$

The two $$\pi_{2p}^{*}$$ electrons are unpaired, so $$O_2$$ is paramagnetic.

• In $$O_2^{+}$$ one electron is removed, obviously from the highest‐energy orbital $$\pi_{2p}^{*}$$. The configuration becomes

$$\sigma_{2p_z}^2\;\pi_{2p_x}^2=\pi_{2p_y}^2\;\pi_{2p_x}^{*\,1}$$

Now $$N_b=10,\;N_a=5$$, giving

$$BO(O_2^{+})=\dfrac{10-5}{2}=2.5$$

There is still one unpaired electron, so the cation remains paramagnetic. Bond order has indeed increased, but magnetic nature is unchanged. Hence this option does not satisfy the requirement.

Option B : $$NO\;\rightarrow\;NO^{+}$$

Because $$NO$$ is heteronuclear, we count electrons directly. Nitrogen contributes $$7$$ electrons, oxygen contributes $$8$$, making $$15$$ in all.

Using the same orbital ordering as $$O_2$$ (a good approximation), the last electron enters a $$\pi_{2p}^{*}$$ orbital, giving one unpaired electron. Thus $$NO$$ is paramagnetic.

Bond-order calculation:

$$N_b=10,\;N_a=5\quad\Rightarrow\quad BO(NO)=\dfrac{10-5}{2}=2.5$$

Now we remove one electron to form $$NO^{+}$$. The electron leaves that singly occupied $$\pi_{2p}^{*}$$ orbital, so the cation has $$14$$ electrons—exactly the same count as $$N_2$$, with all electrons paired.

The numbers now are $$N_b=10,\;N_a=4$$, so

$$BO(NO^{+})=\dfrac{10-4}{2}=3$$

Hence, bond order increases from $$2.5$$ to $$3$$, and because no orbital contains an unpaired electron, $$NO^{+}$$ is diamagnetic. Therefore this process fulfils both required conditions.

Option C : $$O_2\;\rightarrow\;O_2^{-}$$

Adding one electron to $$O_2$$ means the new electron enters a $$\pi_{2p}^{*}$$ orbital, giving

$$\pi_{2p_x}^{*\,2}=\pi_{2p_y}^{*\,1}$$

Now $$N_b=10,\;N_a=7$$, so

$$BO(O_2^{-})=\dfrac{10-7}{2}=1.5$$

Because a $$\pi_{2p}^{*}$$ orbital still holds an unpaired electron, the anion remains paramagnetic, and bond order has in fact decreased. Hence this option fails on both counts.

Option D : $$N_2\;\rightarrow\;N_2^{+}$$

• For $$N_2$$ (total electrons $$=14$$) the ordering up to $$\sigma_{2p_z}$$ is

$$\sigma_{2p_z}^2\;\pi_{2p_x}^2=\pi_{2p_y}^2$$

All electrons are paired, so $$N_2$$ is diamagnetic. The bonding and antibonding electrons are $$N_b=10,\;N_a=4$$, giving

$$BO(N_2)=\dfrac{10-4}{2}=3$$

• Removing one electron to form $$N_2^{+}$$ takes it from the highest occupied bonding orbital $$\pi_{2p}$$, resulting in one unpaired electron.

Now $$N_b=9,\;N_a=4$$, so

$$BO(N_2^{+})=\dfrac{9-4}{2}=2.5$$

The bond order has fallen and the magnetic behaviour has changed from diamagnetic to paramagnetic, i.e.\ in the wrong direction.

After examining every option we see that only the process $$NO\rightarrow NO^{+}$$ simultaneously raises the bond order and converts a paramagnetic molecule into a diamagnetic one.

Hence, the correct answer is Option B.