In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic?

To solve this question, we need to analyze each ionization process and check two conditions: whether the bond energy increases and whether the magnetic behavior changes from paramagnetic to diamagnetic. Paramagnetic species have unpaired electrons and are attracted to magnetic fields, while diamagnetic species have all electrons paired and are repelled by magnetic fields. Bond energy is related to bond order, which can be calculated using molecular orbital theory.

Starting with Option A: $$NO \rightarrow NO^+$$

Nitric oxide (NO) has nitrogen (atomic number 7) and oxygen (atomic number 8), so the total number of valence electrons is 5 (from N) + 6 (from O) = 11. The molecular orbital configuration for NO is:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\pi_{2p})^4, (\sigma_{2p_z})^2, (\pi^*_{2p})^1$$

Here, the unpaired electron in the $$\pi^*_{2p}$$ orbital makes NO paramagnetic. The bond order is calculated as:

Bond order = $$\frac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2}$$

Bonding electrons: $$\sigma_{2s}$$ (2 electrons), $$\pi_{2p}$$ (4 electrons), $$\sigma_{2p_z}$$ (2 electrons) → total 8 bonding electrons.

Antibonding electrons: $$\sigma^*_{2s}$$ (2 electrons), $$\pi^*_{2p}$$ (1 electron) → total 3 antibonding electrons.

Bond order of NO = $$\frac{8 - 3}{2} = \frac{5}{2} = 2.5$$

For $$NO^+$$, one electron is removed from the highest occupied molecular orbital (HOMO), which is the $$\pi^*_{2p}$$ orbital. The configuration becomes:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\pi_{2p})^4, (\sigma_{2p_z})^2$$

All electrons are now paired, so $$NO^+$$ is diamagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\pi_{2p}$$ (4), $$\sigma_{2p_z}$$ (2) → total 8. Antibonding electrons: $$\sigma^*_{2s}$$ (2) → total 2. Bond order of $$NO^+$$ = $$\frac{8 - 2}{2} = \frac{6}{2} = 3$$.

Bond order increases from 2.5 to 3, so bond energy increases. Magnetic behavior changes from paramagnetic (NO) to diamagnetic ($$NO^+$$). Option A satisfies both conditions.

Now, Option B: $$N_2 \rightarrow N_2^+$$

Nitrogen molecule ($$N_2$$) has two nitrogen atoms (atomic number 7 each), so total valence electrons = 5 + 5 = 10. The molecular orbital configuration is:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\pi_{2p})^4, (\sigma_{2p_z})^2$$

All electrons are paired, so $$N_2$$ is diamagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\pi_{2p}$$ (4), $$\sigma_{2p_z}$$ (2) → total 8. Antibonding electrons: $$\sigma^*_{2s}$$ (2) → total 2. Bond order = $$\frac{8 - 2}{2} = 3$$.

For $$N_2^+$$, one electron is removed from the $$\sigma_{2p_z}$$ orbital (HOMO). Configuration becomes:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\pi_{2p})^4, (\sigma_{2p_z})^1$$

This has one unpaired electron, so paramagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\pi_{2p}$$ (4), $$\sigma_{2p_z}$$ (1) → total 7. Antibonding electrons: $$\sigma^*_{2s}$$ (2) → total 2. Bond order = $$\frac{7 - 2}{2} = \frac{5}{2} = 2.5$$.

Bond order decreases from 3 to 2.5, so bond energy decreases. Magnetic behavior changes from diamagnetic to paramagnetic, not paramagnetic to diamagnetic. Option B does not satisfy.

Option C: $$C_2 \rightarrow C_2^+$$

Carbon molecule ($$C_2$$) has two carbon atoms (atomic number 6 each), so total valence electrons = 4 + 4 = 8. The molecular orbital configuration is:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\pi_{2p})^4$$

All electrons are paired, so diamagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\pi_{2p}$$ (4) → total 6. Antibonding electrons: $$\sigma^*_{2s}$$ (2) → total 2. Bond order = $$\frac{6 - 2}{2} = 2$$.

For $$C_2^+$$, one electron is removed from the $$\pi_{2p}$$ orbital. Configuration becomes:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\pi_{2p})^3$$

With three electrons in two degenerate $$\pi_{2p}$$ orbitals, one orbital has two electrons (paired) and the other has one (unpaired), so paramagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\pi_{2p}$$ (3) → total 5. Antibonding electrons: $$\sigma^*_{2s}$$ (2) → total 2. Bond order = $$\frac{5 - 2}{2} = \frac{3}{2} = 1.5$$.

Bond order decreases from 2 to 1.5, so bond energy decreases. Magnetic behavior changes from diamagnetic to paramagnetic. Option C does not satisfy.

Option D: $$O_2 \rightarrow O_2^+$$

Oxygen molecule ($$O_2$$) has two oxygen atoms (atomic number 8 each), so total valence electrons = 6 + 6 = 12. The molecular orbital configuration is:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\sigma_{2p_z})^2, (\pi_{2p})^4, (\pi^*_{2p})^2$$

The two electrons in the degenerate $$\pi^*_{2p}$$ orbitals are unpaired (Hund's rule), so paramagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\sigma_{2p_z}$$ (2), $$\pi_{2p}$$ (4) → total 8. Antibonding electrons: $$\sigma^*_{2s}$$ (2), $$\pi^*_{2p}$$ (2) → total 4. Bond order = $$\frac{8 - 4}{2} = 2$$.

For $$O_2^+$$, one electron is removed from the $$\pi^*_{2p}$$ orbital. Configuration becomes:

$$(\sigma_{2s})^2, (\sigma^*_{2s})^2, (\sigma_{2p_z})^2, (\pi_{2p})^4, (\pi^*_{2p})^1$$

This has one unpaired electron, so paramagnetic. Bonding electrons: $$\sigma_{2s}$$ (2), $$\sigma_{2p_z}$$ (2), $$\pi_{2p}$$ (4) → total 8. Antibonding electrons: $$\sigma^*_{2s}$$ (2), $$\pi^*_{2p}$$ (1) → total 3. Bond order = $$\frac{8 - 3}{2} = \frac{5}{2} = 2.5$$.

Bond order increases from 2 to 2.5, so bond energy increases. However, magnetic behavior remains paramagnetic (both have unpaired electrons). Option D does not satisfy the magnetic behavior change.

Only Option A satisfies both conditions: bond energy increases and magnetic behavior changes from paramagnetic to diamagnetic.

Hence, the correct answer is Option A.