1. Write the thermal decomposition reaction of silver nitrate

For silver nitrate, on heating:

$2\,\mathrm{AgNO_3} \xrightarrow{\Delta} 2\,\mathrm{Ag} + 2\,\mathrm{NO_2} + \mathrm{O_2}$

So the two gaseous products are:

• $\mathrm{NO_2}$
• $\mathrm{O_2}$

Both are paramagnetic.

2. Compare the number of unpaired electrons in the two gases

(a) $\mathrm{NO_2}$

Total valence electrons:

$5 + 2\times 6 = 17$

Since it has an odd number of electrons, it has 1 unpaired electron.

(b) $\mathrm{O_2}$

Using molecular orbital theory, $\mathrm{O_2}$ has the configuration:

$\left(\sigma_{1s}\right)^2\left(\sigma_{1s}^*\right)^2\left(\sigma_{2s}\right)^2\left(\sigma_{2s}^*\right)^2\left(\sigma_{2p_z}\right)^2\left(\pi_{2p_x}\right)^2\left(\pi_{2p_y}\right)^2\left(\pi_{2p_x}^*\right)^1\left(\pi_{2p_y}^*\right)^1$

Thus $\mathrm{O_2}$ has 2 unpaired electrons.

So, the gas with the higher number of unpaired electrons is $\mathrm{O_2}$.

3. Count electrons in antibonding molecular orbitals of $\mathrm{O_2}$

Antibonding MOs occupied in $\mathrm{O_2}$ are:

• $\sigma_{1s}^*$ : $2$ electrons
• $\sigma_{2s}^*$ : $2$ electrons
• $\pi_{2p_x}^*$ and $\pi_{2p_y}^*$ : total $2$ electrons

Hence total electrons in antibonding molecular orbitals:

$2 + 2 + 2 = 6$

4. Final answer

$\boxed{6}$

The derived answer matches the stored correct answer.