Thermal decomposition of AgNO₃ produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is_____.

The correct answer is 6.

Step 1: Understand the thermal decomposition reaction of AgNO₃
When silver nitrate (AgNO₃) undergoes thermal decomposition, it breaks down into silver metal, nitrogen dioxide gas, and oxygen gas:
2AgNO₃(s) → 2Ag(s) + 2NO₂(g) + O₂(g)
The two gases produced in this reaction are nitrogen dioxide (NO₂) and oxygen (O₂).

Step 2: Identify the paramagnetic nature and the number of unpaired electrons in each gas
- Nitrogen dioxide (NO₂) is an odd-electron molecule (total valence electrons = 5 + 6×2 = 17). It contains 1 unpaired electron.
- Oxygen (O₂) molecule has 16 electrons. According to Molecular Orbital (MO) theory, its electronic configuration shows that it contains 2 unpaired electrons in its degenerate antibonding molecular orbitals (${\pi }_{*2p_x}$ and ${\pi }_{*2p_y}$).
Therefore, O₂ is the gas with the higher number of unpaired electrons.

Step 3: Count the total number of electrons in the antibonding molecular orbitals of O₂
The molecular orbital electronic configuration for the O₂ molecule (total 16 electrons) is:
${\sigma }_{1s}{}^{}2$ ${\sigma }_{1s}^{*}{}^{}2$ ${\sigma }_{2s}{}^{}2$ ${\sigma }_{2s}^{*}{}^{}2$ ${\sigma }_{2p_z}{}^{}2$ (${\pi }_{2p_x}{}^{}2$ = ${\pi }_{2p_y}{}^{}2$) (${\pi }_{2p_x}^{*}{}^{}1$ = ${\pi }_{2p_y}^{*}{}^{}1$)

The antibonding molecular orbitals (indicated by an asterisk '*') present in O₂ are:
1. ${\sigma }_{1s}^{*}$ containing 2 electrons
2. ${\sigma }_{2s}^{*}$ containing 2 electrons
3. ${\pi }_{2p_x}^{*}$ containing 1 electron
4. ${\pi }_{2p_y}^{*}$ containing 1 electron

Total number of electrons in antibonding molecular orbitals = 2 + 2 + 1 + 1 = 6.