Antibonding Electrons in Paramagnetic Gas from Decomposition of Silver Nitrate

To find the total number of electrons present in the antibonding molecular orbitals of the specified gas, we proceed step-by-step as follows:

Step 1: Thermal Decomposition of Silver Nitrate

Upon heating, silver nitrate ($\text{AgNO}_3$) undergoes thermal decomposition according to the balanced chemical equation: $2\text{AgNO}_3(\text{s}) \xrightarrow{\Delta} 2\text{Ag}(\text{s}) + 2\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$

The two gases produced during this reaction are nitrogen dioxide ($\text{NO}_2$) and oxygen ($\text{O}_2$).

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Step 2: Identification of Unpaired Electrons in Both Gases

1. Nitrogen Dioxide ($\text{NO}_2$):

   • Total number of electrons = $7 (\text{from N}) + 2 \times 8 (\text{from O}) = 23$ electrons.
   • Being an odd-electron molecule, $\text{NO}_2$ contains $1$ unpaired electron.

2. Oxygen Gas ($\text{O}_2$):

   • Total number of electrons = $16$ electrons.
   • According to Molecular Orbital Theory (MOT), the electronic configuration of $\text{O}_2$ (containing $16$ electrons) is: $\sigma_{1s}^2 \ \sigma_{1s}^{*2} \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ \left(\pi_{2p_x}^2 = \pi_{2p_y}^2\right) \ \left(\pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}\right)$
   • $\text{O}_2$ contains $2$ unpaired electrons present in the degenerate antibonding orbitals ($\pi_{2p_x}^*$ and $\pi_{2p_y}^*$).

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Step 3: Determining the Target Gas

Comparing the number of unpaired electrons:

• Unpaired electrons in $\text{NO}_2 = 1$
• Unpaired electrons in $\text{O}_2 = 2$

Thus, the gas possessing the greater number of unpaired electrons is Oxygen ($\text{O}_2$).

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Step 4: Counting the Antibonding Electrons in $\text{O}_2$

The antibonding molecular orbitals (ABMOs) of $\text{O}_2$ and the electrons present in them are:

• $\sigma_{1s}^* \longrightarrow 2 \text{ electrons}$
• $\sigma_{2s}^* \longrightarrow 2 \text{ electrons}$
• $\pi_{2p_x}^* \longrightarrow 1 \text{ electron}$
• $\pi_{2p_y}^* \longrightarrow 1 \text{ electron}$

$\text{Total number of antibonding electrons} = 2 + 2 + 1 + 1 = 6$

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Final Answer

The total number of electrons present in the antibonding molecular orbitals of the gas possessing the higher number of unpaired electrons is 6.