Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Thermal decomposition of ${\text{AgNO}}_3$ 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 ______.

Let's break down this fascinating problem step by step. The journey begins with a classic inorganic reaction: the thermal decomposition of silver nitrate (${\text{AgNO}}_3$). When we subject silver nitrate to heat, it doesn't just melt; it breaks apart completely.

The balanced chemical equation for this process is:

Notice the products. We get solid silver metal ($\text{Ag}$), and two distinct gases: nitrogen dioxide (${\text{NO}}_2$) and oxygen (${\text{O}}_2$). The problem specifically points our attention toward these two gaseous products.

The question mentions that the decomposition produces two paramagnetic gases. Let's verify this.

First, consider nitrogen dioxide (${\text{NO}}_2$). If you count its valence electrons, you'll find it has an odd number ($5$ from Nitrogen + $2\times 6$ from Oxygen = $17$ valence electrons). Any molecule with an odd number of electrons must have at least one unpaired electron. Therefore, ${\text{NO}}_2$ is paramagnetic with exactly $1$ unpaired electron.

Next, let's look at oxygen (${\text{O}}_2$). While its Lewis dot structure might suggest all electrons are paired in a double bond, Molecular Orbital Theory (MOT) reveals the truth. The ${\text{O}}_2$ molecule has $2$ unpaired electrons residing in its degenerate ${\pi }^{\ast }$ antibonding orbitals.

The question asks us to focus on the gas with the higher number of unpaired electrons. Comparing the two, ${\text{O}}_2$ ($2$ unpaired electrons) beats ${\text{NO}}_2$ ($1$ unpaired electron). So, oxygen is our target gas!

Now that we've isolated ${\text{O}}_2$, the core task is to find the total number of electrons present in its antibonding molecular orbitals.

To do this accurately, we must write out the complete molecular orbital electronic configuration for the ${\text{O}}_2$ molecule, which contains a total of $16$ electrons ($8$ from each oxygen atom):

Antibonding orbitals are denoted by the asterisk ($\ast$) symbol. Let's carefully scan our configuration and tally up the electrons residing in these specific orbitals:

1. Inner Shell ($1s$): The ${\sigma }_{1s}^{\ast }$ orbital is fully occupied with $2$ electrons. 2. Valence $s$-Shell ($2s$): Moving up, the ${\sigma }_{2s}^{\ast }$ orbital is also fully occupied with $2$ electrons. 3. Valence $p$-Shell ($2p$): Finally, we reach the highest occupied molecular orbitals (HOMO). According to Hund's rule of maximum multiplicity, the two degenerate ${\pi }^{\ast }$ orbitals each take one electron. So, ${\pi }_{2p_x}^{\ast }$ has $1$ electron, and ${\pi }_{2p_y}^{\ast }$ has $1$ electron.

Adding them all together:

There are exactly $6$ electrons in the antibonding molecular orbitals of the oxygen molecule. This is a brilliant exercise that tests not only your knowledge of inorganic reactions but also your precision in applying Molecular Orbital Theory!