Animated Solution for Chemistry - Surface Chemistry: When O${}_2$ is adsorbed on a metallic surface, electron transfer occurs from the metal to O${}_2$. The TRUE, statement (s) regarding this adsorption is (are)

Imagine you are observing a microscopic world where a pristine metallic surface lies exposed. This surface is a bustling sea of electrons, highly reactive and ready to interact with anything that comes close. Now, visualize an oxygen molecule ($O_2$) drifting towards this metallic surface.

The problem gives us a massive clue right away: electron transfer occurs from the metal to the oxygen molecule. This single piece of information is the key that unlocks the entire problem. Let's break down what this means step by step.

Physisorption vs. Chemisorption

When a gas molecule interacts with a solid surface, it can do so in two primary ways. The first is physisorption, which is akin to a gentle, temporary handshake. It relies on weak van der Waals forces. There is no sharing or transferring of electrons; the molecules just stick together loosely.

The second way is chemisorption. This is a full-blown chemical reaction at the surface. It involves the actual transfer or sharing of electrons to form strong chemical bonds.

Since the problem explicitly states that an electron is transferred from the metal to the $O_2$ molecule, we are undeniably dealing with chemisorption. The metal and the oxygen are forming a true chemical bond. Therefore, the statement that '$O_2$ is physisorbed' is fundamentally incorrect.

Now, let's think about the energy of this system. Why do molecules adsorb onto surfaces in the first place?

At the surface of the metal, the atoms are not surrounded by other metal atoms on all sides. They have unbalanced, or residual surface forces. These atoms are essentially 'reaching out', looking for something to bond with to lower their energy state.

When the oxygen molecule comes along and bonds with the surface, these residual forces are satisfied. The overall energy of the surface decreases. In the universe of thermodynamics, energy doesn't just disappear; it has to go somewhere. This lost energy is released into the surroundings as heat.

This is a universal rule for adsorption: Adsorption is always an exothermic process. Because the system goes from a higher energy state to a lower, more stable energy state, $\Delta H<0$. Thus, the statement 'heat is released' is absolutely correct.

The Fate of the Transferred Electron

This is where the physics and chemistry get truly beautiful. The metal has handed an electron to the oxygen molecule. But where exactly does this electron go? To answer this, we must look at the Molecular Orbital (MO) Theory of oxygen.

The electronic configuration of an $O_2$ molecule is fascinating. If we fill the molecular orbitals according to Hund's rule and the Pauli exclusion principle, we find that the highest energy electrons reside in the antibonding pi orbitals. Specifically, there are two degenerate orbitals: ${\pi }_{2p_x}^{\ast }$ and ${\pi }_{2p_y}^{\ast }$.

In a neutral $O_2$ molecule, each of these ${\pi }_{2p}^{\ast }$ orbitals contains exactly one unpaired electron. This is why liquid oxygen is paramagnetic and can be held between the poles of a strong magnet!

When the metal transfers an extra electron to the $O_2$ molecule, this electron must enter the lowest available energy level. The lowest unoccupied (or in this case, partially occupied) orbital is the ${\pi }_{2p}^{\ast }$ orbital.

Therefore, the incoming electron pairs up with one of the existing electrons in the ${\pi }_{2p}^{\ast }$ orbital. The occupancy of the ${\pi }_{2p}^{\ast }$ orbital is increased. This makes the statement 'occupancy of ${\pi }_{2p}^{\ast }$ of $O_2$ is increased' perfectly correct.

Bond Order and Bond Length

Finally, we must ask: what does this extra electron do to the bond between the two oxygen atoms?

In Molecular Orbital Theory, the strength of a bond is quantified by its Bond Order. The formula for bond order is beautifully simple:

Here, $N_b$ is the number of electrons in bonding orbitals, and $N_a$ is the number of electrons in antibonding orbitals. Bonding electrons act like glue, holding the atoms together, while antibonding electrons act like wedges, pushing the atoms apart.

By adding an electron to the ${\pi }_{2p}^{\ast }$ orbital, we have increased the number of antibonding electrons ($N_a$). According to our formula, if $N_a$ increases, the overall Bond Order must decrease.

A lower bond order means that the bond between the two oxygen atoms has become weaker. And in chemistry, a weaker bond is almost always a longer bond. The atoms are not held together as tightly, so they drift slightly further apart.

Therefore, the bond length of $O_2$ is increased. This confirms that our final statement is also correct.

In conclusion, by carefully following the journey of a single transferred electron, we have unraveled the thermodynamics, the quantum mechanics, and the structural consequences of this adsorption process. The correct statements are indeed B, C, and D.