Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: According to molecular orbital theory, which of the following will not be a viable molecule?

A Molecular Orbital Theory Adventure

Imagine you are an atomic architect, tasked with building molecules from scratch. You take two atoms, push them together, and hope they stick. But how do you know if they will actually form a stable bond or just bounce off each other? This is where Molecular Orbital Theory (MOT) comes to the rescue.

MOT provides us with a powerful mathematical tool called the Bond Order. Think of bond order as the "glue score" of a molecule. It tells us the net number of bonds holding the atoms together. The formula is beautifully simple:

Here, $N_b$ represents the number of electrons in stabilizing bonding orbitals, and $N_a$ represents the number of electrons in destabilizing anti-bonding orbitals.

The Golden Rule of Viability: For a molecule to exist (to be "viable"), its bond order must be strictly greater than zero. If the bond order is zero or negative, the repulsive forces overpower the attractive forces, and the molecule simply falls apart.

Let's put our four candidates to the test.

Candidate 1: ${\text{He}}_2^{2+}$ A neutral helium atom has 2 electrons. Two helium atoms would give us 4 electrons. However, the $2+$ charge indicates that 2 electrons have been stripped away. This leaves us with exactly 2 electrons. Following the Aufbau principle, these two electrons will occupy the lowest energy molecular orbital, which is the bonding ${\sigma }_{1s}$ orbital. - Configuration: ${\sigma }_{1s}^2$ - Bond Order: $\frac{2-0}{2}=1$ Since the bond order is 1, this ion is perfectly viable.

Candidate 2: ${\text{He}}_2^{+}$ Here, we start with 4 electrons from the two helium atoms and subtract 1 for the positive charge, leaving us with 3 electrons. The first two electrons fill the bonding ${\sigma }_{1s}$ orbital. The third electron has nowhere to go but up into the higher-energy, anti-bonding ${\sigma }_{1s}^{\ast }$ orbital. - Configuration: ${\sigma }_{1s}^2,{\sigma }_{1s}^{\ast 1}$ - Bond Order: $\frac{2-1}{2}=0.5$ A fractional bond order of 0.5 means there is a net bonding effect equivalent to half a bond. It's weak, but it's greater than zero, so the molecule is viable.

Candidate 3: ${\text{H}}_2^{-}$ Each hydrogen atom brings 1 electron, and the negative charge adds 1 more, giving us a total of 3 electrons. Notice something interesting? This is the exact same number of electrons as ${\text{He}}_2^{+}$! Species with the same number of electrons are called isoelectronic. They will have the identical molecular orbital configuration. - Configuration: ${\sigma }_{1s}^2,{\sigma }_{1s}^{\ast 1}$ - Bond Order: $\frac{2-1}{2}=0.5$ Just like before, the bond order is positive, making this species viable.

Candidate 4: ${\text{H}}_2^{2-}$ Finally, we arrive at our last suspect. Two hydrogen atoms provide 2 electrons, and the $2-$ charge adds 2 more. We have a total of 4 electrons to place. The first two electrons fill the bonding ${\sigma }_{1s}$ orbital. The remaining two electrons are forced to completely fill the anti-bonding ${\sigma }_{1s}^{\ast }$ orbital. - Configuration: ${\sigma }_{1s}^2,{\sigma }_{1s}^{\ast 2}$ - Bond Order: $\frac{2-2}{2}=0$

Look at that result! The two electrons in the anti-bonding orbital completely cancel out the stabilizing effect of the two electrons in the bonding orbital. The net attractive force is zero. Without any "glue" to hold them together, the atoms will immediately separate.

Therefore, according to Molecular Orbital Theory, ${\text{H}}_2^{2-}$ is not a viable molecule.