In the context of electron geometry and crystal field theory, what happens to the energy of the metal d orbitals due to an increase in electrostatic repulsion?

• A. The energy of the d orbitals increases.
• B. The energy of the d orbitals remains unchanged.
• C. The energy of the d orbitals decreases.
• D. The energy of the d orbitals becomes zero.

Answer: (A)

Verified step by step guidance

1. Understand that in crystal field theory, the metal d orbitals interact with the surrounding ligands, which are treated as point charges or dipoles creating an electrostatic field.
2. Recognize that electrostatic repulsion occurs between the negatively charged ligands and the electrons in the metal's d orbitals, affecting the energy levels of these orbitals.
3. Recall that an increase in electrostatic repulsion means the ligands exert a stronger repulsive force on the d electrons, which raises the energy of the d orbitals because electrons in these orbitals experience greater repulsion.
4. Note that this increase in energy can lead to splitting of the d orbitals into different energy levels (such as eg and t2g in an octahedral field), but the overall effect of increased repulsion is an increase in the energy of the d orbitals compared to the free ion.
5. Conclude that due to increased electrostatic repulsion from ligands, the energy of the metal d orbitals increases.