Question

Reaction of [Co(H₂O)₆]²⁺ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t_2g-orbitals of the product is _______

Answer 1

Correct Answer: 6

To solve this problem, we need to determine the number of electrons in the t_2g-orbitals for the product of the reaction between [Co(H₂O)₆]²⁺ and excess ammonia in the presence of oxygen, resulting in a diamagnetic product.

Start by considering the initial complex [Co(H₂O)₆]²⁺. Here, cobalt is in the +2 oxidation state, leading to a d⁷ electron configuration for Co²⁺ since cobalt's atomic number is 27 (resulting in an electron configuration of [Ar] 3d⁷ 4s² for the neutral atom, minus two electrons for the Co²⁺ ion).

In the reaction, excess ammonia acts as a ligand, forming a more stable complex. Ammonia is a strong field ligand due to its ability to cause pairing of electrons in the d orbitals. The oxidation of Co²⁺ in the presence of oxygen will form Co³⁺, resulting in the electron configuration d⁶ for the cobalt ion in this state.

Since the produced compound is diamagnetic, all electrons must be paired according to Hund's rule and ligand field theory. The Co³⁺ ion in an octahedral field, influenced by strong field ammonia ligands, will have a low spin configuration:
d⁶ results in the distribution of electrons as t_2g⁶e_g⁰, with all six electrons paired in the lower energy t_2g orbitals.

Thus, the number of electrons in the t_2g orbitals is 6.

This value of 6 falls within the expected range of 6 to 6, confirming the solution's correctness.