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 ___________.

Solution

Chemistry ma’am · 4 steps

Oxygen pushes cobalt from +2 to +3, and ammonia then pairs up all six d electrons.

1. Identify the product. With excess NH₃ all six water ligands are replaced. O₂ then oxidises Co²⁺ to Co³⁺, so the product is [Co(NH₃)₆]³⁺.
2. Count d electrons. Cobalt is [Ar] 3d⁷ 4s², so Co³⁺ is 3d⁶. The complex is octahedral, so the d orbitals split into a lower t_(2g) set of three and an upper e_g set of two.
3. Use the diamagnetism. Diamagnetic means no unpaired electrons. For d⁶ that is possible only as low spin, which fits NH₃ as a strong-field ligand on a 3+ ion: all six electrons pair up in t_(2g).

   Co³⁺ (d⁶, low spin): t_(2g)⁶ e_g⁰

4. So. The t_(2g) orbitals of [Co(NH₃)₆]³⁺ hold 6 electrons.

The common slip: 4. Filling d⁶ as high spin gives t_(2g)⁴ e_g², so 4, but that leaves four unpaired electrons and cannot be diamagnetic.

The idea to keep. Find the oxidation state first, then let the magnetism tell you the spin state. A diamagnetic octahedral d⁶ complex is always t_(2g)⁶ e_g⁰.

Answer from the NTA final answer key. Solution by PrepWiser.