The homoleptic and octahedral complex of Co²⁺ and H₂ O has ___________ unpaired electron(s) in the t_(2g) set of orbitals.

Solution

Chemistry ma’am · 5 steps

Water is a weak-field ligand, so fill Co²⁺ high-spin and count only the t_(2g) electrons.

1. Find the d count. Cobalt is [Ar] 3d⁷ 4s². Removing two electrons from 4s leaves Co²⁺ as 3d⁷. The complex is [Co(H₂O)₆]²⁺.
2. Weak field ligand. H₂O is a weak-field ligand, so the splitting Δₒ is smaller than the pairing energy. The electrons spread into e_g before they pair up in t_(2g). This is the high-spin arrangement.
3. Fill the orbitals. Put one electron in each of the five orbitals first: three in t_(2g), two in e_g. The remaining two electrons pair up in t_(2g), which gives t_(2g)⁵ e_g².
4. Count in t2g. Five electrons in three t_(2g) orbitals means two orbitals are full and one holds a single electron. The two e_g electrons are also unpaired, but they are not in t_(2g).
5. So. The t_(2g) set has 1 unpaired electron. The complex has 3 unpaired electrons in total.

The common slip: 3. Counting every unpaired electron in the complex gives 3, but two of those are in e_g. Only 1 is in t_(2g).

The idea to keep. First decide weak or strong field from the ligand, then fill the orbitals. When a question asks about one set (t_(2g) or e_g), count only that set, not the whole complex.

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