The homoleptic and octahedral complex of Co^2+ and H_2 O has ___________ unpaired electron(s) in the t_2g set of orbitals.

The homoleptic and octahedral complex of Co2 and H2 O has un: null (Chemistry)

When the water ligands coordinate to the

\mathrm{Co}^{2+}

ion in

\mathrm{[Co(H_2O)6]^{2+}}

, the

d

orbitals of the

\mathrm{Co}^{2+}

ion split into two sets of orbitals: the

t{2g}

set, which consists of the

d_{xy}

d_{xz}

, and

d_{yz}

orbitals, and the

e_g

set, which consists of the

d_{z^2}

and

d_{x^2-y^2}

orbitals.

The

t_{2g}

orbitals are lower in energy than the

e_g

orbitals.

The

\mathrm{Co}^{2+}

ion has a d

^7

electron configuration, with three electrons in the

t_{2g}

set and two electrons in the

e_g

set. In an octahedral crystal field, the three

t_{2g}

orbitals will be lower in energy than the two

e_g

orbitals. Therefore, the three electrons in the

t_{2g}

set will occupy all three

t_{2g}

orbitals, and the two electrons in the

e_g

set will occupy the higher energy

e_g

orbitals. As a result, there is only

\boxed{1}

unpaired electron in the

t_{2g}

set of orbitals.