Question

(a) \( \mathrm{[Co(NH_3)_6]^{3+}} \) is diamagnetic whereas \( \mathrm{[CoF_6]^{3-}} \) is paramagnetic. Justify the statement. [Atomic number of Co = 27]

(b) Write the electronic configuration for \( d^4 \) ion if \( \Delta_0 \gt P \) on the basis of crystal field theory.

Hint:

Co3+ is d6. NH3 is a strong field ligand (low spin, no unpaired electron) and F- is a weak field ligand (high spin, 4 unpaired). For d4 with large splitting the electrons pair as t2g4 eg0.

Answer 1

Step 1: Find the metal ion configuration.
Cobalt has atomic number 27, so its configuration is \([\mathrm{Ar}]\,3d^7\,4s^2\).
In both complexes cobalt is in the +3 state. Removing 2 electrons from 4s and 1 from 3d gives \(\mathrm{Co^{3+}}: 3d^6\).

Step 2: Complex with ammonia (part a).
\(\mathrm{NH_3}\) is a strong field ligand. It pushes the 3d electrons to pair up inside the lower set of orbitals.
So the six d electrons occupy the \(t_{2g}\) set fully.
\[ \mathrm{Co^{3+}}\ \text{in}\ \mathrm{[Co(NH_3)_6]^{3+}}: t_{2g}^{6}\, e_g^{0} \]
Here the electrons are all paired. There are 0 unpaired electrons. So the complex is diamagnetic.
In valence bond terms, two 3d, one 4s and three 4p orbitals mix to give \(d^2sp^3\) hybrids. This is an inner orbital, low spin complex.

Step 3: Complex with fluoride (part a).
\(\mathrm{F^-}\) is a weak field ligand. The splitting is small, so electrons prefer to occupy the higher \(e_g\) orbitals before pairing.
\[ \mathrm{Co^{3+}}\ \text{in}\ \mathrm{[CoF_6]^{3-}}: t_{2g}^{4}\, e_g^{2} \]
This leaves 4 unpaired electrons. So the complex is paramagnetic. It is a high spin, outer orbital complex with \(sp^3d^2\) hybrids.

Step 4: Part (b), the \(d^4\) ion with \(\Delta_0 \gt P\).
Here P is the pairing energy. When \(\Delta_0\) is larger than P, it costs less energy to pair an electron in \(t_{2g}\) than to lift it to \(e_g\).
So the fourth electron pairs up in the lower level.
\[ d^4: t_{2g}^{4}\, e_g^{0} \]
This is a low spin arrangement with 2 unpaired electrons.

Final Answer:
(a) \(\mathrm{[Co(NH_3)_6]^{3+}}\) is \(t_{2g}^6 e_g^0\) (no unpaired electron), while \(\mathrm{[CoF_6]^{3-}}\) is \(t_{2g}^4 e_g^2\) (4 unpaired electrons). (b) \[ \boxed{t_{2g}^{4}\, e_g^{0}} \]