(a) WHICH d-ORBITALS ARE RAISED IN AN OCTAHEDRAL CRYSTAL FIELD

ANSWER: The two $e_g$ orbitals, namely $d_{z^2}$ and $d_{x^2-y^2}$, are raised in energy.

Step 1 - Set up the geometry.
In an octahedral complex, the six ligands approach the central metal ion ALONG the three
Cartesian axes - two along $+x$ and $-x$, two along $+y$ and $-y$, and two along $+z$
and $-z$.

Step 2 - Look at the orientation of the five d-orbitals.
$d_{z^2}$ and $d_{x^2-y^2}$ (the $e_g$ set) have their lobes pointing DIRECTLY ALONG the
axes - that is, straight at the incoming ligands.
$d_{xy}$, $d_{yz}$ and $d_{zx}$ (the $t_{2g}$ set) have their lobes pointing BETWEEN the
axes - that is, in between the ligands.

Step 3 - Apply the electrostatic argument (the reason).
The ligands are treated as point negative charges (or as dipoles with the negative end
towards the metal). Electrons in the metal d-orbitals are REPELLED by these ligand
electrons.
Since the $e_g$ orbitals point straight at the ligands, they experience a MUCH GREATER
ELECTROSTATIC REPULSION. Their energy is therefore RAISED.
The $t_{2g}$ orbitals point away from the ligands, experience LESS repulsion, and their
energy is LOWERED.

Step 4 - Quantify the splitting (barycentre rule).
The degeneracy of the five d-orbitals is destroyed. Taking the average (barycentre)
energy as the reference:
$$e_g \ (d_{z^2},\, d_{x^2-y^2}) : \text{raised by } +0.6\,\Delta_0$$
$$t_{2g}\ (d_{xy},\, d_{yz},\, d_{zx}) : \text{lowered by } -0.4\,\Delta_0$$
where $\Delta_0$ is the crystal field splitting energy for an octahedral field.
(Check: $2 \times (+0.6\Delta_0) + 3 \times (-0.4\Delta_0) = 1.2\Delta_0 - 1.2\Delta_0 = 0$.)

(b) ELECTRONIC CONFIGURATIONS USING CRYSTAL FIELD THEORY

The rule: if $\Delta_0 <P> P$, the ligand is STRONG FIELD
and electrons pair up in $t_{2g}$ first (LOW SPIN).

(i) $[CoF_6]^{3-}$

Step 1 - Oxidation state: $x + 6(-1) = -3 \Rightarrow x = +3$, so $Co^{3+}$.
Step 2 - Configuration: $Co(Z=27) = [Ar]3d^7 4s^2$; $Co^{3+} = [Ar]3d^6$
Step 3 - $F^-$ is a WEAK FIELD ligand (it lies low in the spectrochemical series), so
$\Delta_0 <P> P$. All six electrons pair up
in the lower $t_{2g}$ set.
Step 4 - Configuration:
$$\boxed{t_{2g}^6\ e_g^0}$$
This is a LOW SPIN complex with 0 unpaired electrons; diamagnetic, $\mu = 0$ BM.
It is an inner orbital ($d^2sp^3$) complex.

(c) HYBRIDISATION AND MAGNETIC BEHAVIOUR OF $[Fe(CN)_6]^{3-}$

Step 1 - Find the oxidation state of iron.
Let it be $x$. Each $CN^-$ carries a charge of $-1$:
$$x + 6(-1) = -3$$
$$x = +3, \qquad \text{so iron is present as } Fe^{3+}$$

Step 2 - Write the electronic configuration of the free ion.
$$Fe\,(Z=26) = [Ar]3d^6 4s^2$$
$$Fe^{3+} = [Ar]3d^5 4s^0$$
In the free ion the five $3d$ electrons are spread over the five $3d$ orbitals
(all unpaired, by Hund's rule).

Step 3 - Consider the nature of the ligand.
$CN^-$ is a STRONG FIELD (low spin) ligand, high in the spectrochemical series.
It causes the $3d$ electrons to PAIR UP, because here $\Delta_0 &gt; P$.

Step 4 - Rearrange the electrons.
The five $3d$ electrons are forced into the three lower $t_{2g}$ orbitals:
$$t_{2g}^5\ e_g^0 \quad\Longrightarrow\quad (\uparrow\downarrow)\ (\uparrow\downarrow)\ (\uparrow)$$
This leaves TWO INNER $3d$ ORBITALS COMPLETELY VACANT.

Step 5 - Determine the hybridisation.
The two vacant inner $3d$ orbitals, together with one $4s$ and three $4p$ orbitals, are
used for bonding with the six $CN^-$ ligands:
$$\textbf{Hybridisation} = d^2sp^3 \quad \text{(inner orbital / low spin complex)}$$
Geometry: OCTAHEDRAL.

Step 6 - Determine the magnetic behaviour.
Number of unpaired electrons, $n = 1$.
$$\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} = 1.73\ \text{BM}$$

CONCLUSION: $[Fe(CN)_6]^{3-}$ is $d^2sp^3$ hybridised, octahedral, and PARAMAGNETIC
with one unpaired electron ($\mu = 1.73$ BM). It is an inner orbital, low spin complex.

(Compare: $[FeF_6]^{3-}$ has the weak field ligand $F^-$, so no pairing occurs. It is
$sp^3d^2$ hybridised with 5 unpaired electrons, $\mu = 5.92$ BM.)