Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory. [

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[CoF_6]^{3-}

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Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

Imagine you are a negatively charged electron sitting on a metal ion. All around you, the space is perfectly spherical — every direction feels the same. Your energy depends only on how far you are from the nucleus, not on which way you face.

Now imagine that six negative ions (or the negative ends of polar molecules) march in from the $x$ , $y$ , and $z$ axes and stop close to you. Suddenly, the space around you is no longer uniform. If you try to move straight toward one of these approaching ions, you feel a strong repulsion — that path costs extra energy. If you move between the axes (say, along a diagonal), you feel less repulsion because you are farther from the incoming charges.

This is the core intuition: when ligands approach a metal ion, they break the spherical symmetry of the space around the metal. Different directions in space are no longer equivalent. Electrons in orbitals that point directly at the ligands get pushed up in energy; electrons in orbitals that point between the ligands stay lower.

The Precise Statement

d

For the most common geometry — octahedral — here is what happens:

Six ligands sit at the corners of an octahedron, along the $+x$ , $-x$ , $+y$ , $-y$ , $+z$ , $-z$ axes.
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point their lobes directly along these axes. These are the $e_g$ set. They feel maximum repulsion → higher energy.
The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point their lobes between the axes (into the octahedral faces). These are the $t_{2g}$ set. They feel less repulsion → lower energy.

The energy gap between these two sets is denoted by $\Delta_o$ (or $10\,Dq$ ). The $t_{2g}$ set drops by $0.4\Delta_o$ and the $e_g$ set rises by $0.6\Delta_o$ , keeping the average energy unchanged (the "barycentre" rule).

Note

The labels $e_g$ and $t_{2g}$ come from group theory — they describe how the orbitals transform under the symmetry operations of an octahedron. You do not need to memorise the derivation, but the notation is standard in every exam.

Why This Matters

Crystal field splitting explains three things you will see repeatedly:

Colour — electrons can jump from $t_{2g}$ to $e_g$ by absorbing visible light. The gap $\Delta_o$ determines the colour you see.
Magnetism — if $\Delta_o$ is large, electrons pair up in the lower $t_{2g}$ set (low spin). If $\Delta_o$ is small, electrons spread out (high spin). This changes the number of unpaired electrons. …

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split in energy when placed in an electrostatic field created by surrounding ligands (anions or polar molecules). The key result is that five degenerate d-orbitals split into two or more sets with different energies. Let's understand why this happens.

1. The Starting Point: Degenerate d-Orbitals

In a free transition metal ion (no ligands), all five d-orbitals have the same energy (degenerate). Their shapes are:

$d_{xy}$ , $d_{xz}$ , $d_{yz}$ — lobes lie between the x, y, z axes (called $t_{2g}$ set in octahedral symmetry)
$d_{x^2-y^2}$ , $d_{z^2}$ — lobes point directly along the x, y, z axes (called $e_g$ set)

Key idea: The spatial orientation of each orbital determines how it interacts with approaching ligands.

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

Imagine six ligands approaching along the +x, –x, +y, –y, +z, –z axes (octahedral geometry).

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

Their lobes point directly at the ligands.
The negatively charged ligands repel the electron density in these orbitals.
This repulsion raises the energy of these orbitals — they become less stable (higher energy).

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

Their lobes point between the axes (e.g., $d_{xy}$ lobes lie in the xy-plane but at 45° to x and y).
They avoid the ligands — less repulsion.
Their energy is lower than the $e_g$ set.

The Splitting Pattern

\Delta_{\text{oct}} = E(e_g) - E(t_{2g})

Where:

$E(e_g)$ = energy of $d_{x^2-y^2}$ and $d_{z^2}$ (higher)
$E(t_{2g})$ = energy of $d_{xy}$ , $d_{xz}$ , $d_{yz}$ (lower)
$\Delta_{\text{oct}}$ is called the crystal field splitting energy (CFSE)

Why the name? The $e_g$ orbitals are "doubly degenerate" (2 orbitals), $t_{2g}$ are "triply degenerate" (3 orbitals). The letters come from group theory symmetry labels.

3. The Energy Conservation Rule

The total energy of all five d-orbitals must remain constant (no energy is created or destroyed). So:

The center of gravity (average energy) of the split set equals the original degenerate energy.
For octahedral splitting:
- 2 $e_g$ orbitals go up by $+0.6\Delta_{\text{oct}}$ each
- 3 $t_{2g}$ orbitals go down by $-0.4\Delta_{\text{oct}}$ each

Check:

2 \times (+0.6\Delta) + 3 \times (-0.4\Delta) = 1.2\Delta - 1.2\Delta = 0

This conservation of energy is a fundamental constraint — the splitting is not arbitrary.

4. The Tetrahedral Case: Why It's Opposite and Smaller

In a tetrahedral complex, four ligands approach from alternate corners of a cube. The axes are different:

The $d_{xy}$ , $d_{xz}$ , $d_{yz}$ orbitals now point closer to the ligands (more repulsion).
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point away from ligands (less repulsion).

Result:

$e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — lower energy
$t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — higher energy

The splitting is inverted compared to octahedral.

Magnitude:

\Delta_{\text{tet}} \approx \frac{4}{9} \Delta_{\text{oct}}

Why smaller?

Only 4 ligands (vs. 6) → less total repulsion.
Ligands are not directly along axes → weaker interaction. …

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Crystal Field Splitting theory explains how ligands affect d-orbital energies, determining whether a complex is high-spin or low-spin. For $[CoF_6]^{3-}$ : $t_{2g}^4 e_g^2$ (high-spin); for $[Fe(CN)_6]^{4-}$ : $t_{2g}^6 e_g^0$ (low-spin); for $[Cu(NH_3)_6]^{2+}$ : $t_{2g}^6 e_g^3$ (one unpaired electron).

The Core Idea: Why Ligands Matter

Crystal Field Splitting (CFS) is about how the arrangement of ligands around a central metal ion breaks the degeneracy of the d-orbitals. In an octahedral field, the five d-orbitals split into two sets: the lower-energy $t_{2g}$ set ( $d_{xy}, d_{xz}, d_{yz}$ ) and the higher-energy $e_g$ set ( $d_{z^2}, d_{x^2-y^2}$ ). The energy gap between them is called $\Delta_o$ (or 10 Dq).

The key question is: how do electrons fill these orbitals? That depends on two things:

The strength of the ligand (how large $\Delta_o$ is).
The pairing energy (P) — the energy cost to put two electrons in the same orbital.

If $\Delta_o$ is small (weak field ligands like $F^-$ ), electrons prefer to occupy all five orbitals singly first (Hund's rule) — this gives a high-spin configuration. If $\Delta_o$ is large (strong field ligands like $CN^-$ ), electrons pair up in the lower $t_{2g}$ set before occupying $e_g$ — this gives a low-spin configuration.

Watch out

A common mistake is to forget that the oxidation state of the metal changes the d-electron count. Always determine the metal's oxidation state first, then count d-electrons.

Let's work through each complex step by step.

1. $[CoF_6]^{3-}$

Step 1: Find the oxidation state of Co.

The complex has a net charge of $3-$ . Each $F^-$ ligand carries a $-1$ charge. Let the oxidation state of Co be $x$ .

x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3

So, cobalt is in the $+3$ oxidation state: $Co^{3+}$ .

Step 2: Determine the d-electron count.

Cobalt (atomic number 27) has the ground state configuration $[Ar] 3d^7 4s^2$ . For $Co^{3+}$ , we remove three electrons (the two 4s electrons and one 3d electron), leaving $3d^6$ . So, $Co^{3+}$ has 6 d-electrons.

Step 3: Identify the ligand and its field strength.

$F^-$ (fluoride) is a weak field ligand. It lies near the end of the spectrochemical series. This means $\Delta_o$ is small, and $\Delta_o < P$ (pairing energy). Electrons will not pair up unless forced.

Step 4: Fill the orbitals.

With 6 electrons and a small $\Delta_o$ , we follow Hund's rule: fill all five d-orbitals singly first, then pair up.

$t_{2g}$ set (3 orbitals): Put one electron in each → 3 electrons.
$e_g$ set (2 orbitals): Put one electron in each → 2 electrons.
Now we have placed 5 electrons. The 6th electron must go into the $t_{2g}$ set, pairing with an existing electron.

The configuration is: $t_{2g}^4 e_g^2$ .

Tip

For $d^6$ in a weak field, the high-spin configuration always has 4 unpaired electrons. You can quickly check: $t_{2g}^4$ has 2 unpaired (since one orbital is doubly occupied), and $e_g^2$ has 2 unpaired — total 4 unpaired electrons.

Final configuration for $[CoF_6]^{3-}$ : $t_{2g}^4 e_g^2$ (high-spin).

2. $[Fe(CN)_6]^{4-}$

Step 1: Find the oxidation state of Fe.

Net charge = $4-$ . Each $CN^-$ ligand carries a $-1$ charge. Let Fe's oxidation state be $x$ .

x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2

So, iron is $Fe^{2+}$ .

Step 2: Determine the d-electron count.

Iron (atomic number 26) has ground state $[Ar] 3d^6 4s^2$ . For $Fe^{2+}$ , remove two 4s electrons → $3d^6$ . So, $Fe^{2+}$ has 6 d-electrons.

Step 3: Identify the ligand and its field strength.

$CN^-$ (cyanide) is a strong field ligand. It lies near the top of the spectrochemical series. This means $\Delta_o$ is large, and $\Delta_o > P$ . Electrons will pair up in the lower $t_{2g}$ set before occupying $e_g$ .

Step 4: Fill the orbitals.

With 6 electrons and a large $\Delta_o$ , we fill the $t_{2g}$ set completely first.

$t_{2g}$ set: Put 2 electrons in each of the 3 orbitals → 6 electrons.
$e_g$ set: Empty.

The configuration is: $t_{2g}^6 e_g^0$ .

Note

This is a low-spin $d^6$ configuration. All 6 electrons are paired, so the complex is diamagnetic (no unpaired electrons). This is a classic example of how a strong ligand can force pairing.

Final configuration for $[Fe(CN)_6]^{4-}$ : $t_{2g}^6 e_g^0$ (low-spin).

--- …

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Crystal Field Splitting – Electronic Configuration Method

Method Name: Crystal Field Theory (CFT) – Filling of $t_{2g}$ and $e_g$ orbitals based on ligand field strength.

Steps to determine electronic configuration:

Determine the oxidation state of the central metal ion (using charge balance).
Write the d-electron count for the metal ion.
Identify the geometry (here all are octahedral).
Classify the ligand as weak field (high spin) or strong field (low spin) using the spectrochemical series.
Fill the $t_{2g}$ and $e_g$ orbitals following Hund’s rule and the energy gap ( $\Delta_o$ $Δ_{o}$ ):
- Weak field (small $\Delta_o$ ): electrons fill all orbitals singly before pairing (high spin).
- Strong field (large $\Delta_o$ ): electrons pair in $t_{2g}$ first (low spin).

1. $[CoF_6]^{3-}$

Oxidation state: $Co$ : $x + 6(-1) = -3 \Rightarrow x = +3$ → Co(III)
d-count: Co (atomic number 27) → Co $^{3+}$ = $[Ar]\,3d^6$
Ligand: $F^-$ is a weak field ligand → small $\Delta_o$ → high spin
Filling: $t_{2g}$ gets 4 electrons (one pair), $e_g$ gets 2 electrons (unpaired)

Electronic configuration:

\boxed{t_{2g}^4\, e_g^2 \quad (\text{high spin})}

2. $[Fe(CN)_6]^{4-}$

Oxidation state: $Fe$ : $x + 6(-1) = -4 \Rightarrow x = +2$ → Fe(II)
d-count: Fe (atomic number 26) → Fe $^{2+}$ = $[Ar]\,3d^6$
Ligand: $CN^-$ is a strong field ligand → large $\Delta_o$ → low spin
Filling: All 6 electrons pair up in $t_{2g}$ ; $e_g$ is empty

Electronic configuration:

\boxed{t_{2g}^6\, e_g^0 \quad (\text{low spin})}

3. $[Cu(NH_3)_6]^{2+}$

Oxidation state: $Cu$ : $x + 6(0) = +2 \Rightarrow x = +2$ → Cu(II)
d-count: Cu (atomic number 29) → Cu $^{2+}$ = $[Ar]\,3d^9$ …

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Here are the most common mistakes students make when applying Crystal Field Splitting Theory (CFT) to determine electronic configurations, along with how to avoid each.

Mistake 1: Forgetting to Determine the Oxidation State of the Central Metal Ion

This is the most fundamental error. The electronic configuration of the free metal ion is the starting point, and you cannot get it right without the correct charge on the metal.

The Mistake: Students directly use the configuration of the neutral atom (e.g., Fe for $[Fe(CN)_6]^{4-}$ ) instead of the ion ( $Fe^{2+}$ ).
How to Avoid:
1. Always calculate the oxidation state first. Let the charge on the complex = (charge on metal) + (charge on ligands × number of ligands).
2. For $[Fe(CN)_6]^{4-}$ : CN⁻ is a -1 ligand. So, $x + 6(-1) = -4 \implies x = +2$ . The metal ion is $Fe^{2+}$ .
3. Write the configuration of the free ion. $Fe$ (atomic number 26): $[Ar] 3d^6 4s^2$ . For $Fe^{2+}$ , remove the 4s electrons first: $[Ar] 3d^6$ .
4. Key Exam Tip: The 4s orbital is always filled before 3d in the atom, but emptied before 3d when forming ions.

Mistake 2: Confusing Strong-Field and Weak-Field Ligands

The entire splitting pattern (high-spin vs. low-spin) depends on whether the ligand is a strong-field (low-spin) or weak-field (high-spin) ligand.

The Mistake: Assuming all ligands behave the same way, or misidentifying ligands like CN⁻ (strong) and F⁻ (weak).
How to Avoid:
- Memorize the Spectrochemical Series (partial list):
  - Strong field (low spin): $CN^- > CO > en > NH_3$
  - Weak field (high spin): $H_2O > F^- > Cl^- > Br^- > I^-$
- Apply it:
  - $[CoF_6]^{3-}$ : $F^-$ is a weak field ligand → high spin complex.
  - $[Fe(CN)_6]^{4-}$ : $CN^-$ is a strong field ligand → low spin complex.
  - $[Cu(NH_3)_6]^{2+}$ : $NH_3$ is a strong field ligand → low spin complex.

Mistake 3: Incorrectly Filling Electrons in the $t_{2g}$ and $e_g$ Orbitals

This is where the actual configuration is written. The rules of Hund's rule and Pauli's exclusion principle still apply, but the order of filling depends on the ligand field.

The Mistake: For $d^4$ , $d^5$ , $d^6$ , or $d^7$ configurations, students often fill all $t_{2g}$ orbitals first (pairing them) even for weak-field ligands, or they spread electrons out in $e_g$ before pairing in $t_{2g}$ for strong-field ligands.
How to Avoid:
- For weak field (small $\Delta_o$ ): Electrons fill all five orbitals singly first (Hund's rule) before pairing. This gives the maximum number of unpaired electrons.
  - Example: $Co^{3+}$ is $d^6$ . For $[CoF_6]^{3-}$ (weak field): $t_{2g}^4 e_g^2$ (4 unpaired electrons).
- For strong field (large $\Delta_o$ ): Electrons fill the lower-energy $t_{2g}$ $t_{2 g}$ orbitals completely first (pairing them) before any electron goes into the higher-energy $e_g$ $e_{g}$ orbitals. This gives the minimum number of unpaired electrons.
  - Example: $Fe^{2+}$ is $d^6$ . For $[Fe(CN)_6]^{4-}$ (strong field): $t_{2g}^6 e_g^0$ (0 unpaired electrons).

Mistake 4: Forgetting Jahn-Teller Distortions for $d^9$ and $d^4$ (high-spin) / $d^7$ (low-spin)

For $[Cu(NH_3)_6]^{2+}$ , the $Cu^{2+}$ ion is $d^9$ . A perfectly octahedral $d^9$ complex is unstable and undergoes a Jahn-Teller distortion. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2026Set ANNUAL1 mark
$Q. What is the number of unpaired electrons in the complex ion [Fe(CN)6]4-?$
›Reveal solutionSolution

Fe in [Fe(CN)6]4- is Fe2+ (d6); CN- is a strong-field ligand causing all 6 d-electrons to pair up in the lower t2g set (low spin), leaving no unpaired electrons.

In [Fe(CN)6]4-, the overall charge is 4-, and CN- contributes -1 each (6 x -1 = -6), so iron must be Fe2+ (electronic configuration [Ar]3d6). CN- is a strong field ligand (high in the spectrochemical series), so it causes a large crystal field splitting (delta-o) in this octahedral complex. This favours pairing of electrons in the lower-energy t2g orbitals rather than promotion to the higher-energy eg orbitals (low-spin confi …

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Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markMCQ
$Q. For the complex ion [CoF6]3-, the number of unpaired electrons is -- (Co = 27)$
(a) 2
(b) 3
(c) 4
(d) 0
›Reveal solutionSolution

Co in [CoF6]3- is Co(III), a d6 ion; F- is a weak-field ligand giving a high-spin octahedral complex with 4 unpaired electrons.

Step 1 - oxidation state of Co. The complex ion has overall charge -3. Six F- ligands contribute -6. So: Co + 6(-1) = -3, giving Co = +3.

Step 2 - d-electron count. Co (Z = 27) has configuration [Ar]3d^7 4s^2. Co3+ is formed by removing 3 electrons (2 from 4s, 1 from 3d), giving 3d^6, i.e. a d6 ion.

Step 3 - field strength of F-. F- is a weak-field ligand (low in the spectrochemical series), so it cannot force pairing of electrons. The complex is therefore high-spin.

…

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 mark
$Q. [Ti(H2O)6]3+ is coloured, but [Sc(H2O)6]3+ is colourless — explain.$
›Reveal solutionSolution

Colour in transition-metal complexes arises from d-d electronic transitions; this requires the metal ion to have partially filled d-orbitals (neither empty nor completely full). Ti3+ (3d1) has such an electron and is coloured; Sc3+ (3d0) has none and is colourless.

[Ti(H2O)6]3+: Titanium (Z=22) has configuration [Ar]3d2 4s2; Ti3+ loses 3 electrons (2 from 4s, 1 from 3d) to give [Ar]3d1 - i.e., ONE d-electron. In the octahedral aqua complex, the surrounding water ligands split the d-orbitals into lower t2g and higher eg sets. The single d-electron can absorb a specific wavelength of visible light and get excited from t2g to eg (a d-d transition). The complementary colour of the absorbed light is what we observe (violet colour for this ion).

…

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Q.Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory. [CoF6]3−[CoF_6]^{3-}[CoF6​]3−, [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4− and [Cu(NH3)6]2+[Cu(NH_3)_6]^{2+}[Cu(NH3​)6​]2+.

The Core Idea: Why Ligands Matter

1. [CoF6]3−[CoF_6]^{3-}[CoF6​]3−

2. [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−

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Q.Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory.
$[CoF_6]^{3-}$ , $[Fe(CN)_6]^{4-}$ and $[Cu(NH_3)_6]^{2+}$ .

1. $[CoF_6]^{3-}$

2. $[Fe(CN)_6]^{4-}$

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