Using valence bond theory, explain the following in relation to the complexes given below: $[Mn(CN)_6]^{3-}$, $[Co(NH_3)_6]^{3+}$, $[Cr(H_2O)_6]^{3+}$, $[FeCl_6]^{4-}$ (i) Type of hybridisation. (ii) Inner or outer orbital complex. (iii) Magnetic behaviour. (iv) Spin only magnetic moment value.

Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)₆]³⁻, [Co(NH

NCERT Exemplar · Q28

[Mn(CN)_6]^{3-}

(i) Type of hybridisation.

(ii) Inner or outer orbital complex.

(iii) Magnetic behaviour.

(iv) Spin only magnetic moment value.

Yanam BieapLong· 5mImportance★★★★★

75% · 76/101 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Concept: Valence Bond Theory — Hybridisation, Inner/Outer Orbital Complexes, and Magnetic Behaviour

Step 1 – Determine oxidation state and $d$ -electron count

$[Mn(CN)_6]^{3-}$ : Mn is +3 → $d^4$
$[Co(NH_3)_6]^{3+}$ : Co is +3 → $d^6$
$[Cr(H_2O)_6]^{3+}$ : Cr is +3 → $d^3$
$[FeCl_6]^{4-}$ : Fe is +2 → $d^6$

Step 2 – Identify ligand field strength and pairing

CN⁻, NH₃ are strong field → cause pairing (low spin)
H₂O is intermediate; for Cr³⁺ ( $d^3$ ) no pairing possible (all orbitals half-filled)
Cl⁻ is weak field → no pairing (high spin)

Step 3 – Assign hybridisation, orbital type, and magnetic moment

Complex	$d$ -count	Field	Hybridisation	Inner/Outer	Unpaired e⁻	$\mu_s$ (B.M.)
$[Mn(CN)_6]^{3-}$	$d^4$	strong	$d^2sp^3$	inner	2	$\sqrt{2(2+2)} = 2.83$
$[Co(NH_3)_6]^{3+}$	$d^6$	strong	$d^2sp^3$	inner	0	0 (diamagnetic)

Start your 14-day free trial to unlock the full solution →

Using valence bond theory, the hybridisation, orbital type, magnetic behaviour, and spin-only moment of each complex are determined by the oxidation state, ligand field strength, and electron configuration of the central metal ion. The results are: $[Mn(CN)_6]^{3-}$ — $d^2sp^3$ , inner, paramagnetic, $\sqrt{8}$ BM; $[Co(NH_3)_6]^{3+}$ — $d^2sp^3$ , inner, diamagnetic, 0 BM; $[Cr(H_2O)_6]^{3+}$ — $d^2sp^3$ , inner, paramagnetic, $\sqrt{15}$ BM; $[FeCl_6]^{4-}$ — $sp^3d^2$ , outer, paramagnetic, $\sqrt{24}$ BM.

Valence bond theory (VBT) treats bonding in coordination complexes as the overlap of ligand lone pairs with hybridised orbitals on the central metal ion. The key idea is that the metal ion uses empty orbitals (from the $3d$ , $4s$ , and $4p$ sets) to accept electron pairs from ligands. The number and type of hybrid orbitals formed depend on the coordination number (here, 6 for all complexes, so octahedral geometry). But the crucial twist is whether the metal uses its inner $d$ -orbitals ( $(n-1)d$ ) or outer $d$ -orbitals ( $nd$ ) — this decides if the complex is inner orbital (low-spin) or outer orbital (high-spin). The magnetic behaviour follows directly from the number of unpaired electrons left in the $d$ -orbitals after hybridisation.

Let’s work through each complex step by step.

1. $[Mn(CN)_6]^{3-}$

Step 1: Determine the oxidation state and $d$ -electron count.

Mn is in the +3 oxidation state (since each CN⁻ is -1, total ligand charge = -6, complex charge = -3, so Mn must be +3). Mn atomic number = 25, so Mn $^{3+}$ has $25 - 3 = 22$ electrons. The electron configuration of Mn is $[Ar] 3d^5 4s^2$ ; removing three electrons (from 4s first, then 3d) gives $[Ar] 3d^4$ . So Mn $^{3+}$ has 4 $d$ -electrons.

Step 2: Identify ligand strength and decide inner vs. outer.

CN⁻ is a strong field ligand. It causes large crystal field splitting, forcing electrons to pair up in the lower $t_{2g}$ orbitals before occupying $e_g$ . With 4 electrons in the three $t_{2g}$ orbitals: the first three fill singly (Hund's rule), and the fourth pairs up with one of them. That gives 2 unpaired electrons (one orbital has a pair, the other two orbitals have one electron each).

Step 3: Hybridisation.

Since the ligand is strong, the metal uses inner $d$ -orbitals: two of the $3d$ orbitals are empty (the $e_g$ set) and can be hybridised with $4s$ and $4p$ to form $d^2sp^3$ hybridisation. So it’s an inner orbital complex.

Step 4: Magnetic behaviour and spin-only moment.

With 2 unpaired electrons, the complex is paramagnetic. Spin-only moment: $\mu = \sqrt{n(n+2)} = \sqrt{2 \times 4} = \sqrt{8} \approx 2.83$ BM.

Watch out

A common mistake is to think $d^4$ in strong field gives 4 unpaired electrons (like in weak field). Remember: strong field causes pairing, so $t_{2g}^4$ has only 2 unpaired.

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

Step 1: Oxidation state and $d$ -electron count.

NH $_3$ is neutral, so Co must be +3 to balance the 3+ charge on the complex. Co atomic number = 27, Co $^{3+}$ has $27 - 3 = 24$ electrons. Co ground state: $[Ar] 3d^7 4s^2$ ; remove 3 electrons → $[Ar] 3d^6$ . So Co $^{3+}$ is $d^6$ .

Step 2: Ligand strength.

NH $_3$ is a moderate field ligand, but for Co $^{3+}$ it acts as strong field (Co $^{3+}$ has high charge, so it’s a good electron pair acceptor, enhancing splitting). So it’s low-spin: all 6 electrons pair up in $t_{2g}$ ( $t_{2g}^6$ ). That gives 0 unpaired electrons.

Step 3: Hybridisation.

Since it’s low-spin, the $e_g$ orbitals are empty, so inner $d$ -orbitals are used: $d^2sp^3$ hybridisation. Inner orbital complex.

Step 4: Magnetic behaviour.

Diamagnetic (no unpaired electrons). Spin-only moment = 0 BM.

Tip

Co $^{3+}$ is one of the few cases where NH $_3$ (usually intermediate) behaves as a strong field ligand due to the high oxidation state. Always check the metal’s charge — it influences the splitting.

3. $[Cr(H_2O)_6]^{3+}$

Step 1: Oxidation state and $d$ -electron count.

H $_2$ O is neutral, so Cr is +3. Cr atomic number = 24, Cr $^{3+}$ has $24 - 3 = 21$ electrons. Cr ground state: $[Ar] 3d^5 4s^1$ ; remove 3 electrons → $[Ar] 3d^3$ . So $d^3$ .

Step 2: Ligand strength.

H $_2$ O is a weak field ligand. For $d^3$ , regardless of field strength, the three electrons occupy all three $t_{2g}$ orbitals singly (Hund’s rule). So you get 3 unpaired electrons — no pairing possible because you’d need to put two in one orbital, but that’s less stable. So it’s high-spin.

Step 3: Hybridisation.

For $d^3$ , the two $e_g$ orbitals stay completely empty no matter how strong or weak the ligand field is (there are only three electrons, and they occupy the three $t_{2g}$ orbitals singly by Hund's rule). Those two empty inner $(n-1)d$ orbitals are always available to combine with $4s$ and $4p$ , giving $d^2sp^3$ hybridisation. So $[Cr(H_2O)_6]^{3+}$ is an inner orbital complex even though it is high-spin — for $d^1$ , $d^2$ , and $d^3$ ions, "inner orbital" and "high-spin" are not mutually exclusive, because no electron pairing is ever needed to keep the $e_g$ set empty.

Step 4: Magnetic behaviour.

Paramagnetic with 3 unpaired electrons. Spin-only moment: $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM.

Note

…

Start your 14-day free trial to unlock the full solution →

Crystal Field Splitting & Valence Bond Theory Approach

Method Used: Valence Bond Theory (VBT) with Crystal Field Splitting concepts for predicting hybridisation, magnetic behaviour, and spin-only moment.

Step 1: Determine oxidation state and electronic configuration of central metal ion

Complex	Metal ion	Configuration
$[Mn(CN)_6]^{3-}$	$Mn^{3+}$	$3d^4$
$[Co(NH_3)_6]^{3+}$	$Co^{3+}$	$3d^6$
$[Cr(H_2O)_6]^{3+}$	$Cr^{3+}$	$3d^3$
$[FeCl_6]^{4-}$	$Fe^{2+}$	$3d^6$

Step 2: Identify ligand strength and decide pairing (strong vs weak field)

Strong field ligands (CN⁻, NH₃) → cause pairing → low spin complexes
Weak field ligands (H₂O, Cl⁻) → no pairing → high spin complexes

Step 3: Apply VBT — decide hybridisation and orbital type

Rule:

If inner $d$ -orbitals ( $(n-1)d$ ) are used → inner orbital complex (low spin)
If outer $d$ -orbitals ( $nd$ ) are used → outer orbital complex (high spin)

Step 4: Calculate spin-only magnetic moment

Formula:

\mu_s = \sqrt{n(n+2)} \, \text{BM}

Where $n$ = number of unpaired electrons.

Final Results Table

Complex	Hybridisation	Type	Unpaired e⁻	Magnetic behaviour	$\mu_s$ (BM)
$[Mn(CN)_6]^{3-}$	$d^2sp^3$	Inner orbital	2	Paramagnetic	$\sqrt{2(2+2)} = \mathbf{2.83}$
$[Co(NH_3)_6]^{3+}$	$d^2sp^3$	Inner orbital	0	Diamagnetic	$\mathbf{0}$
$[Cr(H_2O)_6]^{3+}$	$d^2sp^3$	Inner orbital	3	Paramagnetic	$\sqrt{3(3+2)} = \mathbf{3.87}$

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make when solving Crystal Field Splitting / Valence Bond Theory problems for coordination complexes, along with how to avoid each.

Mistake 1: Confusing Inner vs. Outer Orbital Complexes

The Mistake:

Students often think “inner orbital” means the complex uses only inner d-orbitals (like 3d) and “outer orbital” means it uses outer d-orbitals (like 4d). In reality, the classification depends on which d-orbitals are used for hybridisation — $(n-1)d$ (inner) vs. $nd$ (outer).

How to Avoid:

Inner orbital complex: Uses $(n-1)d$ orbitals for $d^2sp^3$ hybridisation.
Outer orbital complex: Uses $nd$ orbitals for $sp^3d^2$ hybridisation.
Check the oxidation state and number of d-electrons first. Then see if pairing occurs (strong field ligand → inner) or not (weak field ligand → outer).

Example:

$[Mn(CN)_6]^{3-}$ : Mn is +3, $d^4$ configuration. CN⁻ is strong field → pairing occurs → uses $(n-1)d$ → inner orbital.

Mistake 2: Forgetting to Determine the Oxidation State Correctly

The Mistake:

Students directly count d-electrons from the neutral atom without adjusting for the charge on the complex.

How to Avoid:

Let the metal’s oxidation state be $x$ .
Sum of charges of ligands + $x$ = overall charge on complex.
Then find $d^n$ configuration for that oxidation state.

Example:

$[FeCl_6]^{4-}$ : Cl⁻ has charge −1 each. So $x + 6(-1) = -4 \Rightarrow x = +2$ . Fe²⁺ is $d^6$ .

Mistake 3: Misidentifying Strong vs. Weak Field Ligands

The Mistake:

Students memorise a short list but forget that CN⁻ is strong, NH₃ is intermediate (strong for 3d⁶ Co³⁺), H₂O is weak (except for a few cases), and Cl⁻ is weak.

How to Avoid:

Use the spectrochemical series (partial): $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- < \text{CO}$
Strong field → pairing → low spin.
Weak field → no pairing → high spin.

Example:

$[Co(NH_3)_6]^{3+}$ : NH₃ is strong enough to pair electrons in Co³⁺ ( $d^6$ ) → low spin.

Mistake 4: Incorrect Hybridisation for $d^2sp^3$ vs. $sp^3d^2$

The Mistake:

Students write $d^2sp^3$ for outer orbital complexes or $sp^3d^2$ for inner orbital complexes.

How to Avoid:

Inner orbital: $(n-1)d$ orbitals used → hybridisation is $d^2sp^3$ (note: d comes before s).
Outer orbital: $nd$ orbitals used → hybridisation is $sp^3d^2$ (s before p before d).
Always write the hybridisation in the correct order: $d^2sp^3$ (inner) vs. $sp^3d^2$ (outer).

Mistake 5: Calculating Spin-Only Magnetic Moment Wrong

The Mistake:

Using the wrong number of unpaired electrons ( $n$ ) because of incorrect pairing assumptions.

How to Avoid:

First decide high spin or low spin based on ligand field strength.
Then count unpaired electrons in the $d$ orbitals.
Use formula:

\mu_s = \sqrt{n(n+2)} \, \text{BM}

Example:

$[Cr(H_2O)_6]^{3+}$ : Cr³⁺ is $d^3$ , H₂O is weak → no pairing → $n=3$ → …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ27 — Using crystal field theory, draw energy level diagram, write electroni…→Next questionQ29 — CoSO₄Cl · 5NH₃ exists in two isomeric forms 'A' and 'B'. Isomer 'A' re…

Q.Using valence bond theory, explain the following in relation to the complexes given below: [Mn(CN)6]3−[Mn(CN)_6]^{3-}[Mn(CN)6​]3−, [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+, [Cr(H2O)6]3+[Cr(H_2O)_6]^{3+}[Cr(H2​O)6​]3+, [FeCl6]4−[FeCl_6]^{4-}[FeCl6​]4−

1. [Mn(CN)6]3−[Mn(CN)_6]^{3-}[Mn(CN)6​]3−

2. [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+

3. [Cr(H2O)6]3+[Cr(H_2O)_6]^{3+}[Cr(H2​O)6​]3+

🎓Unlock everything free for 14 days

Q.Using valence bond theory, explain the following in relation to the complexes given below:
$[Mn(CN)_6]^{3-}$ , $[Co(NH_3)_6]^{3+}$ , $[Cr(H_2O)_6]^{3+}$ , $[FeCl_6]^{4-}$

1. $[Mn(CN)_6]^{3-}$

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

3. $[Cr(H_2O)_6]^{3+}$

Unlock everything free for 14 days