Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code: Column I (Complex ion): A. $[Cr(H_2O)_6]^{3+}$ B. $[Co(CN)_4]^{2-}$ C. $[Ni(NH_3)_6]^{2+}$ D. $[MnF_6]^{4-}$ Column II (Hybridisation, number of unpaired electrons): 1. $dsp^2$, 1 2. $sp^3d^2$, 5 3. $d^2sp^3$, 3 4. $sp^3$, 4 5. $sp^3d^2$, 2

> [!TLDR] > The key is to determine the oxidation state, electron configuration, and ligand field strength for each complex, then decide the hybridisation and count unpaired electrons. The correct matching is A-3, B-1, C-5, D-2. Let’s unpack each complex one by one. The central idea is **Crystal Field Theory** — ligands create an electric field that splits the d-orbitals into two energy levels. Strong-field ligands (like CN⁻, NH₃) cause a large splitting, forcing electrons to pair up in lower orbitals (low spin). Weak-field ligands (like H₂O, F⁻) cause a small splitting, so electrons occupy all orbitals singly first (high spin). The hybridisation depends on how many empty orbitals are available after pairing. --- ### 1. Complex A: $[Cr(H_2O)_6]^{3+}$ - **Oxidation state**: Water is neutral, so Cr must be +3. Cr atomic number = 24, so Cr³⁺ has $24 - 3 = 21$ electrons. Electronic configuration of Cr: $[Ar]\,3d^5\,4s^1$. Removing three electrons (first from 4s, then two from 3d) gives Cr³⁺: $[Ar]\,3d^3$. - **Ligand field**: H₂O is a weak-field ligand (small splitting). So electrons remain unpaired as much as possible. With three d-electrons, they occupy three separate t₂g orbitals (Hund’s rule). No pairing occurs. - **Hybridisation**: The complex is octahedral (six ligands). The metal uses two d-orbitals (from inner 3d), one s, and three p orbitals — that’s $d^2sp^3$ hybridisation. Since the d-orbitals used are from the inner shell (3d), it’s inner orbital complex. - **Unpaired electrons**: Three unpaired electrons. > [!WARNING] > A common mistake is to think Cr³⁺ has 4s electrons left. Always remove 4s electrons first when forming cations. So A matches with **3** ($d^2sp^3$, 3). --- ### 2. Complex B: $[Co(CN)_4]^{2-}$ - **Oxidation state**: CN⁻ is −1 each, four of them give −4. Overall charge is −2, so Co must be +2. Co atomic number = 27, Co²⁺ has $27 - 2 = 25$ electrons. Co: $[Ar]\,3d^7\,4s^2$, remove two 4s electrons → $[Ar]\,3d^7$. - **Ligand field**: CN⁻

Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Colu

NCERT Exemplar · Q43

[Cr(H_2O)_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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The key is to determine the oxidation state, electron configuration, and ligand field strength for each complex, then decide the hybridisation and count unpaired electrons. The correct matching is A-3, B-1, C-5, D-2.

Let’s unpack each complex one by one. The central idea is Crystal Field Theory — ligands create an electric field that splits the d-orbitals into two energy levels. Strong-field ligands (like CN⁻, NH₃) cause a large splitting, forcing electrons to pair up in lower orbitals (low spin). Weak-field ligands (like H₂O, F⁻) cause a small splitting, so electrons occupy all orbitals singly first (high spin). The hybridisation depends on how many empty orbitals are available after pairing.

1. Complex A: $[Cr(H_2O)_6]^{3+}$

Oxidation state: Water is neutral, so Cr must be +3. Cr atomic number = 24, so Cr³⁺ has $24 - 3 = 21$ electrons. Electronic configuration of Cr: $[Ar]\,3d^5\,4s^1$ . Removing three electrons (first from 4s, then two from 3d) gives Cr³⁺: $[Ar]\,3d^3$ .
Ligand field: H₂O is a weak-field ligand (small splitting). So electrons remain unpaired as much as possible. With three d-electrons, they occupy three separate t₂g orbitals (Hund’s rule). No pairing occurs.
Hybridisation: The complex is octahedral (six ligands). The metal uses two d-orbitals (from inner 3d), one s, and three p orbitals — that’s $d^2sp^3$ hybridisation. Since the d-orbitals used are from the inner shell (3d), it’s inner orbital complex.
Unpaired electrons: Three unpaired electrons.

Watch out

A common mistake is to think Cr³⁺ has 4s electrons left. Always remove 4s electrons first when forming cations.

So A matches with 3 ( $d^2sp^3$ , 3).

2. Complex B: $[Co(CN)_4]^{2-}$

Oxidation state: CN⁻ is −1 each, four of them give −4. Overall charge is −2, so Co must be +2. Co atomic number = 27, Co²⁺ has $27 - 2 = 25$ electrons. Co: $[Ar]\,3d^7\,4s^2$ , remove two 4s electrons → $[Ar]\,3d^7$ .
Ligand field: CN⁻ is a very strong-field ligand. It causes large splitting, forcing electrons to pair up. For a d⁷ system in a strong field, the configuration becomes $(t_{2g})^6 (e_g)^1$ — six electrons paired in t₂g, one in e_g. That gives one unpaired electron.
Hybridisation: $[Co(CN)_4]^{2-}$ is a well-known exception to the usual four-coordinate geometry: because CN⁻ is such a strong-field ligand, the low-spin $d^7$ configuration is stabilised as square planar rather than tetrahedral, using $dsp^2$ hybridisation ( $d_{x^2-y^2}$ combined with one $s$ and two $p$ orbitals). This leaves one unpaired electron in another d-orbital.
Unpaired electrons: One.

So B matches with 1 ( $dsp^2$ , 1).

Tip

For $[Co(CN)_4]^{2-}$ , remember it’s square planar, not tetrahedral — a classic exam trap. The strong field of CN⁻ overrides the usual tetrahedral preference for four-coordinate Co²⁺.

3. Complex C: $[Ni(NH_3)_6]^{2+}$

Oxidation state: NH₃ is neutral, so Ni is +2. Ni atomic number = 28, Ni²⁺ has $28 - 2 = 26$ electrons. Ni: $[Ar]\,3d^8\,4s^2$ , remove two 4s → $[Ar]\,3d^8$ . …

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Method: Crystal Field Theory (CFT) + Electronic Configuration Approach

This method uses the oxidation state of the central metal ion, its d-electron count, the nature of the ligand (strong or weak field), and the resulting crystal field splitting to determine hybridisation and number of unpaired electrons.

Steps

Find oxidation state of the metal

Use the charge of the complex and known charges of ligands.
Determine d-electron count

From the electronic configuration of the metal in that oxidation state.
Classify ligands as strong or weak field
- Strong field (low spin): CN⁻, CO, NH₃ (for some metals)
- Weak field (high spin): H₂O, F⁻, Cl⁻, etc.
Apply crystal field splitting
- For octahedral complexes:
  - Weak field → electrons fill all five d-orbitals singly first (Hund’s rule) → high spin
  - Strong field → electrons pair up in lower $t_{2g}$ orbitals → low spin
- For tetrahedral complexes: always high spin (small splitting).
Determine hybridisation
- Octahedral: $sp^3d^2$ (outer) or $d^2sp^3$ (inner)
- Square planar: $dsp^2$
- Tetrahedral: $sp^3$
Count unpaired electrons from the d-orbital filling.

Applying to each complex

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

Oxidation state: Cr = +3
d-count: Cr³⁺ = $3d^3$
Ligand: H₂O (weak field)
Splitting: Octahedral, weak field → all three electrons unpaired in $t_{2g}$
Hybridisation: $d^2sp^3$ (inner orbital, uses two 3d orbitals)
Unpaired electrons: 3
Match: 3 in Column II

B. $[Co(CN)_4]^{2-}$

Oxidation state: Co = +2
d-count: Co²⁺ = $3d^7$
Ligand: CN⁻ (very strong field)
Geometry: With CN⁻ this strong, $[Co(CN)_4]^{2-}$ is actually square planar, not tetrahedral — the strong field favours maximum pairing, which square-planar geometry accommodates better than tetrahedral.
Splitting: Square planar, low spin → electrons pair up, leaving one $d$ -orbital singly occupied.
Hybridisation: $dsp^2$
Unpaired electrons: 1
Correct match: 1 ( $dsp^2$ , 1)

C. $[Ni(NH_3)_6]^{2+}$ …

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✗ Mistake 1: Forgetting to check the oxidation state of the central metal ion

Why it happens:

Students jump straight to the electronic configuration of the neutral atom, without adjusting for charge.

Example:

For $[Cr(H_2O)_6]^{3+}$ , many write Cr as $[Ar] 3d^5 4s^1$ and then get confused.

How to avoid:

Always find the oxidation state first.

Cr in $[Cr(H_2O)_6]^{3+}$ : H₂O is neutral → Cr must be +3. Cr³⁺: remove 3 electrons → $[Ar] 3d^3$ .

Now apply CFT:

For a $d^3$ ion, all three electrons occupy the three $t_{2g}$ orbitals singly (Hund's rule), regardless of ligand field strength — there are no $e_g$ electrons to pair up or displace.
Unpaired electrons = 3.
Hybridisation: since the $e_g$ orbitals stay empty for $d^3$ , the complex always uses the inner $3d$ orbitals — $d^2sp^3$ (inner orbital), never $sp^3d^2$ .

✓ Correct match: A → 3 ( $d^2sp^3$ , 3 unpaired electrons).

✗ Mistake 2: Confusing tetrahedral vs square planar geometry

Why it happens:

For $[Co(CN)_4]^{2-}$ , students see CN⁻ (strong field) and assume tetrahedral.

How to avoid:

CN⁻ is a strong field ligand → causes pairing.
Co in $[Co(CN)_4]^{2-}$ :

CN⁻ is −1 each → total −4. Charge on complex is −2.

So Co must be +2.

Co²⁺: $[Ar] 3d^7$ .
For 4-coordinate complexes:
- If strong field and $d^8$ or $d^7$ (low spin), geometry is often square planar (to maximise pairing).
- Square planar uses $dsp^2$ hybridisation.
$d^7$ with strong field:

Pairing occurs → 1 unpaired electron remains.

✓ Correct match: B → 1 ( $dsp^2$ , 1).

✗ Mistake 3: Ignoring ligand field strength for Ni²⁺ complexes

Why it happens:

Students assume all octahedral Ni²⁺ complexes are $sp^3d^2$ with 2 unpaired electrons.

How to avoid:

Ni in $[Ni(NH_3)_6]^{2+}$ :

NH₃ is neutral → Ni is +2.

Ni²⁺: $[Ar] 3d^8$ .
NH₃ is a moderate field ligand — but for Ni²⁺, it’s strong enough to cause pairing?

Actually, for $d^8$ in octahedral field:
- Weak field: $t_{2g}^6 e_g^2$ → 2 unpaired, $sp^3d^2$ .
- Strong field: $t_{2g}^6 e_g^2$ (same — because $d^8$ has no choice: $e_g$ always has 2 electrons, both unpaired). So no pairing occurs regardless of ligand strength for $d^8$ octahedral.
So: 2 unpaired electrons, $sp^3d^2$ hybridisation.

✓ Correct match: C → 5 ( $sp^3d^2$ , 2).

✗ Mistake 4: Misidentifying Mn oxidation state and spin state

Why it happens:

Students forget that F⁻ is a weak field ligand and assume pairing.

How to avoid:

$[MnF_6]^{4-}$ :

F⁻ is −1 each → total −6. Complex charge is −4.

So Mn must be +2.

Mn²⁺: $[Ar] 3d^5$ .
F⁻ is a weak field ligand → high spin. …

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

2. Complex B: [Co(CN)4]2−[Co(CN)_4]^{2-}[Co(CN)4​]2−

3. Complex C: [Ni(NH3)6]2+[Ni(NH_3)_6]^{2+}[Ni(NH3​)6​]2+

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1. Complex A: $[Cr(H_2O)_6]^{3+}$

2. Complex B: $[Co(CN)_4]^{2-}$

3. Complex C: $[Ni(NH_3)_6]^{2+}$

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