(a) Although both $[NiCl_4]^{2-}$ and $[Ni(CO)_4]$ have $sp^3$ hybridisation yet $[NiCl_4]^{2-}$ is paramagnetic and $[Ni(CO)_4]$ is diamagnetic. Give reason. (Atomic no. of Ni = 28) (b) Write the electronic configuration of $d^5$ on the basis of crystal field theory when (i) $\Delta_o P$

> [!TLDR] > The magnetic behaviour difference arises from the **ligand field strength**: weak-field Cl⁻ leaves Ni²⁺ with two unpaired electrons (paramagnetic), while strong-field CO forces pairing in Ni⁰ (diamagnetic). For a $d^5$ ion, $\Delta_o P$ gives low-spin with 1 unpaired electron. --- ### Part (a): Why $[NiCl_4]^{2-}$ is paramagnetic but $[Ni(CO)_4]$ is diamagnetic The key is not just hybridisation — it’s the **electronic configuration of the metal** and the **strength of the ligand field**. **Step 1: Determine the oxidation state and $d$-electron count for each complex.** - In $[NiCl_4]^{2-}$, each Cl⁻ has a −1 charge. Let Ni have oxidation state $x$: $x + 4(-1) = -2 \implies x = +2$. Ni (atomic number 28) has ground-state configuration $[Ar]\,3d^8 4s^2$. Ni²⁺ loses the two 4s electrons: $3d^8$. - In $[Ni(CO)_4]$, CO is a neutral ligand. Let Ni have oxidation state $x$: $x + 4(0) = 0 \implies x = 0$. Ni⁰ has configuration $3d^8 4s^2$ — but in complexes, the 4s electrons often move to 3d, giving $3d^{10}$. > [!NOTE] > CO is a **strong field ligand** (high up in the spectrochemical series). It causes significant pairing of electrons. Cl⁻ is a **weak field ligand** (low in the spectrochemical series) and does not force pairing. **Step 2: Apply crystal field theory for tetrahedral geometry.** Both complexes are tetrahedral ($sp^3$ hybridisation). In a tetrahedral field, the $d$ orbitals split into two sets: - Lower energy: $e$ set ($d_{x^2-y^2}, d_{z^2}$) - Higher energy: $t_2$ set ($d_{xy}, d_{yz}, d_{zx}$) The splitting energy is $\Delta_t$, which is roughly $4/9$ of $\Delta_o$ (octahedral splitting). Because $\Delta_t$ is small, **tetrahedral complexes are almost always high-spin** — except when the ligand is extremely strong. **Step 3: Fill electrons for $[NiCl_4]^{2-}$ (Ni²⁺, $3d^8$, weak field).** For $d^8$ in a tetrahedral field, the filling follows Hund’s r

(a) Although both [NiCl₄]²⁻ and [Ni(CO)₄] have sp³ hybridisation yet [NiCl₄]²⁻ is paramagnetic and [Ni(CO)₄] i

Question

[NiCl_4]^{2-}

(b) Write the electronic configuration of

d^5

on the basis of crystal field theory when

(i)

\Delta_o < P

and

(ii)

\Delta_o > P

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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 magnetic behaviour difference arises from the ligand field strength: weak-field Cl⁻ leaves Ni²⁺ with two unpaired electrons (paramagnetic), while strong-field CO forces pairing in Ni⁰ (diamagnetic). For a $d^5$ ion, $\Delta_o < P$ gives high-spin with 5 unpaired electrons; $\Delta_o > P$ gives low-spin with 1 unpaired electron.

Part (a): Why $[NiCl_4]^{2-}$ is paramagnetic but $[Ni(CO)_4]$ is diamagnetic

The key is not just hybridisation — it’s the electronic configuration of the metal and the strength of the ligand field.

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

In $[NiCl_4]^{2-}$ , each Cl⁻ has a −1 charge. Let Ni have oxidation state $x$ :

$x + 4(-1) = -2 \implies x = +2$ .

Ni (atomic number 28) has ground-state configuration $[Ar]\,3d^8 4s^2$ .

Ni²⁺ loses the two 4s electrons: $3d^8$ .
In $[Ni(CO)_4]$ , CO is a neutral ligand. Let Ni have oxidation state $x$ :

$x + 4(0) = 0 \implies x = 0$ .

Ni⁰ has configuration $3d^8 4s^2$ — but in complexes, the 4s electrons often move to 3d, giving $3d^{10}$ .

Note

CO is a strong field ligand (high up in the spectrochemical series). It causes significant pairing of electrons. Cl⁻ is a weak field ligand (low in the spectrochemical series) and does not force pairing.

Step 2: Apply crystal field theory for tetrahedral geometry.

Both complexes are tetrahedral ( $sp^3$ hybridisation). In a tetrahedral field, the $d$ orbitals split into two sets:

Lower energy: $e$ set ( $d_{x^2-y^2}, d_{z^2}$ )
Higher energy: $t_2$ set ( $d_{xy}, d_{yz}, d_{zx}$ )

The splitting energy is $\Delta_t$ , which is roughly $4/9$ of $\Delta_o$ (octahedral splitting). Because $\Delta_t$ is small, tetrahedral complexes are almost always high-spin — except when the ligand is extremely strong.

Step 3: Fill electrons for $[NiCl_4]^{2-}$ (Ni²⁺, $3d^8$ , weak field).

For $d^8$ in a tetrahedral field, the filling follows Hund’s rule since $\Delta_t$ is small:

The $e$ set (lower) gets 2 electrons (paired).
The $t_2$ set (higher) gets 6 electrons — but with 8 total, we place 2 in $e$ (paired) and 6 in $t_2$ (4 paired, 2 unpaired).

Result: Two unpaired electrons → paramagnetic.

Step 4: Fill electrons for $[Ni(CO)_4]$ (Ni⁰, $3d^{10}$ , strong field).

CO is such a strong field ligand that it forces complete pairing. For Ni⁰, the effective configuration is $3d^{10}$ :

All five $d$ orbitals are completely filled (2 electrons each).
No unpaired electrons.

Result: Zero unpaired electrons → diamagnetic.

Watch out

A common mistake is to think both complexes have the same $d^8$ configuration. They don’t — Ni is in different oxidation states. Always check the oxidation state first.

Tip

For tetrahedral complexes, paramagnetism is the norm unless the metal has a $d^{10}$ configuration (like Zn²⁺, Cd²⁺, or Ni⁰ here) or a very strong field ligand forces pairing — which is rare in tetrahedral geometry.

Part (b): Electronic configuration of $d^5$ in octahedral field

In an octahedral field, the $d$ orbitals split into:

Lower energy: $t_{2g}$ set ( $d_{xy}, d_{yz}, d_{zx}$ ) — three orbitals
Higher energy: $e_g$ set ( $d_{x^2-y^2}, d_{z^2}$ ) — two orbitals

The splitting energy is $\Delta_o$ . The pairing energy $P$ is the energy cost to put two electrons in the same orbital.

Case (i): $\Delta_o < P$ (weak field ligand) …

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

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write any one example of low spin complex.$
›Reveal solutionSolution

A low-spin complex forms when a strong-field ligand causes the d electrons to pair up in the lower-energy t2g set rather than spreading into eg, reducing the number of unpaired electrons.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion [A]: [Ni(CN)4]2- is a square-planar and diamagnetic. Reason [R]: It has no unpaired electrons due to presence of strong field.$
(a) Both [A] and [R] are true and [R] is the correct explanation of [A].
(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].
(c) [A] is true, but [R] is false.
(d) [A] is false, but [R] is true.
›Reveal solutionSolution

$[Ni(CN)_4]^{2-}$ is indeed square planar and diamagnetic, and this is correctly explained by CN⁻ being a strong field ligand that forces electron pairing, leaving no unpaired electrons.

In $[Ni(CN)_4]^{2-}$ , nickel is in the $+2$ oxidation state: $Ni^{2+}$ has configuration $3d^8$ (8 electrons: $t_{2g}^6 e_g^2$ in a free-ion sense, or $3d^8 = \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\ \uparrow$ ).

…

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CBSE 2026Set ANNUAL1 mark
$[Co(NH_3)_6]^{3+}$
›Reveal solutionSolution
Because $NH_3$ is a strong-field ligand, $Co^{3+}$ 's $d$ -electrons pair up and the complex uses the inner $(n-1)d$ orbitals for hybridisation — making $[Co(NH_3)_6]^{3+}$ the inner-orbital complex, unlike $[CoF_6]^{3-}$ .

Analysis

$Co^{3+}$ has the configuration $3d^6$ in both complexes; the difference lies in the field strength of the ligand.

In $[Co(NH_3)_6]^{3+}$ : $NH_3$ is a strong-field ligand. It forces all 6 $d$ -electrons to pair up within three $3d$ orbitals ( $t_{2g}^6$ ), freeing the other two $3d$ orbitals for hybridisation. Cobalt then hybridises as $d^2sp^3$ , using inner $(n-1)d$ , i.e. $3d$ , orbitals — this is an inner-orbital (low-spin) complex, diamagnetic. …
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CBSE 2026Set ANNUAL1 mark
$Q. The oxidation number of all the alkali metals in their compounds is ________.$
›Reveal solutionSolution
[!TLDR]

+1

Method

Alkali metals (Group 1) have one valence electron and invariably show a +1 oxida …
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CBSE 2025Set 56/5/11 markMCQ
$Cu^{+}$
›Reveal solutionSolution
The colour of a transition metal ion in aqueous solution depends on the presence of unpaired d-electrons, which allow d-d transitions. Both $Co^{2+}$ and $Fe^{2+}$ have unpaired d-electrons and are coloured, while $Cu^{+}$ and $Ti^{4+}$ have fully filled or empty d-subshells and are colourless. The correct pair is III and IV, i.e., option (C).

The question asks which two ions among the given four are coloured in aqueous solution. Colour in transition metal ions arises from the absorption of visible light due to electronic transitions between split d-orbitals — the famous d-d transition. But this only happens if the d-subshell is partially filled (i.e., has at least one unpaired electron and at least one vacant orbital). If the d-subshell is completely empty ( $d^0$ ) or completely filled ( $d^{10}$ ), no d-d transition is possible, and the ion is colourless (or white) in solution.

Let’s examine each ion one by one.

$Cu^{+}$ (Copper(I))

Atomic number of Cu = 29. Neutral Cu has configuration $[Ar]\,3d^{10}4s^1$ .

$Cu^{+}$ loses the 4s electron, so its configuration becomes $[Ar]\,3d^{10}$ .

The d-subshell is completely filled. No d-d transitions possible.

Result: Colourless in aqueous solution.

$Ti^{4+}$ (Titanium(IV))

Atomic number of Ti = 22. Neutral Ti has $[Ar]\,3d^{2}4s^2$ .

$Ti^{4+}$ loses all four valence electrons (two from 4s and two from 3d), so its configuration becomes $[Ar]\,3d^{0}$ .

The d-subshell is completely empty. No d-d transitions possible.

Result: Colourless in aqueous solution.

$Co^{2+}$ (Cobalt(II))

Atomic number of Co = 27. Neutral Co has $[Ar]\,3d^{7}4s^2$ .

$Co^{2+}$ loses the two 4s electrons, giving $[Ar]\,3d^{7}$ .

The d-subshell is partially filled (7 electrons in 5 orbitals — there are unpaired electrons). In aqueous solution, $Co^{2+}$ forms the pink $[Co(H_2O)_6]^{2+}$ complex.

Result: Coloured (pink) in aqueous solution. …
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CBSE 2025Set D1 markMCQ
$Q. The structure of complex ion [Ni(CN)4]2- is$
(a) Linear
(b) Tetrahedral
(c) Square planar
(d) Octahedral
›Reveal solutionSolution

Ni2+ (d8) with strong-field CN- gives dsp2 hybridisation -> square planar [Ni(CN)4]2-.

Step 1 - oxidation state: In [Ni(CN)4]2-, four CN- (each -1) give -4; overall charge -2, so Ni is +2.

Step 2 - configuration: Ni2+ is 3d8.

Step 3 - ligand strength: CN- is a strong-field ligand. It pairs up the d electrons, freeing one 3d orbital. …

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CBSE 2025Set ANNUAL1 mark
$Q. CO is stronger ligand than Cl⁻¹. (True / False)$
›Reveal solutionSolution

True — CO lies far above Cl⁻ in the spectrochemical series, so it is a much stronger field ligand.

The spectrochemical series arranges ligands in order of increasing crystal-field splitting ( $\Delta_o$ ) they cause:

$I^- < Br^- < S^{2-} < SCN^- < Cl^- < ... < NH_3 < en < CN^- < CO$

…

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CBSE 2025Set ANNUAL1 mark
$Q. Draw a figure to show the splitting of d-orbitals in an octahedral crystal field.$
›Reveal solutionSolution
Figure — The stem 'Draw a figure to show the splitting of d-orbitals in an octahedral crystal field' needs the t2g/eg e

Ligands approaching along the axes in an octahedral complex raise the energy of orbitals pointing along the axes more than those pointing between the axes, splitting the 5 degenerate d-orbitals into two sets separated by $\Delta_o$ .

Description of the splitting (energy-level diagram in words)

In a free (gaseous) metal ion, all five $d$ -orbitals ( $d_{xy}, d_{yz}, d_{zx}, d_{x^2-y^2}, d_{z^2}$ ) are degenerate (equal energy). When 6 ligands approach the metal ion symmetrically along the $\pm x, \pm y, \pm z$ axes to form an octahedral complex, the orbitals lying along the axes experience more electrostatic repulsion from the approaching ligand electron pairs than the orbitals lying between the axes. This splits the 5 orbitals into two sets:

$e_g$ set (higher energy): $d_{x^2-y^2}$ and $d_{z^2}$ — these point directly at the ligands along the axes, so they are raised in energy above the mean (barycentre) by $+0.6\Delta_o$ (i.e. $+\tfrac{3}{5}\Delta_o$ ).

$t_{2g}$ set (lower energy): $d_{xy}, d_{yz}, d_{zx}$ — these point between the axes (away from the ligand directions), so they are lowered below the barycentre by $-0.4\Delta_o$ (i.e. $-\tfrac{2}{5}\Delta_o$ ).

Schematically (energy increasing upward):

____ ____ <- e_g (d(x2-y2), d(z2)) +0.6(Delta_o) …
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CBSE 2024Set 56/3/11 markMCQ
$Co^{3+}$
›Reveal solutionSolution
The key is to determine the number of unpaired electrons in $Co^{3+}$ ( $3d^6$ ) under each geometry and ligand field. Only the octahedral strong-field (low-spin) case gives zero unpaired electrons, making it diamagnetic. The correct option is (A).

Let’s start with the core idea. A substance is diamagnetic when all its electrons are paired — no unpaired electrons means no net magnetic moment. For transition metal complexes, this depends entirely on how the $d$ -orbitals split in energy under the influence of the surrounding ligands (Crystal Field Splitting) and how electrons fill those orbitals.

Cobalt has atomic number 27. Its ground state electron configuration is $[Ar] 3d^7 4s^2$ . When it forms $Co^{3+}$ , it loses three electrons — typically the two $4s$ electrons and one $3d$ electron. So $Co^{3+}$ has a $3d^6$ configuration.

Now, six $d$ -electrons can arrange themselves in different ways depending on the geometry of the complex and the strength of the ligand field. The geometry determines the splitting pattern of the $d$ -orbitals, and the ligand field strength decides whether electrons pair up in lower orbitals or spread out (Hund’s rule) into higher ones.

Let’s examine each option one by one.

Option (A): Octahedral complex with strong field ligand

In an octahedral field, the five $d$ -orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals). The energy gap $\Delta_o$ is large when the ligand is strong (like CN⁻, CO).

For $3d^6$ , a strong field forces electrons to pair up in the $t_{2g}$ set before any electron goes to $e_g$ . So the filling is: $t_{2g}^6$ — all six electrons paired in three orbitals. That gives zero unpaired electrons.

Tip

Strong field = low spin = maximum pairing. For $d^6$ , low-spin octahedral is always diamagnetic.

Option (B): Octahedral complex with weak field ligand

Here $\Delta_o$ is small. Electrons follow Hund’s rule: they occupy all five orbitals singly before pairing. For $d^6$ , the first five electrons go one each into $t_{2g}$ and $e_g$ (actually $t_{2g}^3 e_g^2$ ), and the sixth electron must pair in a $t_{2g}$ orbital. So the configuration is $t_{2g}^4 e_g^2$ — that’s four electrons in $t_{2g}$ (one pair, two unpaired) and two unpaired in $e_g$ . Total unpaired electrons = 4. Hence paramagnetic.

Option (C): Square planar complex …
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CBSE 2024Set ANNUAL1 mark
$Q. What is crystal field splitting energy?$
›Reveal solutionSolution
When ligands approach a metal ion, electrostatic repulsion splits the previously degenerate d-orbitals into two energy sets; the gap between them is the crystal field splitting energy, $\Delta$ .

In an isolated (gas-phase) transition-metal ion, all five d-orbitals are degenerate (equal energy). When ligands approach to form a complex, their electron pairs create an electric field that repels electrons in the d-orbitals unequally, depending on each orbital's spatial orientation relative to the ligand positions.

In an octahedral field, the d-orbitals split into two sets:

$t_{2g}$ ( $d_{xy}, d_{yz}, d_{xz}$ ) — lower energy, point between the ligand axes

$e_g$ ( $d_{x^2-y^2}, d_{z^2}$ ) — higher energy, point directly at the ligands …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. A coordination compound is colourless due to -$
(a) the absence of ligand
(b) loss of water molecules
(c) d-d transition of the electron
(d) energy of crystal field splitting energy
›Reveal solutionSolution

A coordination compound is colourless when it cannot undergo d-d electronic transitions — either because it has no d electrons or a completely filled d-subshell.

Colour in most coordination compounds arises from d–d transitions, where an electron is excited from a lower-energy $d$ -orbital ( $t_{2g}$ ) to a higher-energy one ( $e_g$ ) after crystal field splitting, absorbing a specific wavelength of visible light (and transmitting/reflecting the complementary colour).

…

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CBSE 2023Set 56/1/11 markMCQ
$Q. Assertion (A) : Low spin tetrahedral complexes are rarely observed. Reason (R) : Crystal field splitting energy is less than pairing energy for tetrahedral complexes. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
The assertion is true — low-spin tetrahedral complexes are rare — and the reason is also true: for tetrahedral complexes, the crystal field splitting energy $\Delta_t$ is much smaller than the pairing energy $P$ , making low-spin configurations energetically unfavourable. The reason correctly explains the assertion, so option (A) is correct.

Why this question hinges on crystal field splitting

In coordination chemistry, the spin state of a complex (high-spin vs low-spin) depends on a tug-of-war between two energies: the crystal field splitting energy ( $\Delta$ ) and the pairing energy ( $P$ ). If $\Delta > P$ , electrons prefer to pair up in the lower-energy orbitals (low-spin). If $\Delta < P$ , electrons spread out to avoid pairing (high-spin).

For tetrahedral complexes, the splitting pattern is the inverse of octahedral — the $d_{xy}, d_{yz}, d_{zx}$ orbitals (called $t_2$ ) are higher in energy, and the $d_{x^2-y^2}, d_{z^2}$ orbitals (called $e$ ) are lower. But the key number is the magnitude of $\Delta_t$ (tetrahedral splitting).
$\Delta_t \approx \frac{4}{9} \Delta_o$
Since $\Delta_o$ itself is often comparable to or smaller than $P$ for many metal ions (especially first-row transition metals), $\Delta_t$ ends up being much smaller than $P$ in almost all cases. That means the energy cost of pairing electrons is never recovered by the splitting — so electrons always occupy orbitals singly before pairing, giving high-spin configurations.

Watch out

A common mistake is to think that low-spin tetrahedral complexes are impossible. They are not — they are just rare. With very strong-field ligands (like CN⁻) and heavy metals (where $\Delta$ is larger), a few examples exist. But for typical exam contexts (first-row transition metals, common ligands), the statement holds.

Step-by-step reasoning

Understand the assertion: "Low spin tetrahedral complexes are rarely observed." This is a factual statement about coordination chemistry. For a tetrahedral complex to be low-spin, the splitting $\Delta_t$ must exceed the pairing energy $P$ . But because $\Delta_t$ is inherently small (about $4/9$ of $\Delta_o$ ), this condition is seldom met. …
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Q.(a) Although both [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− and [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] have sp3sp^3sp3 hybridisation yet [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− is paramagnetic and [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] is diamagnetic. Give reason. (Atomic no. of Ni = 28)

Part (a): Why [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− is paramagnetic but [Ni(CO)4][Ni(CO)_4][Ni(CO)4​] is diamagnetic

Part (b): Electronic configuration of d5d^5d5 in octahedral field

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Q.(a) Although both $[NiCl_4]^{2-}$ and $[Ni(CO)_4]$ have $sp^3$ hybridisation yet $[NiCl_4]^{2-}$ is paramagnetic and $[Ni(CO)_4]$ is diamagnetic. Give reason. (Atomic no. of Ni = 28)

Part (a): Why $[NiCl_4]^{2-}$ is paramagnetic but $[Ni(CO)_4]$ is diamagnetic

Part (b): Electronic configuration of $d^5$ in octahedral field

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