(a) Answer the following questions about the complexes [NiCl₄]²⁻ and [Ni(CN)₄]²⁻ : (i) Write the hybridization

Question

[NiCl_4]^{2-}

(i) Write the hybridization involved in each case.

(ii) Which of them is the inner orbital complex and which one is the outer orbital complex ?

(iii) Compare their magnetic behaviour. [Atomic number : Ni = 28]

(OR)

(b)

(i) Name two coordination compounds which are important in biological systems.

(ii) What is meant by chelate effect ? Give an example.

(iii) Why are low spin tetrahedral complexes rarely formed ?

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Part (a)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. …

Part (b)Concept understanding — Complex Formation

Complex Formation: The Intuition

Imagine you have a metal ion — say, a copper ion ( $\text{Cu}^{2+}$ ) — floating in water. It's positively charged, so it attracts anything negative or electron-rich nearby. Water molecules themselves have lone pairs of electrons on oxygen, so they crowd around the copper ion, each one donating a pair of electrons to form a coordinate bond. That cluster — the metal ion surrounded by water molecules — is already a complex ion: $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ .

Now, suppose you add ammonia ( $\text{NH}_3$ ) to the solution. Ammonia also has a lone pair on nitrogen, and it's a stronger electron donor than water. One by one, the ammonia molecules push the water molecules aside, replacing them. You end up with a deep blue complex: $[\text{Cu}(\text{NH}_3)_4]^{2+}$ .

That process — the stepwise replacement of one set of molecules (or ions) around a central metal atom by another set — is complex formation. The central metal is the Lewis acid (electron-pair acceptor), and the molecules or ions that attach to it are ligands (Lewis bases, electron-pair donors). The resulting species is a coordination compound or complex.

Note

The word "complex" doesn't mean complicated. It just means a central atom (usually a metal) bonded to surrounding molecules or ions.

The Precise Statement

Complex formation is the reversible, stepwise reaction in which a central metal atom or ion (usually a transition metal) accepts electron pairs from one or more ligands to form a coordination entity. Each step has its own equilibrium constant, and the overall stability of the complex is measured by the formation constant ( $K_f$ ) or stability constant.

For a general reaction:

\text{M} + n\text{L} \rightleftharpoons \text{ML}_n

The overall formation constant is:

K_f = \frac{[\text{ML}_n]}{[\text{M}][\text{L}]^n}

A large $K_f$ means the complex is very stable — the ligands bind tightly and are hard to remove.

K_f = \frac{[\text{complex}]}{[\text{metal}][\text{ligand}]^n}

Key Features to Remember

Stepwise nature: Complexes don't form all at once. First one ligand binds, then another, and so on. Each step has its own constant ( $K_1, K_2, \dots$ ). The product of all stepwise constants equals the overall $K_f$ .
Coordination number: The number of ligand donor atoms directly bonded to the metal. Common values are 4 (tetrahedral or square planar) and 6 (octahedral). For $[\text{Cu}(\text{NH}_3)_4]^{2+}$ , the coordination number is 4.
Ligand denticity: A ligand can have one donor atom (monodentate, like $\text{NH}_3$ or $\text{H}_2\text{O}$ ) or multiple donor atoms (polydentate, like EDTA, which wraps around the metal with six donor atoms). Polydentate ligands form especially stable complexes — this is the chelate effect.
Reversibility: Complex formation is an equilibrium. Change the concentration of ligand, pH, or temperature, and the complex can break apart or form a different one. …

Why this formula?

Complex Formation: Understanding the Why Behind the Key Formulas

Complex formation is a fundamental concept in coordination chemistry and equilibrium. Let's build the reasoning step-by-step, starting from the simplest idea.

1. What is Complex Formation?

A complex forms when a central metal ion (Lewis acid) accepts electron pairs from surrounding molecules or ions called ligands (Lewis bases).

Example:

$Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$

The key question: Why do we get a specific formula for the equilibrium constant?

2. The Stepwise Formation (The Core Reason)

Complex formation does not happen in one giant leap. It occurs in successive, reversible steps — each step adding one ligand.

For a metal $M$ and ligand $L$ :

Step 1:

$M + L \rightleftharpoons ML$

Equilibrium constant: $K_1 = \frac{[ML]}{[M][L]}$

Step 2:

$ML + L \rightleftharpoons ML_2$

$K_2 = \frac{[ML_2]}{[ML][L]}$

Step 3:

$ML_2 + L \rightleftharpoons ML_3$

$K_3 = \frac{[ML_3]}{[ML_2][L]}$

... and so on up to $ML_n$ .

Each $K_i$ is called a stepwise formation constant.

3. The Overall Formation Constant (The Key Formula)

Now, what if we want the equilibrium constant for the overall reaction:

$M + nL \rightleftharpoons ML_n$

We can multiply the stepwise equilibria (because when you add reactions, you multiply their equilibrium constants):

$K_f = K_1 \times K_2 \times K_3 \times \dots \times K_n$

So:

\boxed{K_f = \frac{[ML_n]}{[M][L]^n}}

Why this form?

Because each step contributes one $[L]$ in the denominator and one $[ML_i]$ in the numerator, but intermediate species cancel out when multiplied.

4. Why the Denominator Has $[L]^n$ (Not $n[L]$ )

This is a common confusion. Let's derive it explicitly:

From step 1: $[ML] = K_1[M][L]$

From step 2: $[ML_2] = K_2[ML][L] = K_1K_2[M][L]^2$

From step 3: $[ML_3] = K_3[ML_2][L] = K_1K_2K_3[M][L]^3$

Continuing:

$[ML_n] = (K_1K_2\dots K_n)[M][L]^n$

Therefore:

\frac{[ML_n]}{[M][L]^n} = K_1K_2\dots K_n = K_f

The exponent $n$ comes from repeated multiplication, not addition. Each ligand adds one factor of $[L]$ in the denominator.

5. The Stability Connection

A larger $K_f$ means:

The complex is more stable
Equilibrium lies far to the right (products favoured) …

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Part (a)

$Ni = 28 \Rightarrow Ni^{2+} = [Ar]3d^8$ in both complexes.

Complex	Ligand	Geometry	Hybridisation	Orbital type	Unpaired e⁻	Magnetism
$[NiCl_4]^{2-}$	$Cl^-$ (weak)	tetrahedral	$sp^3$	outer	2	paramagnetic
$[Ni(CN)_4]^{2-}$	$CN^-$ (strong)	square planar	$dsp^2$	inner	0	diamagnetic

(i) $[NiCl_4]^{2-}$ : $sp^3$ ; $[Ni(CN)_4]^{2-}$ : $dsp^2$ .
(ii) $[Ni(CN)_4]^{2-}$ is the inner orbital complex ( $3d$ used); $[NiCl_4]^{2-}$ is the outer orbital complex. …

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$[NiCl_4]^{2-}$ is $sp^3$ , tetrahedral, outer-orbital, paramagnetic (2 unpaired e⁻); $[Ni(CN)_4]^{2-}$ is $dsp^2$ , square planar, inner-orbital, diamagnetic.
chlorophyll & haemoglobin are biological complexes; the chelate effect is the extra (entropy-driven) stability of ring-forming ligands; low-spin tetrahedral complexes are rare because $\Delta_t (\approx \tfrac49\Delta_o)$ is smaller than the pairing energy.

Part (a) — $[NiCl_4]^{2-}$ vs $[Ni(CN)_4]^{2-}$

Nickel is atomic number 28: $[Ar]3d^8 4s^2$ . In both complexes the ligand charge is $-1$ each, so $x + 4(-1) = -2 \Rightarrow x = +2$ . Removing the two $4s$ electrons gives $Ni^{2+} = [Ar]3d^8$ .

(i) Hybridisation.

$Cl^-$ is a weak-field ligand. It does not force the $3d^8$ electrons to pair, so no inner $3d$ orbital is freed. Bonding uses one $4s$ + three $4p$ orbitals → $sp^3$ , giving a tetrahedral shape.
$CN^-$ is a strong-field ligand. For a $d^8$ ion it drives pairing so that one $3d$ orbital ( $d_{x^2-y^2}$ ) is emptied; bonding then uses that $3d$ + $4s$ + two $4p$ → $dsp^2$ , giving a square planar shape.

(ii) Inner vs outer orbital.

$[Ni(CN)_4]^{2-}$ uses an inner $(n-1)d$ orbital → inner-orbital (low-spin) complex.
$[NiCl_4]^{2-}$ uses only outer $n=4$ orbitals → outer-orbital (high-spin) complex.

(iii) Magnetic behaviour. Fill the eight $d$ -electrons:

$[NiCl_4]^{2-}$ : weak field, electrons stay maximally unpaired → 2 unpaired electrons → paramagnetic ( $\mu = \sqrt{2(2+2)} = \sqrt8 \approx 2.83\ \text{BM}$ ).
$[Ni(CN)_4]^{2-}$ : strong field, all electrons paired → 0 unpaired electrons → diamagnetic. …

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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 44 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 markMCQ
$Q. Mohr's salt is-$
(a)
(i) Fe₂(SO₄)₃.(NH₄)₂SO₄.6H₂O
(b)
(ii) FeSO₄.(NH₄)₂SO₄.6H₂O
(c)
(iii) MgSO₄.7H₂O
(d)
(iv) FeSO₄.7H₂O
›Reveal solutionSolution
Mohr's salt is ferrous ammonium sulphate hexahydrate, $\text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot 6\text{H}_2\text{O}$ . Correct option: (ii).

Concept. Mohr's salt is a double salt — a stoichiometric combination of two simple salts, ferrous sulphate $\text{FeSO}_4$ and ammonium sulphate $(\text{NH}_4)_2\text{SO}_4$ — that dissolves in water to release all its constituent ions independently ( $\text{Fe}^{2+}$ , $\text{NH}_4^{+}$ , $\text{SO}_4^{2-}$ ).

Why the other options are wrong.

(i) $\text{Fe}_2(\text{SO}_4)_3\cdot(\text{NH}_4)_2\text{SO}_4\cdot 6\text{H}_2\text{O}$ contains ferric iron ( $\text{Fe}^{3+}$ ) — that is ferric alum-type, not Mohr's salt.

(iii) $\text{MgSO}_4\cdot 7\text{H}_2\text{O}$ is Epsom salt. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is a chelating ligand ?$
(a) NH3
(b) H2O
(c) Cl-
(d) C2O4^2-
›Reveal solutionSolution
A chelating ligand grips the metal at more than one point; oxalate binds through two O atoms, forming a five-membered ring. Answer: (d) C2O4^2-.

NH3, H2O and Cl- are all monodentate — each donates through a single atom, so they cannot chelate. …
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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 A1 mark
$Q. Write True or False: The Ca2+ and Mg2+ ions form stable complexes with EDTA.$
›Reveal solutionSolution

EDTA is a hexadentate ligand that forms very stable chelate complexes with both Ca²⁺ and Mg²⁺.

EDTA (ethylenediaminetetraacetate) has six donor atoms (two N and four O, from its two amine groups and four carboxylate groups) that can simultaneously bind a single metal ion, wrapping around it to form a highly stable ring (chelate) structure — this is the chelate effect. Both Ca²⁺ and Mg²⁺ form such stable 1:1 octahedral EDTA complexes; …

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CBSE 2025Set A1 mark
$Q. Write the central metal atom in [Ni(CO)4].$
›Reveal solutionSolution

In [Ni(CO)4], the central metal atom to which all four ligands are directly bonded is nickel.

[Ni(CO)4] (tetracarbonylnickel(0)) is a classic coordination/organometallic compound in which a single nickel atom is surrounded by four neutral carbon monoxide (CO) ligands, each donating a lone pair from carbon to the metal. Since CO is a neutral ligand and the complex overall is neutral, nickel here is in the zero oxidation state, Ni(0), with electron …

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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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Q.(a) Answer the following questions about the complexes [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− and [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2− :

Part (a) — [NiCl4]2−[NiCl_4]^{2-}[NiCl4​]2− vs [Ni(CN)4]2−[Ni(CN)_4]^{2-}[Ni(CN)4​]2−

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Q.(a) Answer the following questions about the complexes $[NiCl_4]^{2-}$ and $[Ni(CN)_4]^{2-}$ :

Part (a) — $[NiCl_4]^{2-}$ vs $[Ni(CN)_4]^{2-}$

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