(a) Answer the following about the complexes [FeF₆]³⁻ and [Fe(CN)₆]⁴⁻ : (i) Write the hybridization involved i

Question

[FeF_6]^{3-}

(i) Write the hybridization involved in each case.

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

(iii) Compare their magnetic behaviour. [Atomic number : Fe = 26]

(OR)

(b)

(i) What happens to the colour of complex

[Ti(H_2O)_6]^{3+}

when heated gradually ?

(ii) Write the electronic configuration for

d^5

ion if

\Delta_o < P

(iii) Write the hybridization and magnetic behaviour of the complex

[Ni(CO)_4]

. [Atomic number : Ni = 28]

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

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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 — Colour of Transition Metal Ions

Most students first meet colour in transition metals as a striking fact: copper sulphate is blue, potassium dichromate is orange, nickel salts are green. The question is why — after all, most elements form colourless compounds. The answer lives inside the d-orbitals.

The intuition: a window that absorbs some colours

Imagine a white light beam hitting a solution. If the substance absorbs nothing, all colours pass through and you see white (or colourless). If it absorbs only red light, the remaining mixture of colours looks blue-green — that’s the complementary colour. So a coloured compound is simply one that absorbs some part of the visible spectrum and transmits the rest.

For transition metal ions, the absorbing "antenna" is the set of five d-orbitals. In a free ion these orbitals all have the same energy. But when the ion sits inside a crystal or solution, surrounding ligands (water, ammonia, chloride, etc.) push on the d-orbitals unevenly. Some d-orbitals point directly at the ligands and feel strong repulsion; others point between them and feel less. This splits the d-orbital energies into two groups — a lower-energy set and a higher-energy set. The energy gap between them is called $\Delta$ (or $10Dq$ ), and it often falls right in the range of visible light.

The precise mechanism: d–d transition

An electron sitting in a lower d-orbital can absorb a photon whose energy exactly matches $\Delta$ . That photon disappears, and the electron jumps to a higher d-orbital. This is called a d–d transition. The colour you see is white light minus the absorbed wavelength.

E_{\text{photon}} = h\nu = \Delta = \frac{hc}{\lambda_{\text{absorbed}}}

The exact colour depends on three things:

The metal ion (more charge → larger $\Delta$ )
The ligand (stronger field → larger $\Delta$ )
The geometry (octahedral, tetrahedral, square planar — each gives a different splitting pattern)

For example, $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ absorbs red-orange light ( $\lambda \approx 600\text{ nm}$ ), so it looks blue. $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ absorbs green-yellow and looks violet.

Why not all d-block ions are coloured

A d–d transition is only possible if there is an empty higher d-orbital to jump into. Ions with a full $d^{10}$ configuration (like $\text{Zn}^{2+}$ , $\text{Cu}^+$ , $\text{Ag}^+$ ) have no vacancy — all d-orbitals are filled, so no d–d transition can happen. They are colourless (unless other processes like charge transfer occur, which is a separate topic).

Similarly, $\text{Sc}^{3+}$ has $d^0$ — no d-electrons at all — so there is nothing to excite. Colourless. …

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Part (a): $[FeF_6]^{3-}$ = $sp^3d^2$ , outer-orbital, paramagnetic (5 unpaired); $[Fe(CN)_6]^{4-}$ = $d^2sp^3$ , inner-orbital, diamagnetic. Part (b): $[Ti(H_2O)_6]^{3+}$ loses colour on heating (water lost); high-spin $d^5$ = $t_{2g}^3 e_g^2$ ; $[Ni(CO)_4]$ = $sp^3$ , diamagnetic.

Part (a)

Fe ( $Z=26$ ) $=[Ar]3d^6 4s^2$ .

$[FeF_6]^{3-}$ : Fe $^{3+}=3d^5$ . $F^-$ is a weak-field ligand ( $\Delta_o<P$ ), so electrons stay unpaired (high spin, $t_{2g}^3e_g^2$ ). No inner $3d$ orbitals are free, so bonding uses the outer $4d$ : $sp^3d^2$ (outer-orbital / high-spin complex). Five unpaired electrons -> paramagnetic, $\mu=\sqrt{5(5+2)}\approx5.92$ BM.
$[Fe(CN)_6]^{4-}$ : Fe $^{2+}=3d^6$ . $CN^-$ is a strong-field ligand ( $\Delta_o>P$ ), forcing pairing ( $t_{2g}^6 e_g^0$ ). Two inner $3d$ orbitals are freed -> $d^2sp^3$ (inner-orbital / low-spin complex). Zero unpaired electrons -> diamagnetic.

Answers: (i) $sp^3d^2$ and $d^2sp^3$ ;

(ii) $[FeF_6]^{3-}$ outer-orbital, $[Fe(CN)_6]^{4-}$ inner-orbital; …

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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 36 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
$Q. Which of the following ion is colourless in aqueous solution-$
(a)
(i) Fe²⁺
(b)
(ii) Mn²⁺
(c)
(iii) Zn²⁺
(d)
(iv) Cu²⁺
›Reveal solutionSolution
$\text{Zn}^{2+}$ has a fully filled $3d^{10}$ configuration, so it cannot undergo d–d transitions and is colourless. Correct option: (iii).

Concept. A transition-metal ion is coloured when it can absorb visible light by promoting an electron between the split $t_{2g}$ and $e_g$ d-orbitals (a d–d transition). This is only possible if the d-subshell is partially filled ( $d^1$ to $d^9$ ).

Steps (configurations of the ions).

$\text{Fe}^{2+}$ : $3d^{6}$ — partly filled $\Rightarrow$ coloured (pale green).

$\text{Mn}^{2+}$ : $3d^{5}$ — partly filled $\Rightarrow$ coloured (pale pink). …
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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 56/4/11 markMCQ
$Fe^{3+}$
›Reveal solutionSolution
Colour in transition metal ions arises from d–d transitions, which require unpaired electrons in the d‑orbitals. $Sc^{3+}$ has a $3d^0$ configuration (no d‑electrons), so it cannot undergo d–d transitions and is colourless. The correct option is (A).

The question asks which of the given trivalent ions is colourless in aqueous solution. Colour in transition metal ions is almost always due to the absorption of visible light by electrons moving between d‑orbitals — the famous d–d transition. For this to happen, the ion must have at least one electron in its d‑orbitals (a partially filled d‑subshell). If the d‑subshell is completely empty ( $d^0$ ) or completely filled ( $d^{10}$ ), no d–d transition is possible, and the ion appears colourless (or white) in solution.

Let’s check the electronic configuration of each ion.

$Sc^{3+}$ (Atomic number 21)

Sc ground state: $[Ar]\, 3d^1 4s^2$ .

Removing three electrons (the two 4s electrons and the one 3d electron) gives $Sc^{3+}$ : $[Ar]\, 3d^0$ .

No d‑electrons at all → no d–d transitions → colourless.

$Fe^{3+}$ (Atomic number 26)

Fe ground state: $[Ar]\, 3d^6 4s^2$ .

Removing three electrons gives $Fe^{3+}$ : $[Ar]\, 3d^5$ .

Five unpaired electrons (half‑filled d‑subshell) → d–d transitions possible → coloured (typically yellow‑brown in aqueous solution).

$Cr^{3+}$ (Atomic number 24)

Cr ground state: $[Ar]\, 3d^5 4s^1$ (exception to the usual filling order).

Removing three electrons gives $Cr^{3+}$ : $[Ar]\, 3d^3$ .

Three d‑electrons → d–d transitions possible → coloured (violet or green depending on ligands). …
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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
$Q. Which of the following compounds can be coloured?$
(a) Ag2SO4
(b) CuF2
(c) Cu2Cl2
(d) MgF2
›Reveal solutionSolution
A compound is coloured only if the metal ion has a partially filled d subshell; CuF2 (Cu2+ = d9) qualifies.

Colour in transition-metal compounds arises from d-d electronic transitions, which require a partially filled d subshell. Examining each cation:

Ag2SO4: Ag+ is [Kr] 4d10 -> filled d -> colourless. …
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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 markMCQ
$Q. Which of the following transition elements does not give coloured salt?$
(a) Cr
(b) Mn
(c) Cu
(d) Zn
›Reveal solutionSolution

Colour in transition-metal ions arises from d-d electronic transitions, which require partially filled d orbitals; Zn2+ has a full d10 shell with no d-d transition possible, so its salts are colourless.

Transition metal ions are typically coloured because of d-d transitions: an electron in a lower-energy d orbital (in the crystal field of surrounding ligands/anions) absorbs visible light and jumps to a higher-energy d orbital; the colour seen is complementary to the wavelength absorbed. This requires the d subshell to be partially filled (neither completely empty nor completely full), so a d-d transition is possible.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Colourless metal ion in aqueous solution is -$
(a) Cu2+
(b) Zn2+
(c) Mn2+
(d) V2+
›Reveal solutionSolution
Colour in transition-metal ions arises from d-d transitions of unpaired electrons; Zn2+ (3d10, fully filled) has none, so it is colourless.

Electronic configurations:

Cu2+: [Ar]3d9 - one unpaired electron -> coloured (blue)

Zn2+: [Ar]3d10 - all d orbitals completely filled, no unpaired electron and no vacant d orbital to promote an electron into -> colourless

Mn2+: [Ar]3d5 - unpaired electrons -> coloured (pale pink)

V2+: [Ar]3d3 - unpaired electrons -> coloured (violet) …
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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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Q.(a) Answer the following about the complexes [FeF6]3−[FeF_6]^{3-}[FeF6​]3− and [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4− :

Part (a)

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Q.(a) Answer the following about the complexes $[FeF_6]^{3-}$ and $[Fe(CN)_6]^{4-}$ :

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