Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal ato

NCERT Exemplar · Q27

Q. Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

(i)

[CoF_6]^{3-}

[Co(H_2O)_6]^{2+}

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

(ii)

[FeF_6]^{3-}

[Fe(H_2O)_6]^{2+}

[Fe(CN)_6]^{4-}

Kerala DhseLong· 5mImportance★★★★★

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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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Crystal field theory explains how ligand field strength splits d-orbital energies, determining whether a complex is high-spin or low-spin. For each complex, we identify the metal ion's d-count, the ligand's field strength, fill the $t_{2g}$ and $e_g$ orbitals accordingly, and compute the magnetic moment using $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

Let’s work through each complex step by step. The key idea: ligands like CN⁻ are strong-field (large $\Delta_o$ , causing pairing), while F⁻ and H₂O are weak-field (small $\Delta_o$ , favouring high-spin). The geometry is octahedral for all.

(i) $[CoF_6]^{3-}$ , $[Co(H_2O)_6]^{2+}$ , $[Co(CN)_6]^{3-}$

1. $[CoF_6]^{3-}$

Cobalt in +3 oxidation state: Co atomic number 27, so Co³⁺ has $[Ar]3d^6$ configuration.
F⁻ is a weak-field ligand → small $\Delta_o$ . Electrons fill according to Hund's rule: high-spin.
In octahedral field, the six d-electrons occupy: $t_{2g}^4 e_g^2$ (four in $t_{2g}$ , two in $e_g$ ).
Number of unpaired electrons: $n = 4$ (two in $e_g$ are unpaired, and two of the $t_{2g}$ electrons are unpaired because of the fourth electron pairing one).
Magnetic moment: $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM.

Watch out

A common mistake is to think Co³⁺ with weak field gives $t_{2g}^3 e_g^3$ — that would be 3 unpaired, but the actual filling for d⁶ high-spin is $t_{2g}^4 e_g^2$ with 4 unpaired electrons. Always apply Hund's rule to the $t_{2g}$ set first.

2. $[Co(H_2O)_6]^{2+}$

Cobalt in +2: Co²⁺ has $[Ar]3d^7$ .
H₂O is intermediate but generally weak-field for Co²⁺ (it is borderline; for Co²⁺, H₂O acts as weak field). So high-spin.
d⁷ high-spin octahedral: $t_{2g}^5 e_g^2$ .
Unpaired electrons: $n = 3$ (the $e_g$ has two unpaired, and the $t_{2g}$ has one unpaired because five electrons in three orbitals give one unpaired).
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM.

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

Again Co³⁺, d⁶.
CN⁻ is a strong-field ligand → large $\Delta_o$ , electrons pair up in $t_{2g}$ before occupying $e_g$ .
Configuration: $t_{2g}^6 e_g^0$ (all six electrons paired in the three $t_{2g}$ orbitals).
Unpaired electrons: $n = 0$ .
$\mu = 0$ BM (diamagnetic).

Tip

For d⁶, strong field gives $t_{2g}^6$ (low-spin, 0 unpaired), weak field gives $t_{2g}^4 e_g^2$ (high-spin, 4 unpaired). The difference is dramatic — magnetic moment changes from 0 to ~4.9 BM.

(ii) $[FeF_6]^{3-}$ , $[Fe(H_2O)_6]^{2+}$ , $[Fe(CN)_6]^{4-}$

4. $[FeF_6]^{3-}$

Iron in +3: Fe³⁺ has $[Ar]3d^5$ .
F⁻ is weak-field → high-spin.
d⁵ high-spin octahedral: $t_{2g}^3 e_g^2$ (Hund's rule: all five orbitals singly occupied).
Unpaired electrons: $n = 5$ .
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.

5. $[Fe(H_2O)_6]^{2+}$

Iron in +2: Fe²⁺ has $[Ar]3d^6$ .
H₂O is weak-field for Fe²⁺ (it is borderline but generally considered weak for Fe²⁺). So high-spin. …

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Method: Crystal Field Theory (CFT) for Octahedral Complexes

This method uses the electrostatic approach to explain how ligands split the $d$ -orbital energies of the central metal ion.

Core Idea

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

Higher energy: $e_g$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — point directly at ligands
Lower energy: $t_{2g}$ set ( $d_{xy}$ , $d_{yz}$ , $d_{zx}$ ) — point between ligands

The energy gap is $\Delta_o$ (or $10 Dq$ ).

Steps for Any Complex

Determine oxidation state of the central metal ion.
Write the $d^n$ configuration of the metal ion.
Identify ligand strength:
- Strong field (e.g., CN⁻) → low spin (electrons pair in $t_{2g}$ first)
- Weak field (e.g., F⁻, H₂O) → high spin (electrons fill all orbitals singly before pairing)
Draw the energy level diagram (split $t_{2g}$ and $e_g$ levels).
Fill electrons according to Hund’s rule and spin state.
Calculate magnetic moment using:

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

where $n$ = number of unpaired electrons.

(i) Cobalt Complexes

1. $[CoF_6]^{3-}$

Oxidation state: Co is +3 (since F⁻ is -1 each, total charge -3)
Electronic config of Co³⁺: $[Ar]\,3d^6$
Ligand: F⁻ is weak field → high spin
Filling: $t_{2g}^4\, e_g^2$ (4 electrons in $t_{2g}$ , 2 in $e_g$ )
Unpaired electrons: $n = 4$
Magnetic moment:

\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \text{BM}

2. $[Co(H_2O)_6]^{2+}$

Oxidation state: Co is +2 (H₂O is neutral)
Electronic config of Co²⁺: $[Ar]\,3d^7$
Ligand: H₂O is weak field → high spin
Filling: $t_{2g}^5\, e_g^2$
Unpaired electrons: $n = 3$
Magnetic moment:

\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, \text{BM}

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

Oxidation state: Co is +3 (CN⁻ is -1 each, total charge -3)
Electronic config of Co³⁺: $[Ar]\,3d^6$
Ligand: CN⁻ is strong field → low spin
Filling: $t_{2g}^6\, e_g^0$ (all 6 electrons paired in $t_{2g}$ )
Unpaired electrons: $n = 0$
Magnetic moment:

\mu = 0 \, \text{BM} \quad \text{(diamagnetic)}

(ii) Iron Complexes

1. $[FeF_6]^{3-}$

Oxidation state: Fe is +3 (F⁻ is -1 each)
Electronic config of Fe³⁺: $[Ar]\,3d^5$
Ligand: F⁻ is weak field → high spin
Filling: $t_{2g}^3\, e_g^2$ (Hund’s rule — all 5 orbitals singly occupied)
Unpaired electrons: $n = 5$
Magnetic moment:

\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{BM}

2. $[Fe(H_2O)_6]^{2+}$

Oxidation state: Fe is +2 (H₂O neutral)
Electronic config of Fe²⁺: $[Ar]\,3d^6$
Ligand: H₂O is weak field → high spin …

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Common Mistakes in Crystal Field Splitting Problems

Students frequently lose marks on these exact complexes. Here are the most common errors and how to avoid each.

Mistake 1: Wrong Oxidation State of the Central Metal

The error: Students often take the given complex ion's charge as the metal's oxidation state directly.

Example: In $[CoF_6]^{3-}$ , writing Co as +3 without calculation.

How to avoid: Always calculate systematically:

Let oxidation state of metal = $x$
Ligand charge: F⁻ = -1 each, CN⁻ = -1 each, H₂O = 0
Complex charge = sum of charges

For $[CoF_6]^{3-}$ :

x + 6(-1) = -3 \implies x = +3

For $[Co(H_2O)_6]^{2+}$ :

x + 6(0) = +2 \implies x = +2

Mistake 2: Confusing Strong vs Weak Field Ligands

The error: Treating H₂O as a strong field ligand or F⁻ as a strong field ligand.

The truth:

Strong field ligands: CN⁻, CO, NH₃ (cause pairing, low spin)
Weak field ligands: F⁻, Cl⁻, Br⁻, I⁻, H₂O (no pairing, high spin)

How to avoid: Memorise the spectrochemical series partially:

\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- < \text{CO}

Left of H₂O → weak field (high spin)
Right of H₂O → strong field (low spin)
H₂O itself → borderline, but for 3d metals, usually weak field

Mistake 3: Wrong d-Orbital Splitting Diagram for Octahedral

The error: Drawing $t_{2g}$ above $e_g$ or incorrect labelling.

Correct diagram for octahedral:

Lower energy: $t_{2g}$ (3 orbitals: $d_{xy}, d_{yz}, d_{zx}$ )
Higher energy: $e_g$ (2 orbitals: $d_{x^2-y^2}, d_{z^2}$ )

How to avoid: Remember: "t₂g below, e₉ above" — the energy gap is $\Delta_o$ (or $10Dq$ ).

Mistake 4: Wrong d-Electron Count

The error: Using the wrong number of d-electrons for the metal ion.

Example: For Fe³⁺ (atomic number 26), students write 5 d-electrons incorrectly.

How to avoid: Use the periodic table:

Fe (Z=26): $[Ar] 3d^6 4s^2$
Fe³⁺: remove 3 electrons → $3d^5$
Co (Z=27): $[Ar] 3d^7 4s^2$
Co³⁺: remove 3 electrons → $3d^6$
Co²⁺: remove 2 electrons → $3d^7$

Mistake 5: Filling Electrons Incorrectly in the Diagram

The error: For weak field (high spin), students pair electrons before filling all orbitals singly.

Correct filling rules:

Weak field (small $\Delta_o$ ): Hund's rule — fill all 5 orbitals singly first, then pair
Strong field (large $\Delta_o$ ): Fill $t_{2g}$ completely first (pairing), then $e_g$

Example — $[CoF_6]^{3-}$ (Co³⁺, $d^6$ , weak field):

Correct: $t_{2g}^4 e_g^2$ (4 in t₂g: 3 singly + 1 paired; 2 in e₉ singly)
Wrong: $t_{2g}^6 e_g^0$ (that's low spin — wrong for F⁻)

Example — $[Co(CN)_6]^{3-}$ (Co³⁺, $d^6$ , strong field):

Correct: $t_{2g}^6 e_g^0$ (all paired in t₂g)
Wrong: $t_{2g}^4 e_g^2$ (that's high spin — wrong for CN⁻)

Mistake 6: Wrong Magnetic Moment Formula

The error: Using $\mu = \sqrt{n(n+2)}$ where $n$ is wrong.

How to avoid:

$n$ = number of unpaired electrons
Formula: $\mu = \sqrt{n(n+2)}$ BM (Bohr magnetons) …

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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 17 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$[Co(NH_3)_6]^{+3}$
›Reveal solutionSolution

A complex absorbs the colour complementary to the one it appears; the violet/purple $[CoCl(NH_3)_5]^{2+}$ absorbs yellow.

Observed colour is complementary to the absorbed colour. A species that absorbs yellow light appears violet/purple. …
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KEAM 2026Set eng-2026-04204 marksMCQ
$NCS^- < CN^- < SCN^- < S^{2-} < OH^-$
›Reveal solutionSolution

Field strength increases: $SCN^{-}<S^{2-}<OH^{-}<NCS^{-}<CN^{-}$ . …

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KEAM 2026Set eng-2026-04214 marksMCQ
$Q. The correct increasing order of wavelength of absorption of the following complexes is$
(i) $[CoCl(NH_3)_5]^{2+}$
(ii) $[Co(NH_3)_5(H_2O)]^{3+}$
(iii) $[Co(NH_3)_6]^{3+}$
(iv) $[Co(CN)_6]^{3-}$ (A)
(i) <
(ii) <
(iii) <
(iv) (B)
(i) <
(ii) <
(iv) <
(iii) (C)
(ii) <
(i) <
(iv) <
(iii) (D)
(iv) <
(iii) <
(ii) <
(i) (E)
(iv) <
(i) <
(ii) < (iii)
›Reveal solutionSolution

Wavelength is inverse to $\Delta_o$ ; increasing $\lambda$ : (iv)<(iii)<(ii)<(i). …

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KEAM 2026Set pha-2026-0420F4 marksMCQ
$Q. Which one of the statements is not the limitation of valence bond theory of complexes? (A) It does not give a quantitative interpretation of the thermodynamic stabilities. (B) It does not give quantitative interpretation of magnetic properties.. (C) It explains the colour exhibited by coordination compounds. (D) It does not distinguish between weak and strong ligands. (E) It does not make exact predictions regarding the tetrahedral structures of 4-coordinated complexes.$
›Reveal solutionSolution

A recognised limitation of VBT is that it fails to explain colour; the statement that it explains colour (C) is therefore not a limitation.

Valence bond theory has several well-known limitations: it gives no quantitative account of thermodynamic stabilities (A), no quantitative account of magnetic properties (B), does not distinguish strong from weak ligands (D), and makes no exact prediction of tetrahedral geometries (E) — all genuine shortcomings. …
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KEAM 2025Set eng-2025-04294 marksMCQ
$[NiCl_4]^{2-}$
(i) Magnetic moment
(ii) Geometry
(iii) Hybridisation of central metal ion
(iv) Oxidation state of nickel (A) (i),
(ii) and
(iv) (B) (i),
(ii) and
(iii) (C) (ii),
(iii) and
(iv) (D)
(ii) and
(iii) (E) (i), (ii),
(iii) and (iv)
›Reveal solutionSolution
Both have Ni in the $+2$ state, so (iv) does not differ. They differ in magnetic moment (i), geometry (ii) and hybridisation (iii) — option (B).

Both complexes have $Ni^{2+}$ ( $d^8$ ), so the oxidation state (iv) is the same in both — it cannot be a point of difference.

$[NiCl_4]^{2-}$ : $Cl^-$ is a weak-field ligand, so no pairing occurs. Configuration $d^8$ with 2 unpaired electrons → $sp^3$ hybridisation → tetrahedral, paramagnetic. …
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KEAM 2025Set eng-2025-04294 marksMCQ
$Q. Four complex ions are given in Column I and the colours of light absorbed are given in Column II. Match the correct answer from the codes given below. Complex:$
(a) $[Ti(H_2O)_6]^{3+}$ ;
(b) $[Cu(H_2O)_4]^{2+}$ ;
(c) $[CoCl(NH_3)_5]^{2+}$ ;
(d) $[Co(NH_3)_6]^{3+}$ Colour of light absorbed:
(i) Blue;
(ii) Yellow;
(iii) Blue green;
(iv) Red (A) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) (B) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) (C) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i) (D) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) (E) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
›Reveal solutionSolution
[!TLDR]

Using crystal-field strength to rank the $d$ - $d$ absorption energies, $[Ti(H_2O)_6]^{3+}$ absorbs blue-green, $[Cu(H_2O)_4]^{2+}$ red, $[CoCl(NH_3)_5]^{2+}$ yellow and $[Co(NH_3)_6]^{3+}$ blue — option (A).

Concept

In a transition-metal complex, $d$ -electrons are excited across the crystal-field gap $\Delta$ . The complex absorbs the wavelength matching $\Delta$ ; a stronger ligand field means a larger $\Delta$ and absorption of higher-energy (shorter-wavelength, toward blue/violet) light. This CFT reasoning is standard in the NCERT/CBSE coordination-chemistry chapter that the KEAM syllabus draws from.

Solution

(b) $[Cu(H_2O)_4]^{2+}$ : aqua ligands give a small $\Delta$ ; the deep-blue solution absorbs low-energy red light $\Rightarrow$ (iv).

(a) $[Ti(H_2O)_6]^{3+}$ : a single $d$ - $d$ band near $500\,\text{nm}$ absorbs blue-green light (complex looks purple) $\Rightarrow$ (iii). …
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KEAM 2025Set pha-2025-0424F4 marksMCQ
$[Ni(CO)_4]$
›Reveal solutionSolution
"Spin free" = high spin. Only $[CoF_6]^{3-}$ , with the weak-field $F^-$ ligand, keeps its d-electrons unpaired (high spin).

A spin-free (high-spin) complex forms with weak-field ligands, where the crystal-field splitting is too small to force pairing.

$[Ni(CO)_4]$ : $CO$ strong field, $Ni(0)$ $d^{10}$ — diamagnetic, not relevant as "spin-free."

$[Co(NH_3)_6]^{3+}$ : $NH_3$ strong-ish field, $Co^{3+}$ $d^6$ — low spin.

$[Ni(CN)_4]^{2-}$ : $CN^-$ strong field — low-spin, square planar, diamagnetic. …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$[MnCl_6]^{3-}$
›Reveal solutionSolution
$[Co(C_2O_4)_3]^{3-}$ is diamagnetic.

Check the unpaired electrons in each ion:

$[MnCl_6]^{3-}$ : Mn $^{3+}$ $d^4$ , weak-field Cl $^-$ $\Rightarrow$ 4 unpaired — paramagnetic.

$[Fe(CN)_6]^{3-}$ : Fe $^{3+}$ $d^5$ low spin $t_{2g}^5$ $\Rightarrow$ 1 unpaired — paramagnetic.

$[Co(C_2O_4)_3]^{3-}$ : Co $^{3+}$ $d^6$ ; oxalate is a fairly strong field ligand giving low spin $t_{2g}^6 e_g^0$ $\Rightarrow$ 0 unpaired — diamagnetic. …
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KEAM 2025Set pha-2025-0429F4 marksMCQ
$[Co(NH_3)_6]^{3+}$
›Reveal solutionSolution
Outer-orbital (high-spin) complexes form with weak-field ligands that use the outer $d$ orbitals ( $sp^3d^2$ ). Among the options only $[MnCl_6]^{3-}$ has a weak-field ligand ( $Cl^-$ ); the rest have strong-field ligands ( $CN^-$ , $NH_3$ , $C_2O_4^{2-}$ ) giving inner-orbital ( $d^2sp^3$ ) complexes.

Reasoning

Whether a complex is inner- or outer-orbital depends on the ligand field strength:

$[Co(NH_3)_6]^{3+}$ — $NH_3$ strong field, $d^2sp^3$ , inner orbital.

$[Mn(CN)_6]^{3-}$ — $CN^-$ strong field, inner orbital.

$[Co(C_2O_4)_3]^{3-}$ — Co(III) with oxalate, low-spin inner orbital. …
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KEAM 2024Set eng-2024-06094 marksMCQ
$[Ti(H_2O)_6]^{3+}$
›Reveal solutionSolution

$Ti^{3+}$ is a $d^1$ ion; the single electron jumps $t_{2g}\to e_g$ .

In $[Ti(H_2O)_6]^{3+}$ , titanium is in the $+3$ state: $Ti$ is $[Ar]3d^24s^2$ , so $Ti^{3+}$ is $[Ar]3d^1$ — a single $d$ electron. In an octahedral field it occupies the lower $t_{2g}$ set: $t_{2g}^1e_g^0$ . Absorption of light (giving the violet col …
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KEAM 2024Set eng-2024-06094 marksMCQ
$CO < H_2O < Cl^{-} < I^{-}$
›Reveal solutionSolution

Increasing field strength: $I^-<Cl^-<H_2O<CO$ .

In the spectrochemical series the ligands are arranged by increasing crystal-field splitting power. The relevant fragment is
$I^-<Br^-<Cl^-<F^-<H_2O<NH_3<en<CN^-\approx CO.$
Hence for the given ligands the increasing field strength is …
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KEAM 2024Set eng-2024-06094 marksMCQ
$[CoF_6]^{3-}$
›Reveal solutionSolution

$[CoF_6]^{3-}$ is a high-spin $d^6$ complex (weak-field $F^-$ ) with 4 unpaired electrons.

Cobalt in $[CoF_6]^{3-}$ is $Co^{3+}$ : $[Ar]3d^6$ . Fluoride is a weak-field ligand, so the splitting $\Delta_o$ is small and the complex is high spin. The six $d$ electrons fill as

$t_{2g}^4\,e_g^2,$ …
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Q.Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

(i) [CoF6]3−[CoF_6]^{3-}[CoF6​]3−, [Co(H2O)6]2+[Co(H_2O)_6]^{2+}[Co(H2​O)6​]2+, [Co(CN)6]3−[Co(CN)_6]^{3-}[Co(CN)6​]3−

(ii) [FeF6]3−[FeF_6]^{3-}[FeF6​]3−, [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}[Fe(H2​O)6​]2+, [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−

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(i) $[CoF_6]^{3-}$ , $[Co(H_2O)_6]^{2+}$ , $[Co(CN)_6]^{3-}$

(ii) $[FeF_6]^{3-}$ , $[Fe(H_2O)_6]^{2+}$ , $[Fe(CN)_6]^{4-}$

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