Atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral

NCERT Exemplar · Q32

Q. Atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

(i)

[Co(NH_3)_6]^{3+}

(ii)

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

(iii)

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

(iv)

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

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Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

Imagine you are a negatively charged electron sitting on a metal ion. All around you, the space is perfectly spherical — every direction feels the same. Your energy depends only on how far you are from the nucleus, not on which way you face.

Now imagine that six negative ions (or the negative ends of polar molecules) march in from the $x$ , $y$ , and $z$ axes and stop close to you. Suddenly, the space around you is no longer uniform. If you try to move straight toward one of these approaching ions, you feel a strong repulsion — that path costs extra energy. If you move between the axes (say, along a diagonal), you feel less repulsion because you are farther from the incoming charges.

This is the core intuition: when ligands approach a metal ion, they break the spherical symmetry of the space around the metal. Different directions in space are no longer equivalent. Electrons in orbitals that point directly at the ligands get pushed up in energy; electrons in orbitals that point between the ligands stay lower.

The Precise Statement

d

For the most common geometry — octahedral — here is what happens:

Six ligands sit at the corners of an octahedron, along the $+x$ , $-x$ , $+y$ , $-y$ , $+z$ , $-z$ axes.
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point their lobes directly along these axes. These are the $e_g$ set. They feel maximum repulsion → higher energy.
The $d_{xy}$ , $d_{xz}$ , and $d_{yz}$ orbitals point their lobes between the axes (into the octahedral faces). These are the $t_{2g}$ set. They feel less repulsion → lower energy.

The energy gap between these two sets is denoted by $\Delta_o$ (or $10\,Dq$ ). The $t_{2g}$ set drops by $0.4\Delta_o$ and the $e_g$ set rises by $0.6\Delta_o$ , keeping the average energy unchanged (the "barycentre" rule).

Note

The labels $e_g$ and $t_{2g}$ come from group theory — they describe how the orbitals transform under the symmetry operations of an octahedron. You do not need to memorise the derivation, but the notation is standard in every exam.

Why This Matters

Crystal field splitting explains three things you will see repeatedly:

Colour — electrons can jump from $t_{2g}$ to $e_g$ by absorbing visible light. The gap $\Delta_o$ determines the colour you see.
Magnetism — if $\Delta_o$ is large, electrons pair up in the lower $t_{2g}$ set (low spin). If $\Delta_o$ is small, electrons spread out (high spin). This changes the number of unpaired electrons. …

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split in energy when placed in an electrostatic field created by surrounding ligands (anions or polar molecules). The key result is that five degenerate d-orbitals split into two or more sets with different energies. Let's understand why this happens.

1. The Starting Point: Degenerate d-Orbitals

In a free transition metal ion (no ligands), all five d-orbitals have the same energy (degenerate). Their shapes are:

$d_{xy}$ , $d_{xz}$ , $d_{yz}$ — lobes lie between the x, y, z axes (called $t_{2g}$ set in octahedral symmetry)
$d_{x^2-y^2}$ , $d_{z^2}$ — lobes point directly along the x, y, z axes (called $e_g$ set)

Key idea: The spatial orientation of each orbital determines how it interacts with approaching ligands.

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

Imagine six ligands approaching along the +x, –x, +y, –y, +z, –z axes (octahedral geometry).

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

Their lobes point directly at the ligands.
The negatively charged ligands repel the electron density in these orbitals.
This repulsion raises the energy of these orbitals — they become less stable (higher energy).

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

Their lobes point between the axes (e.g., $d_{xy}$ lobes lie in the xy-plane but at 45° to x and y).
They avoid the ligands — less repulsion.
Their energy is lower than the $e_g$ set.

The Splitting Pattern

\Delta_{\text{oct}} = E(e_g) - E(t_{2g})

Where:

$E(e_g)$ = energy of $d_{x^2-y^2}$ and $d_{z^2}$ (higher)
$E(t_{2g})$ = energy of $d_{xy}$ , $d_{xz}$ , $d_{yz}$ (lower)
$\Delta_{\text{oct}}$ is called the crystal field splitting energy (CFSE)

Why the name? The $e_g$ orbitals are "doubly degenerate" (2 orbitals), $t_{2g}$ are "triply degenerate" (3 orbitals). The letters come from group theory symmetry labels.

3. The Energy Conservation Rule

The total energy of all five d-orbitals must remain constant (no energy is created or destroyed). So:

The center of gravity (average energy) of the split set equals the original degenerate energy.
For octahedral splitting:
- 2 $e_g$ orbitals go up by $+0.6\Delta_{\text{oct}}$ each
- 3 $t_{2g}$ orbitals go down by $-0.4\Delta_{\text{oct}}$ each

Check:

2 \times (+0.6\Delta) + 3 \times (-0.4\Delta) = 1.2\Delta - 1.2\Delta = 0

This conservation of energy is a fundamental constraint — the splitting is not arbitrary.

4. The Tetrahedral Case: Why It's Opposite and Smaller

In a tetrahedral complex, four ligands approach from alternate corners of a cube. The axes are different:

The $d_{xy}$ , $d_{xz}$ , $d_{yz}$ orbitals now point closer to the ligands (more repulsion).
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals point away from ligands (less repulsion).

Result:

$e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — lower energy
$t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — higher energy

The splitting is inverted compared to octahedral.

Magnitude:

\Delta_{\text{tet}} \approx \frac{4}{9} \Delta_{\text{oct}}

Why smaller?

Only 4 ligands (vs. 6) → less total repulsion.
Ligands are not directly along axes → weaker interaction. …

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The key is to identify which complex has all electrons paired (diamagnetic) by considering the metal’s oxidation state, d-electron count, and the strong-field (low-spin) nature of CN⁻ and NH₃ ligands in an inner orbital (d²sp³) octahedral geometry. The diamagnetic complexes are (i) $[Co(NH_3)_6]^{3+}$ and (iii) $[Fe(CN)_6]^{4-}$ .

Why Crystal Field Splitting Decides Magnetism

In an octahedral complex, the five d-orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals). The size of this splitting — called $\Delta_o$ — depends on the ligand. Strong-field ligands like CN⁻ and NH₃ cause a large $\Delta_o$ , forcing electrons to pair up in the $t_{2g}$ orbitals before occupying $e_g$ . This gives a low-spin configuration.

An inner orbital complex uses the inner $(n-1)d$ orbitals for hybridisation (d²sp³), which is exactly what happens in low-spin cases. A complex is diamagnetic only if all electrons are paired — that is, the d-electron count is such that the low-spin configuration has zero unpaired electrons.

Let’s check each complex one by one.

1. $[Co(NH_3)_6]^{3+}$

Cobalt (atomic number 27) in +3 oxidation state: $Co^{3+}$ has electron configuration $[Ar]\,3d^6$ .
NH₃ is a strong-field ligand, so this is a low-spin d⁶ system.
In low-spin octahedral: all six electrons pair up in the three $t_{2g}$ orbitals → $t_{2g}^6\,e_g^0$ .
Zero unpaired electrons → diamagnetic.

Tip

For d⁶, low-spin gives all paired electrons. This is the classic example of a diamagnetic Co(III) complex.

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

Manganese (atomic number 25) in +3 oxidation state: $Mn^{3+}$ has $[Ar]\,3d^4$ .
CN⁻ is a very strong-field ligand → low-spin.
Low-spin d⁴: four electrons go into $t_{2g}$ , but one must pair up → $t_{2g}^4\,e_g^0$ .
That means two unpaired electrons (since $t_{2g}^4$ has one doubly occupied orbital and two singly occupied ones).
Hence paramagnetic, not diamagnetic.

Watch out

A common mistake is to think d⁴ always has four unpaired electrons. In strong field, pairing occurs, leaving two unpaired — not zero.

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

Iron (atomic number 26) in +2 oxidation state: $Fe^{2+}$ has $[Ar]\,3d^6$ .
CN⁻ is strong-field → low-spin d⁶.
Same as Co³⁺ above: $t_{2g}^6\,e_g^0$ — all electrons paired.
Diamagnetic.

Note

…

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Common Mistakes in Crystal Field Splitting (Inner Orbital / Diamagnetic Complexes)

Mistake 1: Confusing Inner vs. Outer Orbital Complexes

The error: Students think any complex with $\mathrm{CN^-}$ is automatically inner orbital.

Why it’s wrong: Inner orbital (low-spin) requires the metal ion to use $(n-1)d$ orbitals for hybridisation — this happens only when the ligand is strong enough to cause pairing. But the metal’s oxidation state and $d^n$ configuration also matter.

How to avoid:

For octahedral complexes:
- Inner orbital → $d^2sp^3$ hybridisation (uses inner $d$ orbitals) → low-spin
- Outer orbital → $sp^3d^2$ hybridisation (uses outer $d$ orbitals) → high-spin
Strong field ligands ( $\mathrm{CN^-}$ , $\mathrm{CO}$ , $\mathrm{NH_3}$ for some metals) favour low-spin.
Always check the $d^n$ count first.

Mistake 2: Forgetting to Find the Correct Oxidation State

The error: Students directly use the atomic number without calculating the oxidation state of the metal in the complex ion.

Why it’s wrong: The number of $d$ electrons depends on the charge on the metal ion, not the neutral atom.

How to avoid:

Let the complex be $[M(L)_n]^{x}$
Let ligand charge be $q$ (e.g., $\mathrm{CN^-}$ is $-1$ , $\mathrm{NH_3}$ is $0$ )
Then: Oxidation state of $M$ = $x - (n \times q)$
Then: $d^n$ = Atomic number of $M$ — oxidation state — (charge of previous noble gas core)

Example:

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

$\mathrm{CN^-}$ charge = $-1$ , so total ligand charge = $6 \times (-1) = -6$
Let Fe oxidation state = $y$
$y + (-6) = -4 \Rightarrow y = +2$
Fe (atomic no. 26) → $\mathrm{Fe^{2+}}$ = $[Ar]\,3d^6$

Mistake 3: Assuming All $\mathrm{NH_3}$ Complexes Are Low-Spin

The error: Thinking $\mathrm{NH_3}$ always causes pairing.

Why it’s wrong: $\mathrm{NH_3}$ is a moderate field ligand — it can cause pairing only for certain $d^n$ configurations (e.g., $d^6$ in $\mathrm{Co^{3+}}$ ) but not for all.

How to avoid:

$\mathrm{NH_3}$ is stronger than water but weaker than $\mathrm{CN^-}$
For $\mathrm{Co^{3+}}$ ( $d^6$ ), $\mathrm{NH_3}$ is strong enough to cause pairing → low-spin $t_{2g}^6$ → diamagnetic
For $\mathrm{Fe^{3+}}$ ( $d^5$ ), $\mathrm{NH_3}$ is not strong enough → high-spin → paramagnetic

Mistake 4: Confusing Diamagnetic with Paramagnetic

The error: Students think “inner orbital” always means diamagnetic.

Why it’s wrong: Inner orbital means low-spin, but if the $t_{2g}$ set is not completely filled, unpaired electrons remain → paramagnetic.

How to avoid:

Diamagnetic → all electrons paired → $t_{2g}$ fully filled (e.g., $t_{2g}^6$ )
Paramagnetic → at least one unpaired electron
For $d^4$ , $d^5$ , $d^6$ , $d^7$ low-spin configurations, check the $t_{2g}$ occupancy:

$d^n$	Low-spin $t_{2g}$	Unpaired?
$d^4$	$t_{2g}^4$	2 unpaired → paramagnetic
$d^5$	$t_{2g}^5$	1 unpaired → paramagnetic
$d^6$	$t_{2g}^6$	0 unpaired → diamagnetic
$d^7$	$t_{2g}^6 e_g^1$	1 unpaired → paramagnetic

Mistake 5: Ignoring the “Inner Orbital” Condition in the Question

The error: Students check only diamagnetism, forgetting the question asks for inner orbital and diamagnetic.

Why it’s wrong: A complex could be diamagnetic but outer orbital (e.g., $[Ni(CN)_4]^{2-}$ is square planar, not relevant here, but the principle holds).

How to avoid: …

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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.Atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

Concept understanding — Crystal Field Splitting

Crystal Field Splitting: From Intuition to Precision

The Precise Statement

Why This Matters

Why this formula?

Crystal Field Splitting: Why the Energy Splitting Occurs

1. The Starting Point: Degenerate d-Orbitals

2. The Octahedral Case: Why $e_g$ Orbitals Are Higher in Energy

What happens to $d_{x^2-y^2}$ and $d_{z^2}$ ?

What happens to $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ?

The Splitting Pattern

3. The Energy Conservation Rule

4. The Tetrahedral Case: Why It's Opposite and Smaller

Result:

Magnitude:

Why Crystal Field Splitting Decides Magnetism

1. $[Co(NH_3)_6]^{3+}$

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

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

Method: Crystal Field Theory (CFT) — Strong Field / Low Spin Analysis

Why this method?

Steps

Step 1: Determine the oxidation state and $d^n$ configuration

Step 2: Apply strong field (low spin) filling rule

Common Mistakes in Crystal Field Splitting (Inner Orbital / Diamagnetic Complexes)

Mistake 1: Confusing Inner vs. Outer Orbital Complexes

Mistake 2: Forgetting to Find the Correct Oxidation State

Mistake 3: Assuming All $\mathrm{NH_3}$ Complexes Are Low-Spin

Mistake 4: Confusing Diamagnetic with Paramagnetic

Mistake 5: Ignoring the “Inner Orbital” Condition in the Question

Concept

Solution

Q.Atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

Why Crystal Field Splitting Decides Magnetism

1. [Co(NH3)6]3+[Co(NH_3)_6]^{3+}[Co(NH3​)6​]3+

2. [Mn(CN)6]3−[Mn(CN)_6]^{3-}[Mn(CN)6​]3−

3. [Fe(CN)6]4−[Fe(CN)_6]^{4-}[Fe(CN)6​]4−

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1. $[Co(NH_3)_6]^{3+}$

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

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

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