Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following m

NCERT Exemplar · Q51

Q. Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?

(i)

Co^{2+}

(ii)

Cr^{2+}

(iii)

Mn^{2+}

(iv)

Cr^{3+}

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Concept understanding — Magnetic Moment Calculation

From a Paperclip to a Magnet: The Intuition

You already know that a magnet can pick up iron nails. But what is actually happening inside that nail when it gets near the magnet? And why does a plastic comb, rubbed on hair, pick up tiny bits of paper — but never iron filings?

The answer lies in magnetization — the process by which a material becomes magnetic.

Think of a piece of iron as a chaotic crowd of tiny compass needles. Each needle is an atomic magnetic moment (a tiny magnet, arising from the spin of electrons). In unmagnetized iron, these needles point in random directions. Their magnetic effects cancel out, so the iron as a whole shows no net magnetism.

Now bring a strong magnet close. Its magnetic field acts like a command: "Line up!" The tiny compass needles inside the iron start rotating, aligning themselves with the external field. The more they align, the stronger the iron's own magnetic field becomes. This alignment is magnetization.

Note

Magnetization is not the same as inducing a current. It is a purely magnetic reorientation of atomic dipoles inside a material.

The Precise Definition

Magnetization ( $\vec{M}$ ) is the net magnetic dipole moment per unit volume of a material. It tells you how strongly a material is magnetized — how many tiny atomic magnets are aligned, and in which direction.

If a material has $N$ atoms per unit volume, each with an average magnetic moment $\vec{\mu}_{\text{avg}}$ , then:

\vec{M} = N \vec{\mu}_{\text{avg}}

The SI unit of $\vec{M}$ is amperes per metre (A/m). Why? Because a magnetic dipole moment has units of A·m², and dividing by volume (m³) gives A/m.

\vec{M} = \frac{\text{total magnetic dipole moment}}{\text{volume}}

How Magnetization Connects to the Magnetic Field

When a material gets magnetized, it produces its own magnetic field. The total magnetic field $\vec{B}$ inside the material is the sum of:

The external applied field $\vec{H}$ (caused by free currents, like the current in a solenoid)
The material's response — the magnetization $\vec{M}$

The fundamental relation is:

\vec{B} = \mu_0 (\vec{H} + \vec{M})

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

Watch out

Do not confuse $\vec{H}$ (magnetic field intensity, or "magnetizing field") with $\vec{B}$ (magnetic flux density). $\vec{H}$ is what you apply; $\vec{M}$ is what the material does; $\vec{B}$ is the total field you measure.

The Three Kinds of Magnetic Materials

Not all materials respond the same way to an external field. The magnetization $\vec{M}$ is proportional to $\vec{H}$ for most materials (at least for small fields):

\vec{M} = \chi_m \vec{H}

where $\chi_m$ is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.

Material Type	$\chi_m$	Behaviour	Example
Diamagnetic	Small and negative ( $\approx -10^{-5}$ )	Weakly repelled by a magnet; $\vec{M}$ opposes $\vec{H}$	Water, copper, bismuth
Paramagnetic	Small and positive ( $\approx 10^{-5}$ to $10^{-3}$ )	Weakly attracted; $\vec{M}$ aligns with $\vec{H}$	Aluminium, oxygen gas
Ferromagnetic	Large and positive ( $\gg 1$ )	Strongly attracted; $\vec{M}$ can be huge and persists even after $\vec{H}$ is removed	Iron, nickel, cobalt

Why this formula?

Magnetic Moment Calculation: Why the Formula Holds

Let's build this from first principles — understanding the why before the formula.

1. What is Magnetic Moment?

A magnetic moment ( $\vec{\mu}$ ) is a measure of the strength and orientation of a magnet or current loop. It tells us how strongly an object will interact with an external magnetic field.

The core idea: any moving charge creates a magnetic field. A loop of current is like a tiny bar magnet — its magnetic moment quantifies this.

2. The Fundamental Formula: Current Loop

The Setup

Consider a planar loop of wire carrying a steady current $I$ , enclosing an area $A$ .

Why $\mu = I A$ ?

Step 1: Force on a moving charge

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences:

\vec{F} = q (\vec{v} \times \vec{B})

Step 2: Torque on a current loop

For a rectangular loop of sides $a$ and $b$ ( $A = ab$ ), placed in a uniform $\vec{B}$ :

Current $I$ means charge flows. On side of length $a$ , the force magnitude is $F = I a B$ (since $I = \frac{q}{t}$ and $v = \frac{a}{t}$ ).
These forces on opposite sides form a couple (equal, opposite, not collinear).
Torque $\tau = \text{force} \times \text{perpendicular distance} = (I a B) \times (b \sin\theta)$

Step 3: Recognize the pattern

\tau = I (ab) B \sin\theta = I A B \sin\theta

This looks exactly like:

\tau = \mu B \sin\theta

Comparing, we identify:

\boxed{\mu = I A}

Why this works: The torque on a current loop is proportional to the current and the area — this product naturally defines the magnetic moment.

3. For a Single Moving Charge (Orbital Magnetic Moment)

The Setup

An electron of charge $-e$ moves in a circular orbit of radius $r$ with speed $v$ .

Why $\mu = \frac{e v r}{2}$ ?

Step 1: Treat orbit as a current loop

Time for one revolution: $T = \frac{2\pi r}{v}$
Current (charge per unit time): $I = \frac{e}{T} = \frac{e v}{2\pi r}$

Step 2: Apply $\mu = I A$

Area of orbit: $A = \pi r^2$
So: $\mu = \left(\frac{e v}{2\pi r}\right) (\pi r^2) = \frac{e v r}{2}$

Step 3: Express in terms of angular momentum

Orbital angular momentum: $L = m v r$
Therefore: $\boxed{\mu = \frac{e}{2m} L}$

Why this matters: The magnetic moment is directly proportional to angular momentum. The factor $\frac{e}{2m}$ is called the gyromagnetic ratio — it links mechanics to magnetism.

4. For a Solenoid (Many Turns)

The Setup

A solenoid of $N$ turns, length $l$ , carrying current $I$ , cross-sectional area $A$ .

Why $\mu = N I A$ ? …

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The spin-only magnetic moment depends on the number of unpaired electrons. $Co^{2+}$ and $Cr^{3+}$ both have 3 unpaired electrons, giving them nearly identical magnetic moments of about $3.87 \, \mu_B$ .

The magnetic moment of a transition metal ion arises primarily from the spin of unpaired electrons. While orbital motion also contributes, in many first-row transition metal ions the orbital contribution is "quenched" by the crystal field, so the spin-only formula works very well. The spin-only magnetic moment is given by:

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

where $n$ is the number of unpaired electrons and $\mu_B$ is the Bohr magneton. This formula is derived from the quantum mechanical spin angular momentum. The key point: ions with the same number of unpaired electrons will have the same spin-only magnetic moment.

Let’s find the number of unpaired electrons for each ion by writing their electronic configurations.

$Co^{2+}$

Cobalt (atomic number 27): $[Ar]\, 3d^7 4s^2$

For $Co^{2+}$ , remove two 4s electrons: $[Ar]\, 3d^7$

In an octahedral field, the 3d orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy). For $d^7$ , the configuration is $t_{2g}^5 e_g^2$ (high-spin, since pairing energy is high for first-row metals).

The $t_{2g}$ set has 3 orbitals and the $e_g$ set has 2 orbitals. Filling $d^7$ by Hund's rule (each orbital singly occupied before any pairing): the three $t_{2g}$ orbitals take 3 electrons singly first, then the remaining electrons pair up starting in $t_{2g}$ (since $t_{2g}$ is lower in energy) — giving $t_{2g}$ 2 paired electrons and 1 unpaired electron. The 2 electrons in $e_g$ each occupy a separate orbital, both unpaired.

Unpaired electrons: 1 (from $t_{2g}$ ) + 2 (from $e_g$ ) = 3 unpaired electrons.

So $n = 3$ for $Co^{2+}$ .
$Cr^{2+}$

Chromium (atomic number 24): $[Ar]\, 3d^5 4s^1$ (exception: half-filled stability).

For $Cr^{2+}$ , remove two electrons: first the 4s electron, then one 3d electron → $[Ar]\, 3d^4$

For $d^4$ , high-spin: $t_{2g}^3 e_g^1$ (all 4 electrons unpaired, since pairing energy is high).

Unpaired electrons: 4.
$Mn^{2+}$

Manganese (atomic number 25): $[Ar]\, 3d^5 4s^2$

For $Mn^{2+}$ , remove two 4s electrons → $[Ar]\, 3d^5$

Half-filled $d^5$ : all 5 orbitals singly occupied → 5 unpaired electrons.
$Cr^{3+}$

Chromium (atomic number 24): $[Ar]\, 3d^5 4s^1$

For $Cr^{3+}$ , remove three electrons: the 4s electron and two 3d electrons → $[Ar]\, 3d^3$ …

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Step 1: Understand the Concept

The spin-only magnetic moment is given by:

\mu_s = \sqrt{n(n+2)} \, \mu_B

where $n$ = number of unpaired electrons.

For first-row transition metals, orbital contribution is often quenched (especially in octahedral complexes), so spin-only is a good approximation.

Step 2: Find $n$ for each ion

We need the d-electron count for each ion:

Ion	Electronic Configuration	$d^n$	Unpaired electrons ( $n$ )
$Co^{2+}$	$[Ar]\,3d^7$	$d^7$	3 (high spin)
$Cr^{2+}$	$[Ar]\,3d^4$	$d^4$	4 (high spin)
$Mn^{2+}$	$[Ar]\,3d^5$	$d^5$	5 (high spin)
$Cr^{3+}$	$[Ar]\,3d^3$	$d^3$	3

So:

$Co^{2+}$ → $n = 3$
$Cr^{2+}$ → $n = 4$
$Mn^{2+}$ → $n = 5$
$Cr^{3+}$ → $n = 3$

Same spin-only magnetic moment means same $n$ .

Thus, $Co^{2+}$ and $Cr^{3+}$ both have $n = 3$ → same $\mu_s$ .

Common Mistakes & How to Avoid Them

✗ Mistake 1: Forgetting to account for high spin vs low spin

In weak field (high spin), electrons fill singly before pairing.
In strong field (low spin), pairing occurs earlier.
For $d^4$ , $d^5$ , $d^6$ , $d^7$ — the spin state can change.
How to avoid: Always check if the problem specifies octahedral and weak field (common for aqueous ions). If not stated, assume high spin for first-row transition metals unless strong field ligands are mentioned.

✗ Mistake 2: Confusing $d^n$ count with unpaired electrons

Example: $d^7$ does not mean 7 unpaired electrons — it means 3 unpaired (in high spin).
How to avoid: Write the Hund's rule filling for each $d^n$ explicitly.

✗ Mistake 3: Mixing up $Cr^{2+}$ and $Cr^{3+}$ …

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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 14 on this concept.

COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{Fe}^{3+}$
›Reveal solutionSolution
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. A value of 5.9 BM corresponds to $n = 5$ unpaired electrons. Among the options, only $\mathrm{Fe}^{3+}$ has 5 unpaired electrons, so the correct option is (A).

The key concept here is the spin-only magnetic moment formula:
$\mu = \sqrt{n(n+2)} \ \text{BM}$
where $n$ is the number of unpaired electrons. This formula works well for first-row transition metal ions because orbital angular momentum is often "quenched" by the crystal field. The experimental value 5.9 BM is very close to the theoretical value for $n=5$ , which is $\sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM. So we need to find which ion has exactly 5 unpaired electrons in its ground state.

Let’s check each option step by step:

Determine the electronic configuration of each ion

$\mathrm{Fe}^{3+}$ : Atomic number of Fe = 26. Neutral Fe: $[\mathrm{Ar}]\, 3d^6 4s^2$ . Removing three electrons (first from 4s, then from 3d) gives $[\mathrm{Ar}]\, 3d^5$ .

$\mathrm{Mn}^{3+}$ : Atomic number of Mn = 25. Neutral Mn: $[\mathrm{Ar}]\, 3d^5 4s^2$ . Removing three electrons gives $[\mathrm{Ar}]\, 3d^4$ .

$\mathrm{Ni}^{2+}$ : Atomic number of Ni = 28. Neutral Ni: $[\mathrm{Ar}]\, 3d^8 4s^2$ . Removing two electrons gives $[\mathrm{Ar}]\, 3d^8$ .

$\mathrm{Co}^{3+}$ : Atomic number of Co = 27. Neutral Co: $[\mathrm{Ar}]\, 3d^7 4s^2$ . Removing three electrons gives $[\mathrm{Ar}]\, 3d^6$ .

Count unpaired electrons using Hund’s rule

For $\mathrm{Fe}^{3+}$ ( $3d^5$ ): All five d-orbitals are half-filled, each with one electron → 5 unpaired electrons.

For $\mathrm{Mn}^{3+}$ ( $3d^4$ ): Four electrons occupy four orbitals singly (Hund’s rule) → 4 unpaired electrons.

For $\mathrm{Ni}^{2+}$ ( $3d^8$ ): The d-orbitals fill as: ↑↓ ↑↓ ↑ ↑ ↑ (two paired, two singly, one singly) → 2 unpaired electrons. …
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KCET 2026Set D31 markMCQ
$Cr^{2+}$
›Reveal solutionSolution

The spin-only magnetic moment depends on the number of unpaired electrons in the ion's d-configuration.

Step 1 — Electron configuration of Cr²⁺

Neutral Cr (Z = 24) is $[Ar]3d^5 4s^1$ . Forming $Cr^{2+}$ removes the single 4s electron first, then one 3d electron, giving $Cr^{2+} = [Ar]3d^4$ .

Step 2 — Counting unpaired electrons …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{A}=\mathrm{Cr}^{2+}(\mathrm{Z}=24) \mathrm{B}=\mathrm{Cu}^{2+}(\mathrm{Z}=29) \quad \mathrm{C}=\mathrm{Ni}^{2+}(\mathrm{Z}=28) \mathrm{D}=\mathrm{Fe}^{3+}(\mathrm{Z}=26)$
›Reveal solutionSolution
The number of unpaired electrons depends on the electronic configuration of each ion, considering the d-orbital filling and Hund's rule. The correct order is B (Cu²⁺, 1 unpaired) < C (Ni²⁺, 2 unpaired) < A (Cr²⁺, 4 unpaired) < D (Fe³⁺, 5 unpaired), which corresponds to option (A).

Concept & Intuition

The key is to write the ground-state electron configurations for each ion, focusing on the d-electrons. Transition metal ions lose s-electrons first, then d-electrons if needed. Unpaired electrons arise from half-filled or partially filled d-orbitals following Hund's rule (each orbital gets one electron before pairing). The more unpaired electrons, the higher the magnetic moment. Here, we count them directly.

Step-by-step reasoning

Cr²⁺ (Z = 24)

Neutral Cr: [Ar] 4s¹ 3d⁵ (exception: half-filled d gives stability).

Remove two electrons: first from 4s, then from 3d → Cr²⁺: [Ar] 3d⁴.

Hund's rule: four d-orbitals each get one electron, one orbital empty → 4 unpaired electrons.

Cu²⁺ (Z = 29)

Neutral Cu: [Ar] 4s¹ 3d¹⁰ (exception: fully filled d).

Remove two electrons: one from 4s, one from 3d → Cu²⁺: [Ar] 3d⁹.

One d-orbital has a single electron (the rest are paired) → 1 unpaired electron.

Ni²⁺ (Z = 28)

Neutral Ni: [Ar] 4s² 3d⁸.

Remove two 4s electrons → Ni²⁺: [Ar] 3d⁸.

Filling: five orbitals get 2, 2, 2, 1, 1 (two unpaired) → 2 unpaired electrons.

Fe³⁺ (Z = 26)

Neutral Fe: [Ar] 4s² 3d⁶. …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Co}=27$
›Reveal solutionSolution
The spin-only magnetic moment depends on the number of unpaired electrons, which is determined by the metal’s oxidation state and the ligand field (strong-field CN⁻ vs. weak-field Cl⁻). The correct order is $\left[\mathrm{MnCl}_4\right]^{2-} > \left[\mathrm{CoCl}_4\right]^{2-} > \left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ , corresponding to option (B).

Concept & Intuition

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ Bohr magnetons, where $n$ is the number of unpaired electrons. So the problem reduces to finding $n$ for each complex. The key twist: ligands like CN⁻ are strong-field (low-spin), causing maximum pairing, while Cl⁻ is weak-field (high-spin), favoring unpaired electrons. Also, geometry matters — here all complexes are either octahedral or tetrahedral, which affects the d-orbital splitting and thus the electron configuration.

Step-by-step reasoning

Determine oxidation states and d-electron counts

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ : CN⁻ is −1 each, so Fe must be +2 (since 6×(−1) + x = −4 → x = +2). Fe (Z=26): [Ar]3d⁶. Fe²⁺: 3d⁶.

$\left[\mathrm{MnCl}_4\right]^{2-}$ : Cl⁻ is −1 each, so Mn is +2 (4×(−1) + x = −2 → x = +2). Mn (Z=25): [Ar]3d⁵. Mn²⁺: 3d⁵.

$\left[\mathrm{CoCl}_4\right]^{2-}$ : Cl⁻ is −1 each, so Co is +2 (4×(−1) + x = −2 → x = +2). Co (Z=27): [Ar]3d⁷. Co²⁺: 3d⁷.

Analyze $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ — octahedral, strong-field

CN⁻ is a strong-field ligand, causing large splitting. For Fe²⁺ (d⁶) in an octahedral field, strong-field means low-spin: all electrons pair in the $t_{2g}$ set first. Configuration: $t_{2g}^6 e_g^0$ . All 6 electrons are paired → 0 unpaired electrons.

$\mu = \sqrt{0(0+2)} = 0$ BM.

Analyze $\left[\mathrm{MnCl}_4\right]^{2-}$ — tetrahedral, weak-field

Cl⁻ is a weak-field ligand. Tetrahedral splitting is smaller than octahedral, so always high-spin. Mn²⁺ is d⁵. In tetrahedral geometry, the $e$ set is lower in energy than $t_2$ . High-spin d⁵: each of the five d-orbitals gets one electron (Hund’s rule) → 5 unpaired electrons.

$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.

Analyze $\left[\mathrm{CoCl}_4\right]^{2-}$ — tetrahedral, weak-field …
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COMEDK 2025Set 2025-E1 markMCQ
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
›Reveal solutionSolution
The magnetic moment depends on the number of unpaired electrons, which is determined by the metal’s oxidation state, ligand field strength (weak vs. strong field), and the resulting high-spin or low-spin configuration. The correct increasing order is B < C < A < D, corresponding to option (B).

Concept & Intuition

Magnetic moment (μ) for a transition metal complex is given by the spin-only formula:
$\mu = \sqrt{n(n+2)} \ \text{BM}$
where $n$ is the number of unpaired electrons. So the order of magnetic moments is simply the order of number of unpaired electrons.

To find $n$ , we need:

The oxidation state of iron in each complex.

Whether the ligand is weak-field (high-spin) or strong-field (low-spin).

The d-electron count and how electrons fill the $t_{2g}$ and $e_g$ orbitals.

Let’s work through each complex step by step.

1. Determine oxidation states and d-electron counts

A: $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ — Water is neutral, so Fe is +2. Fe²⁺ has electron configuration [Ar] 3d⁶.

B: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ — CN⁻ is -1 each, total −6 from ligands, so Fe must be +2 to give overall −4. Again, Fe²⁺, 3d⁶.

C: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ — CN⁻ gives −6, so Fe is +3. Fe³⁺ is 3d⁵.

D: $\left[\mathrm{FeF}_6\right]^{3-}$ — F⁻ is -1 each, total −6, so Fe is +3. Also 3d⁵.

So we have two Fe²⁺ (d⁶) and two Fe³⁺ (d⁵) complexes.

2. Identify ligand field strength

H₂O is a weak-field ligand (causes small splitting).

F⁻ is also a weak-field ligand.

CN⁻ is a strong-field ligand (causes large splitting).

For weak-field ligands, electrons fill according to Hund’s rule (high-spin). For strong-field ligands, pairing occurs before occupying higher orbitals (low-spin).

3. Determine unpaired electrons for each

A: Fe²⁺ (d⁶), weak field (H₂O) → high-spin.

In an octahedral field: $t_{2g}^4 e_g^2$ → 4 unpaired electrons.

$n = 4$ .

B: Fe²⁺ (d⁶), strong field (CN⁻) → low-spin.

All 6 electrons pair in $t_{2g}$ : $t_{2g}^6 e_g^0$ → 0 unpaired electrons.

$n = 0$ .

C: Fe³⁺ (d⁵), strong field (CN⁻) → low-spin. …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{M}^{\mathrm{n}+}$
›Reveal solutionSolution
The key is to determine the oxidation state, geometry, and number of unpaired electrons for each complex, then match the hybridisation and magnetic moment. The correct matches are A→R, B→S, C→Q, D→P, so option (A) is correct.

We need to match each coordination compound with its hybridisation and magnetic moment. The magnetic moment (in Bohr magnetons, BM) is given by $\mu = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons. Let’s analyse each complex step by step.

Complex A: $\mathrm{Ni(CO)_4}$

Nickel is in zero oxidation state (CO is neutral).

CO is a strong field ligand, causing pairing of electrons.

Ni(0) has atomic number 28: $[Ar] 3d^8 4s^2$ . In the presence of strong field CO, the 4s electrons are used for bonding, and the 3d electrons pair up.

The complex is tetrahedral (4 ligands), so hybridisation is $sp^3$ .

All electrons are paired → $\mu = 0$ BM.

So A matches R ( $sp^3, \mu=0$ ).

Complex B: $\left[\mathrm{Ni(CN)_4}\right]^{2-}$

CN⁻ is a strong field ligand.

Oxidation state: Ni + 4(−1) = −2 → Ni is +2.

Ni²⁺: $[Ar] 3d^8$ . Strong field CN⁻ causes pairing of 3d electrons, leaving one d-orbital empty.

Geometry is square planar (4 ligands), hybridisation is $dsp^2$ .

All electrons paired → $\mu = 0$ BM.

So B matches S ( $dsp^2, \mu=0$ ).

Complex C: $\left[\mathrm{Ni(Cl)_4}\right]^{2-}$

Cl⁻ is a weak field ligand.

Ni is again +2 (same as B).

Weak field → no pairing; Ni²⁺ has 8 d-electrons, which in a tetrahedral field give 2 unpaired electrons (since $t_2$ orbitals are higher energy, electrons occupy them singly first).

Geometry is tetrahedral → hybridisation $sp^3$ .

$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$ BM.

So C matches Q ( $sp^3, \mu=2.84$ BM).

Complex D: $\left[\mathrm{MnBr_4}\right]^{2-}$

Br⁻ is a weak field ligand. …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathbf{P}$
›Reveal solutionSolution
Pyrolusite ( $\mathrm{MnO_2}$ ) fused with KOH and $\mathrm{KNO_3}$ gives the manganate(VI) ion in $\mathbf{P}=\mathrm{K_2MnO_4}$ . The metal ion $\mathrm{Mn^{6+}}$ is $d^1$ , so $\mu=\sqrt{1(1+2)}=1.732$ BM. Permanganate then oxidises thiosulphate in neutral medium to sulphate, in which sulphur is $+6$ . The correct option is (D).

Concept

In an oxidising alkaline fusion, $\mathrm{Mn(IV)}$ in pyrolusite is raised to $\mathrm{Mn(VI)}$ , giving the green manganate ion $\mathrm{MnO_4^{2-}}$ . The spin-only magnetic moment depends only on the number of unpaired electrons through $\mu=\sqrt{n(n+2)}$ BM. Separately, the oxidation number of an atom is read from the formula of the species it ends up in.

Solution

Identify P. $\mathrm{MnO_2}+2\mathrm{KOH}+\mathrm{KNO_3}\rightarrow \mathrm{K_2MnO_4}+\mathrm{KNO_2}+\mathrm{H_2O}$ . So $\mathbf{P}=\mathrm{K_2MnO_4}$ and the metal ion is $\mathrm{Mn^{6+}}$ .

Magnetic moment of $\mathrm{Mn^{6+}}$ . Configuration $[\mathrm{Ar}]3d^1$ gives $n=1$ , so $\mu=\sqrt{1(1+2)}=\sqrt{3}=1.732$ BM.

Fate in acid. $3\mathrm{MnO_4^{2-}}+4\mathrm{H^+}\rightarrow 2\mathrm{MnO_4^-}+\mathrm{MnO_2}+2\mathrm{H_2O}$ (disproportionation), giving permanganate and $\mathrm{MnO_2}$ . …
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KCET 2024Set B-21 markMCQ

\mathrm{Zn}^{2+}

›Reveal solution

Write the $d^n$ configuration of each ion, then read off colour (needs a partly-filled $d$ shell) and magnetic moment from the spin-only formula.

Step 1 — Get the $d$ -configurations.

For a transition-metal cation, remove the $4s$ electrons first, then the $3d$ electrons.

Ion	Atomic no.	Neutral atom	$M^{2+}$ configuration
$\mathrm{Zn}^{2+}$	30	$[\mathrm{Ar}]3d^{10}4s^2$	$[\mathrm{Ar}]\,3d^{10}$
$\mathrm{Cu}^{2+}$	29	$[\mathrm{Ar}]3d^{10}4s^1$	$[\mathrm{Ar}]\,3d^{9}$
$\mathrm{Ni}^{2+}$	28	$[\mathrm{Ar}]3d^{8}4s^2$	$[\mathrm{Ar}]\,3d^{8}$

Step 2 — Match III: $\mathrm{Ni}^{2+}$ → (i) $d^8$ .

Read straight off the table above. This is the most direct match, and it already eliminates options (A) and (C), which both send $d^8$ to $\mathrm{Zn}^{2+}$ .

Step 3 — Match I: $\mathrm{Zn}^{2+}$ → (ii) colourless.

Colour in transition-metal ions arises from $d$ – $d$ electronic transitions: an electron absorbs visible light and is promoted from the lower ( $t_{2g}$ ) to the higher ( $e_g$ ) set. This demands a partially filled $d$ subshell. $\mathrm{Zn}^{2+}$ is $d^{10}$ — completely full, no vacancy to promote into — so no $d$ – $d$ transition is possible and the ion is colourless.

Step 4 — Match II: $\mathrm{Cu}^{2+}$ → (iii) $\mu = 1.73$ BM.

Use the spin-only formula, where $n$ = number of unpaired electrons: …

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COMEDK 2024Set 2024-A1 markMCQ
$Q. The metallic ions that have almost same spin only magnetic moment are :$
(i) $\mathrm{Co}^{2+}$
(ii) $\mathrm{Mn}^{2+}$
(iii) $\mathrm{Cr}^{2+}$
(iv) $\mathrm{Cr}^{3+}$ (A) $$ \text {
(iii) and
(iv) } $(B)$ \text {
(i) and
(ii) } $(C)$ \text {
(i) and
(iv) } $(D)$ \text {
(i) and
(iii) } $$
›Reveal solutionSolution
Counting unpaired electrons: Co $^{2+}$ (d $^7$ ) = 3, Mn $^{2+}$ (d $^5$ ) = 5, Cr $^{2+}$ (d $^4$ ) = 4, Cr $^{3+}$ (d $^3$ ) = 3. Co $^{2+}$ and Cr $^{3+}$ share $n=3$ , hence the same $\mu=3.87$ BM → (i) and (iv).

Spin-only moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons:

(i) Co $^{2+}$ : $3d^7$ → 3 unpaired → $\mu=\sqrt{15}=3.87$ BM

(ii) Mn $^{2+}$ : $3d^5$ → 5 unpaired → $\mu=\sqrt{35}=5.92$ BM …
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$\left[\mathrm{Fe} \mathrm{F}_6\right]^{3-}$
›Reveal solutionSolution
[!TLDR]

Using d-electron counts and strong/weak-field ligands, the unpaired electrons are 5, 0, 4, 1 for A-D, matching option (C).

Concept

In Valence Bond Theory a strong-field ligand (like CN-) forces electron pairing, giving a low-spin inner-orbital complex, while a weak-field ligand (F-, H2O) leaves the electrons unpaired (high spin). Always count the d-electrons on the metal ion first (CBSE/NCERT Class 12, Coordination Compounds).

Solution

(A) $[\mathrm{FeF_6}]^{3-}$ : Fe $^{3+}$ is $3d^5$ ; F $^-$ is weak-field $\Rightarrow$ high spin $\Rightarrow$ 5 unpaired $\Rightarrow$ R.

(B) $[\mathrm{Fe(CN)_6}]^{4-}$ : Fe $^{2+}$ is $3d^6$ ; CN $^-$ strong-field $\Rightarrow$ low spin ( $t_{2g}^6$ ) $\Rightarrow$ 0 unpaired $\Rightarrow$ P. …
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$\mathrm{X}(\mathrm{Z}=26)$
›Reveal solutionSolution
The metal is iron (Z=26); in $[\mathrm{Fe(CN)_2(CO)_4}]^+$ it is Fe $^{3+}$ (d $^5$ ), and since CN $^-$ and CO are both strong-field ligands, the complex is low-spin with 1 unpaired electron, giving $\mu = 1.73$ BM — option (B).

Step-by-step reasoning

Identify the metal. $Z=26$ is iron (Fe).

Find the oxidation state.

CN $^-$ contributes $-1$ each (×2), CO is neutral (×4), overall complex charge $=+1$ :

$x + 2(-1) + 4(0) = +1 \quad\Rightarrow\quad x = +3$
So Fe is in the $+3$ state.

Find the d-electron count.

Neutral Fe is $[Ar]3d^64s^2$ ; Fe $^{3+}$ is $[Ar]3d^5$ .

Apply the ligand field. …
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COMEDK 2024Set 2024-M1 markMCQ
$Q. The permanganate ion in acid medium acts as an oxidant and gets converted to its lower oxidation state. What would be the spin only magnetic moment of such reduced manganese ion? (A) 2.84 BM (B) 4.90 BM (C) 3.87 BM (D) 5.92 BM$
›Reveal solutionSolution
In acidic medium, permanganate ( $\text{MnO}_4^-$ ) is reduced to $\text{Mn}^{2+}$ , which has five unpaired $3d$ electrons; the spin-only magnetic moment is $\sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM, so the correct option is (D).

The key concept here is the relationship between oxidation state, electronic configuration, and magnetic moment. The permanganate ion ( $\text{MnO}_4^-$ ) contains manganese in the +7 oxidation state. In acidic medium, it is a powerful oxidant and gets reduced to a lower oxidation state — specifically, to the manganous ion ( $\text{Mn}^{2+}$ ). The question asks for the spin-only magnetic moment of this reduced ion. The spin-only formula is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. So we need to find $n$ for $\text{Mn}^{2+}$ .

Determine the electronic configuration of the reduced manganese ion.

Manganese (atomic number 25) has the ground-state configuration $[\text{Ar}]\, 3d^5 4s^2$ .

In the $\text{Mn}^{2+}$ ion, two electrons are removed — first from the $4s$ orbital (as is standard for transition metals), giving $[\text{Ar}]\, 3d^5$ .

Find the number of unpaired electrons in $\text{Mn}^{2+}$ .

The $3d$ subshell has five orbitals. For a $d^5$ configuration, Hund’s rule tells us that electrons occupy all five orbitals singly with parallel spins before any pairing occurs. Thus, all five $3d$ electrons are unpaired.

So $n = 5$ .

Apply the spin-only magnetic moment formula.

The formula is:

$\mu = \sqrt{n(n+2)} \ \text{BM}$
Substituting $n = 5$ :

$\mu = \sqrt{5 \times 7} = \sqrt{35} \ \text{BM}$ …
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Q.Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?

Concept understanding — Magnetic Moment Calculation

From a Paperclip to a Magnet: The Intuition

The Precise Definition

How Magnetization Connects to the Magnetic Field

The Three Kinds of Magnetic Materials

Why this formula?

Magnetic Moment Calculation: Why the Formula Holds

1. What is Magnetic Moment?

2. The Fundamental Formula: Current Loop

The Setup

Why $\mu = I A$ ?

3. For a Single Moving Charge (Orbital Magnetic Moment)

The Setup

Why $\mu = \frac{e v r}{2}$ ?

4. For a Solenoid (Many Turns)

The Setup

Why $\mu = N I A$ ? …

Method: Spin-Only Magnetic Moment Calculation

Steps

Applying to the given ions

Step 1: Understand the Concept

Step 2: Find $n$ for each ion

Common Mistakes & How to Avoid Them

✗ Mistake 1: Forgetting to account for high spin vs low spin

✗ Mistake 2: Confusing $d^n$ count with unpaired electrons

✗ Mistake 3: Mixing up $Cr^{2+}$ and $Cr^{3+}$ …

Step 1 — Electron configuration of Cr²⁺

Step 2 — Counting unpaired electrons …

Concept

Solution

Concept

Solution

Q.Transition elements show magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost same spin only magnetic moment?

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