The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magne

NCERT Exemplar · Q15

Cr^{3+}

(i) 2.87 B.M.

(ii) 3.87 B.M.

(iii) 3.47 B.M.

(iv) 3.57 B.M.

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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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$Cr^{3+}$ is $[Ar]3d^3$ with 3 unpaired electrons, so $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ B.M. — option (ii).

The spin-only magnetic moment is

\mu_s = \sqrt{n(n+2)}\ \text{B.M.}

where $n$ is the number of unpaired electrons.

Configuration. Neutral Cr ( $Z=24$ ) is $[Ar]3d^5\,4s^1$ . Removing three electrons (the $4s$ electron first, then two $3d$ electrons) gives

$Cr^{3+}:\ [Ar]3d^3.$ …

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🧠 The Core Concept First

The spin-only magnetic moment formula is:

\mu_s = \sqrt{n(n+2)} \ \text{B.M.}

Where:

$n$ = number of unpaired electrons
B.M. = Bohr Magneton

For $\text{Cr}^{3+}$ :

Atomic number of Cr = 24
Electronic configuration: $[\text{Ar}]\, 3d^5\, 4s^1$
$\text{Cr}^{3+}$ loses 3 electrons → $[\text{Ar}]\, 3d^3$
So $n = 3$ unpaired electrons

\mu_s = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ \text{B.M.}

✓ Correct answer: (ii) 3.87 B.M.

✗ Common Mistake #1: Using the wrong $n$ (counting all d-electrons instead of unpaired ones)

What students do wrong:

They see $3d^3$ and think $n = 3$ is correct — but sometimes they mistakenly count total d-electrons (5 for Cr atom) or use the wrong ion.

How to avoid:

Always write the exact electronic configuration of the ion (not the atom).
Count only unpaired electrons in the outermost d-orbital.
For $\text{Cr}^{3+}$ : $3d^3$ → all three are unpaired → $n = 3$ .

✗ Common Mistake #2: Using the orbital contribution formula

What students do wrong:

They try to include orbital angular momentum using $\mu = \sqrt{4S(S+1) + L(L+1)}$ for first-row transition metals.

How to avoid:

For first-row transition metal ions (like Cr), the orbital angular momentum is quenched by the crystal field.
Always use the spin-only formula unless the question explicitly says “including orbital contribution.”
In exam questions, “magnetic moment” for these ions almost always means spin-only.

✗ Common Mistake #3: Calculation error in $\sqrt{n(n+2)}$

What students do wrong:

They compute $3 \times 5 = 15$ correctly, but then approximate $\sqrt{15}$ as 3.47 or 3.57.
They confuse $\sqrt{15}$ with $\sqrt{12}$ (which is for $n=2$ ) or $\sqrt{24}$ (for $n=4$ ).

How to avoid:

Memorise common values:
- $n=1$ → $\sqrt{3} \approx 1.73$
- $n=2$ → $\sqrt{8} \approx 2.83$
- $n=3$ → $\sqrt{15} \approx 3.87$
- $n=4$ → $\sqrt{24} \approx 4.90$
- $n=5$ → $\sqrt{35} \approx 5.92$
Do not approximate — use the exact square root or the given options to match.

--- …

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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.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Which of the following ion has the maximum theoretical magnetic moment? [Fe (Z=26), Cr (Z=24), Ti (Z=22), Co (Z=27)]$
(a) Fe3+
(b) Cr3+
(c) Ti3+
(d) Co3+
›Reveal solutionSolution
Magnetic moment mu = sqrt(n(n+2)) BM rises with the number of unpaired electrons n; the d5 configuration allows the maximum possible unpaired electrons.

Find the d-electron configuration of each 3+ ion:

Fe (Z=26): [Ar]3d⁶4s² → Fe3+ = [Ar]3d⁵ → 5 unpaired electrons (high spin) → μ = √(5×7) = √35 ≈ 5.92 BM

Cr (Z=24): [Ar]3d⁵4s¹ → Cr3+ = [Ar]3d³ → 3 unpaired electrons → μ = √(3×5) = √15 ≈ 3.87 BM

Ti (Z=22): [Ar]3d²4s² → Ti3+ = [Ar]3d¹ → 1 unpaired electron → μ = √3 ≈ 1.73 BM …
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GUJCET 2025Set 031 markMCQ
$\sqrt{35}$
›Reveal solutionSolution
[!TLDR]

$\sqrt{35}$ BM corresponds to 5 unpaired electrons, matching $\text{Mn}^{2+}$ ( $3d^5$ ).

Concept

The spin-only magnetic moment is $\mu=\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

Solution

Solve for $n$ :
$\sqrt{n(n+2)}=\sqrt{35}\ \Rightarrow\ n(n+2)=35\ \Rightarrow\ n=5$
Now find the divalent ion with 5 unpaired $d$ electrons:

$\text{Cr}^{2+}: 3d^4$ → 4 unpaired …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. What is the value of magnetic moment of divalent ion having atomic number 30 in aqueous solution.$
(a) 0 BM
(b) 2.84 BM
(c) 1.73 BM
(d) 5.92 BM
›Reveal solutionSolution

Atomic number 30 is Zn; its divalent ion Zn2+ has a fully filled 3d10 configuration with zero unpaired electrons, so its magnetic moment is zero.

Element with Z = 30 is Zinc, electronic configuration [Ar] 3d10 4s2.

Zn2+ is formed by removing the two 4s electrons, giving [Ar] 3d10 - a completely filled d-subshell with no unpaired electrons.

…

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GUJCET 2024Set 131 markMCQ
$Co^{2+}$
›Reveal solutionSolution
Spin-only moment $\mu=\sqrt{n(n+2)}$ BM increases with the number of unpaired electrons $n$ ; $Mn^{2+}$ ( $d^5$ ) has the most.

Concept: Count $d$ -electrons and unpaired electrons:

$Ti^{2+}$ : $d^2$ , $n=2$ .

$Mn^{2+}$ : $d^5$ , $n=5$ $\Rightarrow\mu=\sqrt{5\times7}=5.92$ BM.

$Fe^{2+}$ : $d^6$ , $n=4$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. What is the magnetic moment of a divalent ion in aqueous solution if its atomic number is 28?$
(a) 3.87 BM
(b) 2.84 BM
(c) 1.73 BM
(d) 4.90 BM
›Reveal solutionSolution

Atomic number 28 is Nickel; its divalent ion Ni2+ has electronic configuration 3d8, giving 2 unpaired electrons, and the spin-only magnetic moment formula gives 2.84 BM.

Ni (Z=28): [Ar] 3d8 4s2

Ni2+: loses the two 4s electrons -> [Ar] 3d8

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. How many number of unpaired electrons are there in complex ion [Ni(CN)4]2-?$
(a) 4
(b) 3
(c) 2
(d) 0
›Reveal solutionSolution

[Ni(CN)4]2- has Ni2+ (d8) with the strong-field ligand CN-, which forces pairing of the d-electrons into a square planar, diamagnetic (dsp2) arrangement.

Ni2+: [Ar] 3d8

CN- is a very strong field ligand (high in the spectrochemical series), causing the 3d8 electrons to pair up, freeing one 3d orbital for dsp2 hybridisation (square planar geometry). …

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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Which compound has magnetic moment equal to 4.90 BM?$
(a) Cr2(SO4)3
(b) NiSO4
(c) FeSO4
(d) MnSO4
›Reveal solutionSolution
mu = 4.90 BM means 4 unpaired electrons; Fe2+ (d6) fits, so FeSO4.

Spin-only moment mu = sqrt(n(n+2)) BM. For mu = 4.90: n(n+2) = 24 -> n = 4 unpaired electrons.

Check the metal ions:

Cr3+ (Cr2(SO4)3): d3 -> 3 unpaired -> 3.87 BM.

Ni2+ (NiSO4): d8 -> 2 unpaired -> 2.83 BM. …
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GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. Among the following, which compound has the highest magnetic moment?$
(a) MnSO4
(b) CrCl3
(c) Ni(NO3)2
(d) FeSO4
›Reveal solutionSolution

Magnetic moment μ = √[n(n+2)] BM, where n = number of unpaired electrons; more unpaired electrons means a higher moment.

MnSO4: Mn2+ is d5 (high spin) → 5 unpaired electrons → μ = √35 = 5.92 BM.

FeSO4: Fe2+ is d6 → 4 unpaired electrons → μ = 4.90 BM.

CrCl3: Cr3+ is d3 → 3 unpaired electrons → μ = 3.87 BM. …

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GUJCET 2021Set 151 markMCQ
$Q. If atomic number of element is 26, then magnetic moment is ___ BM of its divalent aqueous ion? (A) 1.73 (B) 3.87 (C) 2.83 (D) 4.90$
›Reveal solutionSolution

Fe²⁺ ( $3d^6$ ) has 4 unpaired electrons, giving spin-only $\mu=\sqrt{n(n+2)}=\sqrt{24}=4.90$ BM.

Concept: Atomic number 26 = Fe. The divalent ion Fe²⁺ has configuration $[Ar]3d^6$ , with 4 unpaired electrons. …
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GUJCET 2020Set 071 markMCQ
$Q. The divalent ion of which of the following element in aqueous solution has magnetic moment 5.92 BM? (A) Fe (B) Cr (C) Co (D) Mn$
›Reveal solutionSolution

$\mu=\sqrt{n(n+2)}=5.92\Rightarrow n=5$ unpaired; the $d^5$ divalent ion is Mn²⁺.

Concept — spin-only magnetic moment. $\mu=\sqrt{n(n+2)}$ BM. $5.92=\sqrt{5\cdot7}=\sqrt{35}$ , so $n=5$ unpaired electrons. Mn²⁺ is $[\text{Ar}]3d^5$ (5 …
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GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Magnetic moment of a divalent ion in aqueous solution if its atomic number is 25 ______.$
(a) 4.90 BM
(b) 5.92 BM
(c) 2.84 BM
(d) 3.87 BM
›Reveal solutionSolution

Element 25 is manganese; its divalent ion Mn2+ has a 3d5 configuration with 5 unpaired electrons (a spin-only 'half-filled, maximally paramagnetic' case), giving the largest common magnetic moment among first-row transition ions.

…

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GUJCET 2019Set 131 markMCQ
$\text{Ni}^{2+}, \text{Co}^{2+}$
›Reveal solutionSolution
Magnetic moment depends only on the number of unpaired electrons; $Fe^{3+}$ and $Mn^{2+}$ are both $3d^5$ .

Concept: Spin-only magnetic moment $\mu = \sqrt{n(n+2)}\,BM$ , where $n$ = number of unpaired electrons. Two ions have the same moment if they have the same $n$ .

Counting unpaired d-electrons:

$Ni^{2+}$ : $3d^8$ → 2 ; $Co^{2+}$ : $3d^7$ → 3 (not equal).

$Fe^{2+}$ : $3d^6$ → 4 ; $Mn^{2+}$ : $3d^5$ → 5 (not equal). …
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Q.The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of $Cr^{3+}$ ion is ___________.

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 Formula

Steps

🧠 The Core Concept First

✗ Common Mistake #1: Using the wrong $n$ (counting all d-electrons instead of unpaired ones)

✗ Common Mistake #2: Using the orbital contribution formula

✗ Common Mistake #3: Calculation error in $\sqrt{n(n+2)}$

Concept

Solution

Q.The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr3+Cr^{3+}Cr3+ ion is ___________.

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Q.The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of $Cr^{3+}$ ion is ___________.

Unlock everything free for 14 days