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 15 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write the formula for calculating 'spin only' magnetic moment.$
›Reveal solutionSolution

The spin-only formula estimates a transition metal ion's magnetic moment purely from its number of unpaired electrons, ignoring orbital contribution.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Give one example of a complex having tetrahedral geometry and paramagnetic in nature.$
›Reveal solutionSolution

$[NiCl_4]^{2-}$ is the standard example of a tetrahedral, paramagnetic complex, arising from $sp^3$ hybridisation of $Ni^{2+}$ with the weak-field $Cl^-$ ligand.

Why $[NiCl_4]^{2-}$ fits

Ni has configuration $[Ar]3d^84s^2$ ; in $Ni^{2+}$ , this becomes $3d^8$ . $Cl^-$ is a weak-field ligand (low in the spectrochemical series), so it does not force pairing of the $3d$ electrons. With four ligands and no $d$ -orbital freed by pairing, nickel uses one $4s$ and three $4p$ orbitals — $sp^3$ hybridisation — giving a **tetrahedr …

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CBSE 2026Set ANNUAL1 mark
$[Cr(H_2O)_6]^{3+}$
›Reveal solutionSolution

Counting unpaired $d$ -electrons for each ion under VBT shows $Fe^{2+}$ ( $d^6$ , high-spin, 4 unpaired) is more paramagnetic than $Cr^{3+}$ ( $d^3$ , always 3 unpaired).

$[Cr(H_2O)_6]^{3+}$

$Cr$ ( $Z=24$ ) is $[Ar]3d^54s^1$ ; $Cr^{3+}$ removes 3 electrons to give $3d^3$ . With only 3 electrons for the three $t_{2g}$ orbitals, Hund's rule places one electron in each — $t_{2g}^3$ — giving 3 unpaired electrons, regardless of whether the ligand is weak- or strong-field (there's no way to pair up 3 electrons across 3 orbitals to reduce this further).

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

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Value of magnetic moment of a divalent ion in aqueous solution having atomic number 25, will be 5.92 B.M.$
›Reveal solutionSolution

Mn2+ has 5 unpaired electrons, giving a spin-only moment of 5.92 B.M., so the statement is true.

Atomic number 25 = manganese, [Ar] 3d5 4s2. The divalent ion Mn2+ = [Ar] 3d5, which has 5 unpaired electrons.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which one of the following metal ions is likely to have a magnetic moment of 1.73 BM?$
(a) Fe²⁺
(b) Mn²⁺
(c) Cr²⁺
(d) Cu²⁺
›Reveal solutionSolution
Using μ = √(n(n+2)) BM, 1.73 BM means n = 1 unpaired electron; Cu²⁺ (d⁹) is the only ion with one unpaired electron — option (D).

The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. A value of $1.73$ BM gives $\sqrt{1(1+2)} = \sqrt{3} = 1.73$ , so $n = 1$ unpaired electron.

Now count unpaired electrons for each ion:

$Fe^{2+}$ : $3d^6$ → 4 unpaired ( $\mu \approx 4.9$ BM). …
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CBSE 2025Set JZ1 markMCQ
$Q. Magnetic moment of a bivalent ion in aqueous solution will be, if its atomic number is 25$
(a) 1.73 BM
(b) 2.83 BM
(c) 4.96 BM
(d) 5.92 BM
›Reveal solutionSolution

Mn $^{2+}$ ( $3d^5$ ) has 5 unpaired electrons, so $\mu=\sqrt{5(5+2)}=5.92$ BM — option (d).

Concept. The magnetic moment of a transition-metal ion depends only on the number of unpaired d-electrons ( $n$ ), through the spin-only formula $\mu=\sqrt{n(n+2)}$ BM.

Step 1 — identify the ion. Atomic number 25 → manganese (Mn), configuration $[Ar]3d^5 4s^2$ . A bivalent ion Mn $^{2+}$ loses the two $4s$ electrons: Mn $^{2+}=[Ar]3d^5$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The spin magnetic moment of Co3+ ion is:$
(a) sqrt(3) BM
(b) sqrt(8) BM
(c) sqrt(15) BM
(d) sqrt(24) BM
›Reveal solutionSolution

Co3+ has the configuration [Ar]3d6; in the high-spin (free-ion) state this places 4 electrons unpaired, giving a spin-only magnetic moment of √(n(n+2)) = √24 BM.

Cobalt (Z = 27) has ground state configuration [Ar]3d7 4s2. Removing 3 electrons to form Co3+ removes the two 4s electrons first and then one 3d electron, giving Co3+: [Ar]3d6.

Filling the five d orbitals with 6 electrons by Hund's rule (maximum multiplicity, i.e., high-spin, as would apply to the free gaseous ion or in a weak field):

↑↓ ↑ ↑ ↑ ↑ → one orbital doubly occupied, four orbitals singly occupied → 4 unpaired electrons (n = 4).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is a paramagnetic complex?$
(a) [Ni(H2O)6]2+
(b) [Ni(CO)4]
(c) [Zn(NH3)4]2+
(d) [Co(NH3)6]
›Reveal solutionSolution

Ni2+ (d8) with the weak-field ligand H2O keeps 2 electrons unpaired; the other three complexes all have a d10 or strong-field-paired d-count and are diamagnetic.

[Ni(H₂O)₆]²⁺: Ni²⁺ is d⁸; H₂O is a weak-field ligand and cannot force pairing, so 2 electrons remain unpaired — paramagnetic (octahedral, sp³d² outer-orbital complex).

[Ni(CO)₄]: here nickel is in the zero oxidation state, Ni(0), configuration 3d¹⁰4s⁰ — a completely filled d-subshell regardless of ligand field, so it is diamagnetic (sp³, tetrahedral).

[Zn(NH₃)₄]²⁺: Zn²⁺ is always 3d¹⁰ (fully filled) in its only common oxidation state, so it is diamagnetic (sp³, tetrahedral) irrespective of the ligand.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Mn^{+2}$
(a) 2.84 B.M
(b) 3.87 B.M
(c) 4.90 B.M
(d) 5.92 B.M
›Reveal solutionSolution

Mn²⁺ has a half-filled d⁵ configuration with 5 unpaired electrons, and the spin-only formula gives a magnetic moment of 5.92 B.M.

$Mn^{2+}$ has the configuration $[Ar]3d^5$ — a half-filled d-subshell, with all 5 electrons unpaired (by Hund's rule, each of the 5 d-orbitals holds one electron).

Using the spin-only formula:

$\mu = \sqrt{n(n+2)}\ B.M., \quad n = 5$

…

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CBSE 2023Set ANNUAL1 mark
$M^{2+}(aq)$
›Reveal solutionSolution

$Z=27$ corresponds to cobalt; $Co^{2+}(aq)$ has the configuration $3d^7$ with 3 unpaired electrons, giving a spin-only magnetic moment of $\sqrt{15} \approx 3.87\ BM$ .

Identify the ion: $Z = 27$ is cobalt (Co), with ground-state configuration $[Ar]\,3d^7 4s^2$ . Removing 2 electrons (always from $4s$ first) to form $Co^{2+}$ gives:

$Co^{2+}: [Ar]\,3d^7$

Count unpaired electrons: Distributing 7 electrons among the five $3d$ orbitals following Hund's rule (each orbital singly filled first, before pairing) for the aqua ion (a weak-field, high-spin case):

$\underbrace{\uparrow\downarrow}_{}\ \underbrace{\uparrow\downarrow}_{}\ \underbrace{\uparrow}_{}\ \underbrace{\uparrow}_{}\ \underbrace{\uparrow}_{}$

…

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CBSE 2020Set 56/2/11 markMCQ
$Co^{3+}$
›Reveal solutionSolution
Cobalt loses three electrons to form $\text{Co}^{3+}$ , leaving an electronic configuration of $[\text{Ar}] \, 3d^6$ . In the $d^6$ configuration, pairing depends on ligand field strength, but the question asks for the ground-state free ion, which follows Hund's rule and has 4 unpaired electrons.

The number of unpaired electrons in a transition metal ion determines its magnetic properties. To find this, we need the electronic configuration of the ion and then apply Hund's rule of maximum multiplicity.

Understanding the Configuration

Cobalt has atomic number 27. The neutral atom's electronic configuration is:
$[\text{Ar}] \, 3d^7 \, 4s^2$
When cobalt forms $\text{Co}^{3+}$ , it loses three electrons. Electrons are always removed from the outermost shell first—both $4s$ electrons go first, then one $3d$ electron:
$\text{Co}^{3+}: [\text{Ar}] \, 3d^6$
Applying Hund's Rule

The five $3d$ orbitals can hold up to 10 electrons. With 6 electrons to place, Hund's rule tells us to:

Maximize unpaired electrons first by placing one electron in each orbital with parallel spin.

Then pair up any remaining electrons.

Let me show the filling pattern for $3d^6$ :

$d_{xy}$ $d_{yz}$ $d_{zx}$ $d_{x^2-y^2}$ $d_{z^2}$
↑↓ ↑ ↑ ↑ ↑

The first five electrons occupy all five orbitals singly (all spin-up). The sixth electron must pair with one of them.

Result: 4 unpaired electrons and 1 paired set. …
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CBSE 2019Set ANNUAL1 mark
$Q. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.$
›Reveal solutionSolution

The divalent ion of element 25 is $Mn^{2+}$ , a $3d^5$ ion with all five d-orbitals singly occupied; the spin-only formula then gives $\mu \approx 5.92\ BM$ .

Element with atomic number 25 is manganese (Mn): $[Ar]3d^5 4s^2$ .

Forming the divalent ion $Mn^{2+}$ removes the two $4s$ electrons first:

$Mn^{2+}:\ [Ar]3d^5$

By Hund's rule, all five 3d electrons occupy the five d-orbitals singly (maximum multiplicity), giving $n = 5$ unpaired electrons.

…

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$d_{xy}$	$d_{yz}$	$d_{zx}$	$d_{x^2-y^2}$	$d_{z^2}$
↑↓	↑	↑	↑	↑

←Previous questionQ14 — Interstitial compounds are formed when small atoms are trapped inside …→Next questionQ16 — KMnO₄ acts as an oxidising agent in alkaline medium. When alkaline KMn…

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 $\sqrt{n(n+2)}$

Understanding the Configuration

Applying Hund's Rule

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