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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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 $Cr^{3+}$ …

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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