Why do compounds having similar geometry have different magnetic moment?

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Q. Why do compounds having similar geometry have different magnetic moment?

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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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Magnetic moment depends on the number of unpaired electrons, which is determined by the ligand field strength and the metal ion's electron configuration — not just the geometry. Two compounds with identical geometry can have different magnetic moments if the metal ion or the ligand field strength differs, leading to different numbers of unpaired electrons.

The magnetic moment of a transition metal complex is a direct measure of its number of unpaired electrons. For first-row transition metals, the spin-only formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ is the number of unpaired electrons) works very well. So when two compounds share the same geometry — say, both are octahedral — but have different magnetic moments, the reason must lie in a different number of unpaired electrons.

Why would the same geometry give different $n$ ? The answer lies in the ligand field strength and the metal ion's electron configuration. In octahedral geometry, the $d$ orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy). For a given $d^n$ configuration, the electrons can arrange themselves in two ways:

High-spin: Electrons occupy all five $d$ orbitals singly before pairing, maximizing unpaired electrons. This happens when the ligand field splitting $\Delta_o$ is small (weak field ligands like $\ce{H2O}$ , $\ce{F-}$ ).
Low-spin: Electrons pair up in the lower $t_{2g}$ orbitals before occupying $e_g$ , minimizing unpaired electrons. This happens when $\Delta_o$ is large (strong field ligands like $\ce{CN-}$ , $\ce{CO}$ ).

Thus, two octahedral complexes of the same metal ion (same $d^n$ ) can have different magnetic moments if one has weak-field ligands (high-spin) and the other has strong-field ligands (low-spin). Even if the metal ion is different, the $d^n$ count and ligand strength together determine the moment.

Let's work through a concrete example to see this clearly.

Identify the metal ion and its $d^n$ configuration.

Consider $\ce{[CoF6]^{3-}}$ and $\ce{[Co(NH3)6]^{3+}}$ . Both are octahedral. Cobalt in the +3 oxidation state has the electron configuration $[\ce{Ar}]\,3d^6$ . So $d^6$ for both.
Determine the ligand field strength.

$\ce{F-}$ is a weak field ligand (low in the spectrochemical series). $\ce{NH3}$ is a moderate field ligand, but for $\ce{Co^{3+}}$ , it is strong enough to cause pairing. In fact, $\ce{Co^{3+}}$ is a $d^6$ ion that is particularly prone to low-spin configurations with ligands like $\ce{NH3}$ , $\ce{CN-}$ , etc.
Predict the electron configuration in the $t_{2g}$ and $e_g$ orbitals.

For $\ce{[CoF6]^{3-}}$ (weak field, high-spin):
- $\Delta_o$ is small.
- Electrons fill according to Hund's rule: $t_{2g}^4 e_g^2$ (four electrons in $t_{2g}$ with one pair, two unpaired in $e_g$ ).
- Number of unpaired electrons $n = 4$ .
- Magnetic moment $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM.
For $\ce{[Co(NH3)6]^{3+}}$ (strong field, low-spin):
- $\Delta_o$ is large.
- Electrons pair up in $t_{2g}$ : $t_{2g}^6 e_g^0$ (all six electrons paired in $t_{2g}$ ).
- Number of unpaired electrons $n = 0$ .
- Magnetic moment $\mu = 0$ BM (diamagnetic).
Compare the magnetic moments.

The fluoride complex has a moment of about 4.9 BM, while the ammine complex is diamagnetic. Both are octahedral, but the magnetic moments are drastically different — purely because of the ligand field strength.

Watch out

A common mistake is to assume that geometry alone determines the magnetic moment. In fact, the same geometry can yield high-spin or low-spin configurations depending on the ligand. Always check the spectrochemical series and the metal ion's $d^n$ count. …

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Here is a breakdown of the common mistakes students make when tackling the question: "Why do compounds having similar geometry have different magnetic moment?" — along with the conceptual fixes to avoid each.

1. Forgetting that Magnetic Moment depends on Number of Unpaired Electrons

The Mistake:

Students assume that if two complexes have the same geometry (e.g., both octahedral), they must have the same magnetic moment. They ignore the fact that magnetic moment is directly proportional to the number of unpaired electrons ( $n$ ).

The Correct Concept:

Magnetic moment ( $\mu$ ) is given by the spin-only formula:

\mu = \sqrt{n(n+2)} \ \text{BM}

where $n$ = number of unpaired electrons.

Two complexes with identical geometry can have different $n$ if they contain different metal ions or different oxidation states.

How to Avoid:

Always count unpaired electrons using the electronic configuration of the metal ion in that geometry (considering crystal field splitting). Never assume geometry alone determines $\mu$ .

2. Ignoring the Effect of Crystal Field Splitting (Strong vs Weak Field Ligands)

The Mistake:

Students think that for a given metal ion, the number of unpaired electrons is fixed. They forget that ligand field strength can change the pairing of electrons.

The Correct Concept:

In octahedral complexes:

Weak field ligands (high spin) → electrons fill all five $d$ orbitals singly before pairing → more unpaired electrons → higher $\mu$ .
Strong field ligands (low spin) → electrons pair in lower $t_{2g}$ orbitals → fewer unpaired electrons → lower $\mu$ .

Example:

$[\text{CoF}_6]^{3-}$ (weak field) → $n = 4$ → $\mu \approx 4.90$ BM

$[\text{Co(NH}_3)_6]^{3+}$ (strong field) → $n = 0$ → $\mu = 0$ BM

Both are octahedral, but $\mu$ differs drastically.

How to Avoid:

Always check the spectrochemical series of ligands. For a given metal ion and geometry, identify whether the complex is high-spin or low-spin.

3. Confusing Geometry with Coordination Number

The Mistake:

Students think that if two complexes have the same coordination number (e.g., 6), they must have the same geometry (e.g., octahedral). They then assume magnetic moment must be same.

The Correct Concept:

Coordination number 6 can give octahedral, trigonal prismatic, or distorted octahedral geometries. Different geometries produce different crystal field splitting patterns, leading to different numbers of unpaired electrons.

How to Avoid:

First confirm the actual geometry (e.g., using VSEPR or crystal field theory). Then apply the appropriate splitting diagram for that geometry.

4. Forgetting to Account for Oxidation State of the Metal

The Mistake:

Students compare two complexes with the same metal but different oxidation states and wonder why magnetic moments differ — even though geometry is same.

The Correct Concept:

Different oxidation states mean different $d^n$ configurations. For example:

$\text{Fe}^{2+}$ ( $d^6$ ) in octahedral weak field → $n = 4$
$\text{Fe}^{3+}$ ( $d^5$ ) in octahedral weak field → $n = 5$

Both can be octahedral, but $\mu$ differs because $n$ differs.

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

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Q.Why do compounds having similar geometry have different 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 Formula

Steps

Why Similar Geometry → Different Magnetic Moment?

Example: Octahedral $d^4$ complexes …

1. Forgetting that Magnetic Moment depends on Number of Unpaired Electrons

2. Ignoring the Effect of Crystal Field Splitting (Strong vs Weak Field Ligands)

3. Confusing Geometry with Coordination Number

4. Forgetting to Account for Oxidation State of the Metal

Understanding the Configuration

Applying Hund's Rule

Q.Why do compounds having similar geometry have different magnetic moment?

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