Explain why $[Fe(H_2O)_6]^{3+}$ has magnetic moment value of 5.92 BM whereas $[Fe(CN)_6]^{3-}$ has a value of only 1.74 BM.

> [!TLDR] > The difference in magnetic moments arises from the different ligand field strengths of $H_2O$ and $CN^-$. $H_2O$ is a weak field ligand, leaving $Fe^{3+}$ in a high-spin $d^5$ configuration with 5 unpaired electrons ($\mu = 5.92$ BM). $CN^-$ is a strong field ligand, causing pairing and giving a low-spin $d^5$ configuration with only 1 unpaired electron ($\mu = 1.74$ BM). The magnetic moment of a transition metal complex tells us how many unpaired electrons are present. The formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ is the number of unpaired electrons) connects the experimental value directly to the electronic configuration. So when you see 5.92 BM and 1.74 BM, you're really being asked: why does the same metal ion, $Fe^{3+}$, have a different number of unpaired electrons in these two complexes? The answer lies in the **crystal field theory** and the nature of the ligands. 1. **Identify the metal ion and its $d$-electron count.** Iron in both complexes is in the +3 oxidation state. Atomic number of Fe is 26. $Fe^{3+}$ means we remove 3 electrons: $[Ar]3d^5$. So in both cases, we have a $d^5$ system. 2. **Recall the crystal field splitting in an octahedral field.** In an octahedral complex, the five $d$-orbitals split into two sets: the lower-energy $t_{2g}$ set (three orbitals) and the higher-energy $e_g$ set (two orbitals). The energy gap between them is called $\Delta_o$ (or $10 Dq$). 3. **The key: how do electrons fill these orbitals?** For a $d^5$ ion, there are two possible arrangements: - **High-spin**: Electrons fill all five orbitals singly first (Hund's rule), giving 5 unpaired electrons. This happens when $\Delta_o$ is small — it costs less energy to put an electron in a higher $e_g$ orbital than to pair up in a $t_{2g}$ orbital. - **Low-spin**: Electrons pair up in the $t_{2g}$ set first, giving only 1 unpaired electron. This happens when $\Delta_o$ is large — pairing energy is less than the energy needed to promote

Explain why [Fe(H₂O)₆]³⁺ has magnetic moment value of 5.92 BM whereas [Fe(CN)₆]³⁻ has a value of only 1.74 BM.

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[Fe(H_2O)_6]^{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 difference in magnetic moments arises from the different ligand field strengths of $H_2O$ and $CN^-$ . $H_2O$ is a weak field ligand, leaving $Fe^{3+}$ in a high-spin $d^5$ configuration with 5 unpaired electrons ( $\mu = 5.92$ BM). $CN^-$ is a strong field ligand, causing pairing and giving a low-spin $d^5$ configuration with only 1 unpaired electron ( $\mu = 1.74$ BM).

The magnetic moment of a transition metal complex tells us how many unpaired electrons are present. The formula $\mu = \sqrt{n(n+2)}$ BM (where $n$ is the number of unpaired electrons) connects the experimental value directly to the electronic configuration. So when you see 5.92 BM and 1.74 BM, you're really being asked: why does the same metal ion, $Fe^{3+}$ , have a different number of unpaired electrons in these two complexes?

The answer lies in the crystal field theory and the nature of the ligands.

Identify the metal ion and its $d$ -electron count.

Iron in both complexes is in the +3 oxidation state. Atomic number of Fe is 26. $Fe^{3+}$ means we remove 3 electrons: $[Ar]3d^5$ . So in both cases, we have a $d^5$ system.
Recall the crystal field splitting in an octahedral field.

In an octahedral complex, the five $d$ -orbitals split into two sets: the lower-energy $t_{2g}$ set (three orbitals) and the higher-energy $e_g$ set (two orbitals). The energy gap between them is called $\Delta_o$ (or $10 Dq$ ).
The key: how do electrons fill these orbitals?

For a $d^5$ ion, there are two possible arrangements:
- High-spin: Electrons fill all five orbitals singly first (Hund's rule), giving 5 unpaired electrons. This happens when $\Delta_o$ is small — it costs less energy to put an electron in a higher $e_g$ orbital than to pair up in a $t_{2g}$ orbital.
- Low-spin: Electrons pair up in the $t_{2g}$ set first, giving only 1 unpaired electron. This happens when $\Delta_o$ is large — pairing energy is less than the energy needed to promote an electron to $e_g$ .
Now, the ligands decide $\Delta_o$ .

Ligands are arranged in the spectrochemical series in order of increasing field strength:

$I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < NO_2^- < CN^- < CO$
- $H_2O$ is a weak field ligand. It produces a small $\Delta_o$ .
- $CN^-$ is a strong field ligand. It produces a large $\Delta_o$ .
Apply to $[Fe(H_2O)_6]^{3+}$ :

Weak field → small $\Delta_o$ → high-spin configuration.

The five electrons occupy all five orbitals singly: $t_{2g}^3 e_g^2$ .

Number of unpaired electrons, $n = 5$ .

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

This matches the given value exactly. …

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Method: Crystal Field Theory (CFT) – Spin-Only Magnetic Moment Calculation

This method uses the spin-only formula to predict magnetic moment ( $\mu$ ) based on the number of unpaired electrons ( $n$ ):

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

Step-by-step reasoning for both complexes

Step 1: Determine the oxidation state and electron configuration of Fe

Fe atomic number = 26
Ground state: $[Ar]\, 3d^6 4s^2$

For $[Fe(H_2O)_6]^{3+}$ :

Ligand charge: $H_2O$ is neutral → overall +3 charge comes from Fe
Fe is in +3 oxidation state → $Fe^{3+}$
Electron configuration of $Fe^{3+}$ : $[Ar]\, 3d^5$

For $[Fe(CN)_6]^{3-}$ :

Ligand charge: $CN^-$ is −1 each → total −6 from ligands
Overall charge −3 → Fe must be +3 to balance: $Fe^{3+}$
Same $3d^5$ configuration

Step 2: Identify ligand field strength and predict pairing

Complex	Ligand	Field strength	Effect on $3d^5$
$[Fe(H_2O)_6]^{3+}$	$H_2O$	Weak field	No pairing → High spin
$[Fe(CN)_6]^{3-}$	$CN^-$	Strong field	Forces pairing → Low spin

Step 3: Write the $d$ -orbital filling (octahedral splitting)

High spin $Fe^{3+}$ (weak field):

$t_{2g}$ : ↑ ↑ ↑ (3 electrons)
$e_g$ : ↑ ↑ (2 electrons)
Unpaired electrons = 5

Low spin $Fe^{3+}$ (strong field):

$t_{2g}$ : ↑↓ ↑ ↑ (5 electrons, all paired except 1 unpaired)
$e_g$ : empty
Unpaired electrons = 1

Step 4: Apply spin-only formula

For $[Fe(H_2O)_6]^{3+}$ ( $n = 5$ ): …

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Here is a breakdown of the common mistakes students make on this exact question, and how to avoid them.

The Core Concept (Why the difference?)

The magnetic moment ( $\mu$ ) depends on the number of unpaired electrons ( $n$ ). The formula is:

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

$[Fe(H_2O)_6]^{3+}$ : $Fe^{3+}$ has 5 electrons in the 3d orbital. $H_2O$ is a weak field ligand. It does not force pairing. The electrons remain unpaired (high spin). $n = 5$ .

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

$[Fe(CN)_6]^{3-}$ : $Fe^{3+}$ still has 5 electrons. $CN^-$ is a strong field ligand. It forces pairing. The electrons pair up, leaving only 1 unpaired electron (low spin). $n = 1$ .

\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.74 \text{ BM}

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to find the oxidation state of the metal first.

The Error: Students directly count electrons from the neutral atom (e.g., $Fe$ has 8 electrons in 3d and 4s) without adjusting for the charge on the complex.
Example of Error: Saying $Fe$ in $[Fe(CN)_6]^{3-}$ has 8 d-electrons.
How to Avoid:
1. Always calculate the oxidation state first.
2. For $[Fe(CN)_6]^{3-}$ : Let $x$ be the oxidation state of Fe. $CN^-$ has a -1 charge. So: $x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3$ .
3. $Fe^{3+}$ has lost 3 electrons. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$ . For $Fe^{3+}$ , remove the 4s electrons first, then one 3d electron: $[Ar] 3d^5$ .
4. Result: You always start with 5 d-electrons for $Fe^{3+}$ .

Mistake 2: Confusing strong and weak field ligands.

The Error: Thinking $H_2O$ is a strong field ligand or $CN^-$ is a weak field ligand.
Example of Error: Predicting $[Fe(H_2O)_6]^{3+}$ will have paired electrons.
How to Avoid:
- Memorize the Spectrochemical Series (partial list is enough):
  - Strong field (low spin): $CN^- > CO > en > NH_3$
  - Weak field (high spin): $H_2O > F^- > Cl^- > Br^- > I^-$
- Rule of thumb: $CN^-$ and $CO$ are almost always strong. $H_2O$ is borderline but usually weak for $3d$ metals. Halides are always weak.

Mistake 3: Forgetting the "Hund's Rule" logic for d5 configuration.

The Error: Students correctly identify 5 electrons but then pair them incorrectly, leading to a wrong $n$ value.
Example of Error: For $Fe^{3+}$ (d5), placing 2 electrons in one orbital and 3 in others, giving 3 unpaired electrons.
How to Avoid:
- Draw the d-orbital splitting diagram.
- For weak field ( $H_2O$ ): The energy gap ( $\Delta_o$ ) is small. Electrons fill all 5 orbitals singly first (Hund's rule). Result: 5 unpaired electrons.
- For strong field ( $CN^-$ ): The energy gap ( $\Delta_o$ ) is large. Electrons pair up in the lower energy $t_{2g}$ orbitals before going to the higher $e_g$ orbitals. For d5: 3 electrons fill $t_{2g}$ singly, then the next 2 pair up. Result: 1 unpaired electron.

Mistake 4: Using the wrong formula or miscalculating the square root.

The Error: Using $\mu = \sqrt{n(n+1)}$ (which is for orbital angular momentum) or making arithmetic errors.
Example of Error: For $n=5$ , calculating $\sqrt{5 \times 7} = \sqrt{35} \approx 5.91$ (correct), but then writing 5.92 without checking. Or for $n=1$ , calculating $\sqrt{1 \times 3} = \sqrt{3} \approx 1.73$ (correct), but writing 1.74.
How to Avoid:
- Use the spin-only formula: $\mu = \sqrt{n(n+2)}$ .
- Memorize the common values:
  - $n=1 \implies \mu = \sqrt{3} \approx 1.73 \text{ BM}$
  - $n=2 \implies \mu = \sqrt{8} \approx 2.83 \text{ BM}$ …

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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 questionQ21 — Give the electronic configuration of the following complexes on the ba…→Next questionQ23 — Arrange following complex ions in increasing order of crystal field sp…

Q.Explain why [Fe(H2O)6]3+[Fe(H_2O)_6]^{3+}[Fe(H2​O)6​]3+ has magnetic moment value of 5.92 BM whereas [Fe(CN)6]3−[Fe(CN)_6]^{3-}[Fe(CN)6​]3− has a value of only 1.74 BM.

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Q.Explain why $[Fe(H_2O)_6]^{3+}$ has magnetic moment value of 5.92 BM whereas $[Fe(CN)_6]^{3-}$ has a value of only 1.74 BM.

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