Which of the following complexes formed by Cu²⁺ ions is most stable? (i) Cu²⁺ + 4NH₃ leftharpoons [Cu(NH₃)₄]²⁺

NCERT Exemplar · Q1

Cu^{2+}

(i)

Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}

\log K = 11.6

(ii)

Cu^{2+} + 4CN^{-} \rightleftharpoons [Cu(CN)_4]^{2-}

\log K = 27.3

(iii)

Cu^{2+} + 2en \rightleftharpoons [Cu(en)_2]^{2+}

\log K = 15.4

(iv)

Cu^{2+} + 4H_2O \rightleftharpoons [Cu(H_2O)_4]^{2+}

\log K = 8.9

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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

Important

Ferromagnetic materials have spontaneous magnetization — their atomic moments align even without an external field, forming magnetic domains. Magnetization in these materials is not linear; it saturates and shows hysteresis.

A Simple Example

Take a long iron rod placed inside a solenoid carrying current $I$ . The solenoid produces a uniform field $\vec{H}$ inside. The iron rod becomes magnetized: its atomic moments align, producing $\vec{M}$ in the same direction as $\vec{H}$ .

If $\chi_m$ for iron is about $5000$ , then $\vec{M} = 5000 \, \vec{H}$ . The total field inside the rod becomes:

\vec{B} = \mu_0 (\vec{H} + 5000 \vec{H}) = \mu_0 (5001) \vec{H}

That is why an iron core can amplify the magnetic field of a solenoid by thousands of times.

The Bottom Line

Magnetization is the measure of how much a material becomes magnetic when placed in an external field. It arises from the alignment of atomic magnetic dipoles. For linear materials, $\vec{M} = \chi_m \vec{H}$ . For ferromagnets, the response is nonlinear, strong, and can be permanent — that is how you get a bar magnet from a piece of iron.

Magnetic moment calculation using the spin-only formula is a numerically important topic in the NCERT/CBSE Class 12 Chemistry chapters on d-Block Elements and Coordination Compounds, and ‘spin only formula magnetic moment’ is a frequently searched important-question topic for board exams, JEE Main and NEET. Calculating the number of unpaired electrons correctly is a skill tested repeatedly in competitive-exam chemistry numericals.

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 $\mu = N I A$ ?

Each turn contributes $\mu_{\text{turn}} = I A$ . For $N$ identical turns in series:

\mu_{\text{total}} = N \cdot (I A) = N I A

If the solenoid has $n = N/l$ turns per unit length:

\mu = (n l) I A

Key insight: The magnetic moment adds linearly for multiple turns because each turn's torque contribution adds.

5. Summary Table of Key Results

System	Formula	Why
Single current loop	$\mu = I A$	Torque on loop $\propto I A$
Orbiting electron	$\mu = \frac{e}{2m} L$	Current from orbital motion
Solenoid	$\mu = N I A$	Sum of individual loop moments
General definition	$\vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \, dV$	For continuous current distributions

6. Exam-Relevant Takeaway

Always remember:

Magnetic moment always involves current × area (or equivalent)
For particles, it's charge-to-mass ratio × angular momentum
Direction: given by right-hand rule (curl fingers along current, thumb points along $\vec{\mu}$ )

The formula isn't arbitrary — it emerges naturally from the torque a current loop experiences in a magnetic field.

The stability of a complex is directly measured by its formation constant $K$ ; the larger the $\log K$ , the more stable the complex. Here, the complex with $\log K = 27.3$ is the most stable.

The question asks which complex is most stable. In coordination chemistry, the stability of a complex is quantified by its formation constant (also called stability constant) $K$ . The reaction given is the formation of the complex from the metal ion and ligands. A larger $K$ means the equilibrium lies further to the right — the complex is more stable and less likely to dissociate.

The values are given as $\log K$ , so we compare these directly. No conversion is needed: the highest $\log K$ corresponds to the highest $K$ , hence the most stable complex.

Let’s go through each option:

Option (i): $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$ , $\log K = 11.6$

This is a moderately stable complex. Ammonia is a good ligand, but not exceptionally strong for copper(II).
Option (ii): $Cu^{2+} + 4CN^{-} \rightleftharpoons [Cu(CN)_4]^{2-}$ , $\log K = 27.3$

Cyanide ion is a very strong ligand (high field strength, forms strong $\sigma$ and $\pi$ bonds). The $\log K$ is dramatically higher than the others — over 10 orders of magnitude larger in $K$ than the next closest.
Option (iii): $Cu^{2+} + 2en \rightleftharpoons [Cu(en)_2]^{2+}$ , $\log K = 15.4$

Ethylenediamine (en) is a bidentate ligand, which gives a chelate effect — this usually increases stability compared to monodentate ligands like NH $_3$ . Indeed, $\log K = 15.4$ is higher than for NH $_3$ (11.6), but still far below CN $^-$ .
Option (iv): $Cu^{2+} + 4H_2O \rightleftharpoons [Cu(H_2O)_4]^{2+}$ , $\log K = 8.9$

Water is a weak ligand. This is the least stable complex here.

Watch out

A common mistake is to think that chelating ligands (like en) always form the most stable complexes. While the chelate effect does enhance stability, the intrinsic ligand strength matters more. Here, CN $^-$ is such a powerful ligand that it overcomes the chelate advantage.

Tip

You don’t need to calculate $K$ from $\log K$ — just compare the $\log K$ values directly. The largest $\log K$ means the largest $K$ , hence the most stable complex.

✓Final answer

The most stable complex is formed with cyanide ions, option (ii) .

Common Mistakes & How to Avoid Them

Mistake 1: Confusing $\log K$ with $K$

The error: Students compare $\log K$ values directly and think the largest $\log K$ means the least stable complex.

Why it's wrong:

A higher $\log K$ means a larger equilibrium constant $K$ , which indicates greater stability of the complex.

How to avoid:

Remember:

$\log K \uparrow \implies K \uparrow \implies$ more stable complex
For this question: $27.3 > 15.4 > 11.6 > 8.9$ , so option (ii) is most stable.

Mistake 2: Forgetting that $\log K$ is directly proportional to stability

The error: Some students think a lower $\log K$ means the reaction "goes more to completion" — this is backwards.

Why it's wrong:

The equilibrium constant $K$ for complex formation is:

K = \frac{[\text{complex}]}{[\text{Cu}^{2+}][\text{ligand}]^n}

A larger $K$ means the equilibrium lies far to the right — more complex formed, hence more stable.

How to avoid:

Write the expression for $K$ and reason:

Big $K$ → products favoured → stable complex
Small $K$ → reactants favoured → unstable complex

Mistake 3: Ignoring the denticity of ligands

The error: Students compare $\log K$ values without considering that en (ethylenediamine) is bidentate, while NH $_3$ , CN $^-$ , and H $_2$ O are monodentate.

Why it matters:

A bidentate ligand like en forms a chelate ring, which gives extra stability (chelate effect). Even though $\log K$ for en (15.4) is less than for CN $^-$ (27.3), the chelate effect is already included in the given $\log K$ value.

How to avoid:

The $\log K$ values already account for denticity — compare them directly.
Do not try to "adjust" the values manually.

Mistake 4: Overthinking magnetic moment or geometry

The error: Students try to use magnetic moment or crystal field theory to decide stability.

Why it's wrong:

The question gives experimental $\log K$ values — these are the direct measure of stability. Magnetic moment tells you about unpaired electrons, not thermodynamic stability.

How to avoid:

When $\log K$ (or $K$ ) is given, use it directly.
Save magnetic moment reasoning for questions about geometry, spin state, or colour.

Mistake 5: Misreading the question as "least stable"

The error: Students accidentally pick the smallest $\log K$ (option iv) because they read "most stable" as "least stable".

How to avoid:

Circle the word "most" or "least" in the question.
Double-check: largest $\log K$ = most stable.

Final Answer

Most stable complex: Option (ii) $[\text{Cu(CN)}_4]^{2-}$ with $\log K = 27.3$

Quick check:

(ii) $\log K = 27.3$ → largest → most stable ✓
(iv) $\log K = 8.9$ → smallest → least stable

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.

μs = √[n(n+2)] Bohr Magneton (BM), where n is the number of unpaired electrons in the ion. For example, for n = 1, μs = √3 = 1.73 BM; for n = 5, μs = √35 = 5.92 BM.

✓Final answer
$μ(spin only) = \sqrt[n(n+2)] BM.$
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 tetrahedral geometry.

The $3d^8$ configuration in this arrangement retains 2 unpaired electrons, so the complex is paramagnetic.

✓Final answer
$[NiCl_4]^{2-}$
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).

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

$Fe$ ( $Z=26$ ) is $[Ar]3d^64s^2$ ; $Fe^{2+}$ gives $3d^6$ . $H_2O$ is a weak-field ligand, so no forced pairing occurs — the complex is high-spin, using outer $sp^3d^2$ hybridisation. The $3d^6$ electrons distribute as $t_{2g}^4e_g^2$ : the $t_{2g}$ set (3 orbitals, 4 electrons) has one pair + 2 unpaired, and the $e_g$ set (2 orbitals, 2 electrons) has 2 unpaired — total 4 unpaired electrons.

Comparison

Magnetic moment rises with the number of unpaired electrons ( $\mu=\sqrt{n(n+2)}$ BM). With $n=4$ for $Fe^{2+}$ versus $n=3$ for $Cr^{3+}$ , $[Fe(H_2O)_6]^{2+}$ has the higher magnetic moment and is the more strongly paramagnetic of the two.

✓Final answer
$[Fe(H_2O)_6]^{2+}$
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.

The spin-only magnetic moment is mu = sqrt(n(n+2)) B.M., where n = number of unpaired electrons.

mu = sqrt(5(5+2)) = sqrt(5 x 7) = sqrt(35) = 5.92 B.M.

✓Final answer
$True (mu = 5.92 B.M. for Mn2+).$
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).

$Mn^{2+}$ : $3d^5$ → 5 unpaired ( $\mu \approx 5.92$ BM).

$Cr^{2+}$ : $3d^4$ → 4 unpaired ( $\mu \approx 4.9$ BM).

$Cu^{2+}$ : $3d^9$ → 1 unpaired ( $\mu \approx 1.73$ BM).

✓Final answer
$(D) Cu²⁺ — it has one unpaired 3d electron, giving μ = 1.73 BM.$
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 $^{2+}=[Ar]3d^5$ .

Step 2 — count unpaired electrons. The five $3d$ electrons each occupy a separate d-orbital (Hund's rule) → $n=5$ unpaired electrons.

Step 3 — apply the formula.

$\mu=\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35}=5.92\ \text{BM}.$

✓Final answer
$^{2+}$
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).

Spin-only magnetic moment: μ = √(n(n+2)) BM = √(4 × 6) = √24 BM.

(Note: in a strong-field octahedral complex such as [Co(NH3)6]3+, Co3+ becomes low-spin with 0 unpaired electrons and is diamagnetic — but for the bare ion, as asked here, the high-spin value √24 BM applies.)

✓Final answer
$(d) \sqrt24 BM.$
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.

[Co(NH₃)₆] (as [Co(NH₃)₆]³⁺): Co³⁺ is d⁶; NH₃ is a reasonably strong-field ligand for a +3 ion and induces low-spin pairing (t₂g⁶eg⁰), leaving 0 unpaired electrons — diamagnetic (octahedral, d²sp³).

So [Ni(H₂O)₆]²⁺ is the only paramagnetic complex of the four.

✓Final answer
$(a) [Ni(H₂O)₆]²⁺.$
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:

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

$\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\ B.M.$

✓Final answer
$Mn^{2+}$
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):

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

Two orbitals are doubly occupied (paired) and three orbitals hold single (unpaired) electrons, so the number of unpaired electrons $n = 3$ .

Spin-only magnetic moment formula:

$\mu_{s} = \sqrt{n(n+2)}\ BM$

Substituting $n=3$ :

$\mu_s = \sqrt{3(3+2)} = \sqrt{15} = 3.873\ BM \approx 3.87\ BM$

✓Final answer
$Co^{2+}$
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.

Watch out

A common mistake is to assume $\text{Co}^{3+}$ always has a specific number of unpaired electrons. In reality, when $\text{Co}^{3+}$ is in a complex with ligands, strong-field ligands (like CN⁻) can force pairing to give a low-spin $d^6$ configuration with zero unpaired electrons. However, this question asks about the free ion in its ground state, which is high-spin with maximum unpaired electrons.

Note

The question as stated has no correct option among the choices given. The actual answer is 4 unpaired electrons, which does not appear. This may be a printing error in the original question, or the question intended to ask about a different oxidation state or complex. Based on fundamental principles, $\text{Co}^{3+}$ in its free-ion ground state has 4 unpaired electrons.

✓Final answer
$\text{Co}^{3+}$
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.

Using the spin-only magnetic moment formula:

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

✓Final answer
$Mn^{2+}$

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

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Q.Which of the following complexes formed by $Cu^{2+}$ ions is most stable?

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

A Simple Example

The Bottom Line

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 $\mu = N I A$ ?

5. Summary Table of Key Results

6. Exam-Relevant Takeaway

Method: Stability Constant Comparison

Steps

Final Answer

Common Mistakes & How to Avoid Them

Mistake 1: Confusing $\log K$ with $K$

Mistake 2: Forgetting that $\log K$ is directly proportional to stability

Mistake 3: Ignoring the denticity of ligands

Mistake 4: Overthinking magnetic moment or geometry

Mistake 5: Misreading the question as "least stable"

Final Answer

Understanding the Configuration

Applying Hund's Rule

Q.Which of the following complexes formed by Cu2+Cu^{2+}Cu2+ ions is most stable?

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Q.Which of the following complexes formed by $Cu^{2+}$ ions is most stable?

Unlock everything free for 14 days