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

Gujarat GsebMCQ· 1mImportance★★★★★

49% · 49/101 Questions

✓ Free question

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 17 on this concept.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. Which of the following ion has the maximum theoretical magnetic moment? [Fe (Z=26), Cr (Z=24), Ti (Z=22), Co (Z=27)]$
(a) Fe3+
(b) Cr3+
(c) Ti3+
(d) Co3+
›Reveal solutionSolution
Magnetic moment mu = sqrt(n(n+2)) BM rises with the number of unpaired electrons n; the d5 configuration allows the maximum possible unpaired electrons.

Find the d-electron configuration of each 3+ ion:

Fe (Z=26): [Ar]3d⁶4s² → Fe3+ = [Ar]3d⁵ → 5 unpaired electrons (high spin) → μ = √(5×7) = √35 ≈ 5.92 BM

Cr (Z=24): [Ar]3d⁵4s¹ → Cr3+ = [Ar]3d³ → 3 unpaired electrons → μ = √(3×5) = √15 ≈ 3.87 BM

Ti (Z=22): [Ar]3d²4s² → Ti3+ = [Ar]3d¹ → 1 unpaired electron → μ = √3 ≈ 1.73 BM

Co (Z=27): [Ar]3d⁷4s² → Co3+ = [Ar]3d⁶ → 4 unpaired electrons (high spin) → μ = √(4×6) = √24 ≈ 4.90 BM

The d⁵ configuration (Fe3+) gives the maximum number of unpaired electrons achievable across the d-block (5), and therefore the highest theoretical magnetic moment.

✓Final answer
$(a) Fe3+ (μ \approx 5.92 BM, the highest among these ions).$
GUJCET 2025Set 031 markMCQ
$\sqrt{35}$
›Reveal solutionSolution
[!TLDR]

$\sqrt{35}$ BM corresponds to 5 unpaired electrons, matching $\text{Mn}^{2+}$ ( $3d^5$ ).

Concept

The spin-only magnetic moment is $\mu=\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

Solution

Solve for $n$ :
$\sqrt{n(n+2)}=\sqrt{35}\ \Rightarrow\ n(n+2)=35\ \Rightarrow\ n=5$
Now find the divalent ion with 5 unpaired $d$ electrons:

$\text{Cr}^{2+}: 3d^4$ → 4 unpaired

$\text{Mn}^{2+}: 3d^5$ → 5 unpaired ✓

$\text{Fe}^{2+}: 3d^6$ → 4 unpaired

$\text{Co}^{2+}: 3d^7$ → 3 unpaired

Only $\text{Mn}^{2+}$ has 5 unpaired electrons.

[!ANSWER]

(B) Mn
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. What is the value of magnetic moment of divalent ion having atomic number 30 in aqueous solution.$
(a) 0 BM
(b) 2.84 BM
(c) 1.73 BM
(d) 5.92 BM
›Reveal solutionSolution

Atomic number 30 is Zn; its divalent ion Zn2+ has a fully filled 3d10 configuration with zero unpaired electrons, so its magnetic moment is zero.

Element with Z = 30 is Zinc, electronic configuration [Ar] 3d10 4s2.

Zn2+ is formed by removing the two 4s electrons, giving [Ar] 3d10 - a completely filled d-subshell with no unpaired electrons.

Magnetic moment (spin-only) = sqrt[n(n+2)] BM, where n = number of unpaired electrons. Here n = 0, so magnetic moment = sqrt(0) = 0 BM.

✓Final answer
$(a) 0 BM - Zn2+ (3d10) has no unpaired electrons, so it is diamagnetic with zero magnetic moment.$
GUJCET 2024Set 131 markMCQ
$Co^{2+}$
›Reveal solutionSolution
Spin-only moment $\mu=\sqrt{n(n+2)}$ BM increases with the number of unpaired electrons $n$ ; $Mn^{2+}$ ( $d^5$ ) has the most.

Concept: Count $d$ -electrons and unpaired electrons:

$Ti^{2+}$ : $d^2$ , $n=2$ .

$Mn^{2+}$ : $d^5$ , $n=5$ $\Rightarrow\mu=\sqrt{5\times7}=5.92$ BM.

$Fe^{2+}$ : $d^6$ , $n=4$ .

$Co^{2+}$ : $d^7$ , $n=3$ .

$Mn^{2+}$ has the highest number of unpaired electrons and hence the largest moment.

✓Final answer
$Mn^{2+}$

ANSWER: (B)
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. What is the magnetic moment of a divalent ion in aqueous solution if its atomic number is 28?$
(a) 3.87 BM
(b) 2.84 BM
(c) 1.73 BM
(d) 4.90 BM
›Reveal solutionSolution

Atomic number 28 is Nickel; its divalent ion Ni2+ has electronic configuration 3d8, giving 2 unpaired electrons, and the spin-only magnetic moment formula gives 2.84 BM.

Ni (Z=28): [Ar] 3d8 4s2

Ni2+: loses the two 4s electrons -> [Ar] 3d8

3d8 configuration in an octahedral field (or free ion) has 2 unpaired electrons (t2g6 eg2 in high-spin, or simply counting 3d8 as 4 orbitals doubly-filled + 2 singly-filled).

Spin-only magnetic moment: mu = sqrt(n(n+2)) BM, with n = 2

mu = sqrt(2 x 4) = sqrt(8) = 2.83 approx 2.84 BM

✓Final answer
$(b) 2.84 BM.$
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. How many number of unpaired electrons are there in complex ion [Ni(CN)4]2-?$
(a) 4
(b) 3
(c) 2
(d) 0
›Reveal solutionSolution

[Ni(CN)4]2- has Ni2+ (d8) with the strong-field ligand CN-, which forces pairing of the d-electrons into a square planar, diamagnetic (dsp2) arrangement.

Ni2+: [Ar] 3d8

CN- is a very strong field ligand (high in the spectrochemical series), causing the 3d8 electrons to pair up, freeing one 3d orbital for dsp2 hybridisation (square planar geometry).

With all 8 d-electrons paired into 4 filled orbitals, there are 0 unpaired electrons - the complex is diamagnetic.

✓Final answer
$(d) 0 - all electrons are paired (square planar, dsp2, strong-field CN- ligand).$
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Which compound has magnetic moment equal to 4.90 BM?$
(a) Cr2(SO4)3
(b) NiSO4
(c) FeSO4
(d) MnSO4
›Reveal solutionSolution
mu = 4.90 BM means 4 unpaired electrons; Fe2+ (d6) fits, so FeSO4.

Spin-only moment mu = sqrt(n(n+2)) BM. For mu = 4.90: n(n+2) = 24 -> n = 4 unpaired electrons.

Check the metal ions:

Cr3+ (Cr2(SO4)3): d3 -> 3 unpaired -> 3.87 BM.

Ni2+ (NiSO4): d8 -> 2 unpaired -> 2.83 BM.

Fe2+ (FeSO4): d6 -> 4 unpaired -> 4.90 BM. (match)

Mn2+ (MnSO4): d5 -> 5 unpaired -> 5.92 BM.

✓Final answer
$(c) FeSO4.$
GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$Q. Among the following, which compound has the highest magnetic moment?$
(a) MnSO4
(b) CrCl3
(c) Ni(NO3)2
(d) FeSO4
›Reveal solutionSolution

Magnetic moment μ = √[n(n+2)] BM, where n = number of unpaired electrons; more unpaired electrons means a higher moment.

MnSO4: Mn2+ is d5 (high spin) → 5 unpaired electrons → μ = √35 = 5.92 BM.

FeSO4: Fe2+ is d6 → 4 unpaired electrons → μ = 4.90 BM.

CrCl3: Cr3+ is d3 → 3 unpaired electrons → μ = 3.87 BM.

Ni(NO3)2: Ni2+ is d8 → 2 unpaired electrons → μ = 2.83 BM.

Mn2+, with a half-filled d5 configuration, has the most unpaired electrons and hence the highest magnetic moment.

✓Final answer
$(a) MnSO4.$
GUJCET 2021Set 151 markMCQ
$Q. If atomic number of element is 26, then magnetic moment is ___ BM of its divalent aqueous ion? (A) 1.73 (B) 3.87 (C) 2.83 (D) 4.90$
›Reveal solutionSolution

Fe²⁺ ( $3d^6$ ) has 4 unpaired electrons, giving spin-only $\mu=\sqrt{n(n+2)}=\sqrt{24}=4.90$ BM.

Concept: Atomic number 26 = Fe. The divalent ion Fe²⁺ has configuration $[Ar]3d^6$ , with 4 unpaired electrons.
$\mu=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}=4.90\text{ BM}$
✓Final answer
$(D) 4.90$

ANSWER: (D)
GUJCET 2020Set 071 markMCQ
$Q. The divalent ion of which of the following element in aqueous solution has magnetic moment 5.92 BM? (A) Fe (B) Cr (C) Co (D) Mn$
›Reveal solutionSolution

$\mu=\sqrt{n(n+2)}=5.92\Rightarrow n=5$ unpaired; the $d^5$ divalent ion is Mn²⁺.

Concept — spin-only magnetic moment. $\mu=\sqrt{n(n+2)}$ BM. $5.92=\sqrt{5\cdot7}=\sqrt{35}$ , so $n=5$ unpaired electrons. Mn²⁺ is $[\text{Ar}]3d^5$ (5 unpaired). Fe²⁺ ( $d^6$ ) has 4, Co²⁺ ( $d^7$ ) has 3, Cr²⁺ ( $d^4$ ) has 4.

✓Final answer
$(D) Mn$

ANSWER: (D)
GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQ
$Q. Magnetic moment of a divalent ion in aqueous solution if its atomic number is 25 ______.$
(a) 4.90 BM
(b) 5.92 BM
(c) 2.84 BM
(d) 3.87 BM
›Reveal solutionSolution

Element 25 is manganese; its divalent ion Mn2+ has a 3d5 configuration with 5 unpaired electrons (a spin-only 'half-filled, maximally paramagnetic' case), giving the largest common magnetic moment among first-row transition ions.

Manganese (Z = 25) has ground-state configuration [Ar]3d5 4s2. Forming the divalent ion Mn2+ removes the two 4s electrons, leaving [Ar]3d5 - a half-filled d-subshell with all 5 electrons unpaired (one in each of the five d orbitals, by Hund's rule). Spin-only magnetic moment: μ = sqrt[n(n+2)] BM, where n = number of unpaired electrons = 5. So μ = sqrt(5 x 7) = sqrt(35) ≈ 5.92 BM.

✓Final answer
$(b) 5.92 BM.$
GUJCET 2019Set 131 markMCQ
$\text{Ni}^{2+}, \text{Co}^{2+}$
›Reveal solutionSolution
Magnetic moment depends only on the number of unpaired electrons; $Fe^{3+}$ and $Mn^{2+}$ are both $3d^5$ .

Concept: Spin-only magnetic moment $\mu = \sqrt{n(n+2)}\,BM$ , where $n$ = number of unpaired electrons. Two ions have the same moment if they have the same $n$ .

Counting unpaired d-electrons:

$Ni^{2+}$ : $3d^8$ → 2 ; $Co^{2+}$ : $3d^7$ → 3 (not equal).

$Fe^{2+}$ : $3d^6$ → 4 ; $Mn^{2+}$ : $3d^5$ → 5 (not equal).

$Fe^{3+}$ : $3d^5$ → 5 ; $Mn^{2+}$ : $3d^5$ → 5 → equal.

$Cr^{3+}$ : $3d^3$ → 3 ; $Mn^{3+}$ : $3d^4$ → 4 (not equal).

So the matching pair is $Fe^{3+}, Mn^{2+}$ .

✓Final answer
$Fe^{3+}, Mn^{2+}$

ANSWER: (C)

←Previous section · answersAnswers to Intext Questions →Next questionQ2 — The colour of the coordination compounds depends on the crystal field …

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

Concept

Solution

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

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days