Key Fact
Magnetic properties
Paramagnetic: unpaired electrons. Magnetic moment μ = √(n(n+2)) BM (spin only), where n = unpaired e⁻. Diamagnetic: all e⁻ paired (e.g. Zn²⁺ d¹⁰).
-- NCERT Class 12 Chemistry, Ch. 4, p. 102Magnetic Properties
8 MCQs1 revision card9-step worked example
Source: NCERT The d and f Block ElementsPYQ coverage: NEET 2020, 2024, 2025Official key: NTA-verifiedLast updated: 27 Sep 2026
Magnetic Properties, explained for NEET
The magnetic behaviour of transition metal ions comes down to one question: how many unpaired electrons does the ion have? Get that count wrong, and you pick the wrong magnetic moment from the spin-only formula — a common source of lost marks.
Classification. Substances with unpaired electrons are paramagnetic (attracted into a magnetic field). Substances with all electrons paired are diamagnetic (weakly repelled). Among transition metals, paramagnetism dominates because partially filled d-orbitals readily hold unpaired electrons.
The spin-only formula. NCERT Class 12 Chemistry Chapter 4 (page 104) gives:
μ = √(n(n+2)) BM
where n = number of unpaired electrons. This approximation ignores orbital angular momentum contribution — valid for first-row d-block ions where crystal field quenching suppresses orbital contribution.
Counting unpaired electrons. Write the electronic configuration of the ion (not the atom). Remove electrons from 4s first, then from 3d. For example: Fe²⁺ is [Ar] 3d⁶ — four unpaired electrons in a free ion (high-spin). Fe³⁺ is [Ar] 3d⁵ — five unpaired electrons.
Watch-out. The trap is counting unpaired electrons from the neutral atom configuration rather than the ion configuration. Always strip 4s electrons first, then count d-orbital occupancy in the ionic state. A second common error: confusing high-spin and low-spin counts without being told the ligand field strength — in NEET, assume high-spin (weak field) unless the question explicitly states otherwise.
Can you answer these Magnetic Properties MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following ions is diamagnetic?
Show answer and why every option is right or wrong
Answer: A. Zn²⁺ has a completely filled 3d¹⁰ configuration — zero unpaired electrons — making it diamagnetic (NCERT Class 12 Chemistry Chapter 4, page 104).
Why B is wrong: Cu²⁺ has one unpaired electron in 3d⁹, so it is paramagnetic, not diamagnetic.
Why C is wrong: Fe³⁺ has five unpaired electrons (3d⁵, half-filled), strongly paramagnetic.
Why D is wrong: Cr³⁺ has three unpaired electrons in 3d³, so it is paramagnetic.
MCQ 2Direct ApplicationPractice
The spin-only magnetic moment of Mn²⁺ (3d⁵) is:
Show answer and why every option is right or wrong
Answer: C. Mn²⁺ has 5 unpaired electrons. μ = √(5×7) = √35 = 5.92 BM.
Why A is wrong: 4.90 BM corresponds to n = 4 (√(4×6) = √24 = 4.90). This would be the case for Fe²⁺ in high-spin, not Mn²⁺.
Why B is wrong: 3.87 BM corresponds to n = 3 (√(3×5) = √15 = 3.87). Student likely miscounted unpaired electrons.
Why D is wrong: 1.73 BM corresponds to n = 1 (√(1×3) = √3 = 1.73). Gross undercount of unpaired electrons in a half-filled d⁵ ion.
MCQ 3Direct ApplicationPractice
Among Ti³⁺, V³⁺, Cr³⁺, and Mn²⁺, the ion with the highest magnetic moment is:
Show answer and why every option is right or wrong
Answer: C. Ti³⁺ (3d¹) has 1 unpaired e⁻; V³⁺ (3d²) has 2; Cr³⁺ (3d³) has 3; Mn²⁺ (3d⁵) has 5. More unpaired electrons → higher μ. Mn²⁺ gives √35 = 5.92 BM, the highest.
Why A is wrong: Ti³⁺ has only 1 unpaired electron (μ = 1.73 BM) — the lowest in this series.
Why B is wrong: V³⁺ has 2 unpaired electrons (μ = 2.83 BM), less than Cr³⁺ or Mn²⁺.
Why D is wrong: Cr³⁺ has 3 unpaired electrons (μ = 3.87 BM). Higher than Ti³⁺ and V³⁺ but less than Mn²⁺ with 5 unpaired electrons.
MCQ 4Easy RecallPractice
Cu⁺ is diamagnetic while Cu²⁺ is paramagnetic. This is because:
Show answer and why every option is right or wrong
Answer: A. Cu⁺ loses one electron from neutral Cu ([Ar] 3d¹⁰ 4s¹) giving [Ar] 3d¹⁰ — all paired, diamagnetic. Cu²⁺ loses two electrons giving [Ar] 3d⁹ — one unpaired electron, paramagnetic.
Why B is wrong: This reverses the configurations. Cu⁺ has the fully filled 3d¹⁰, not 3d⁹. Student confused which oxidation state retains the full d-shell.
Why C is wrong: Both cannot have 3d¹⁰ — Cu²⁺ has lost two electrons total from Cu atom and ends at 3d⁹.
Why D is wrong: Ionic size difference exists but does not explain diamagnetic vs paramagnetic behaviour — that depends on unpaired electron count.
MCQ 5Direct ApplicationPractice
A transition metal ion has a magnetic moment of 3.87 BM. The number of unpaired electrons in the ion is:
Show answer and why every option is right or wrong
Answer: B. μ = √(n(n+2)). For n = 3: √(3×5) = √15 = 3.87 BM. So the ion has 3 unpaired electrons.
Why A is wrong: For n = 2: μ = √(2×4) = √8 = 2.83 BM, not 3.87 BM.
Why C is wrong: For n = 4: μ = √(4×6) = √24 = 4.90 BM, not 3.87 BM.
Why D is wrong: For n = 5: μ = √(5×7) = √35 = 5.92 BM, not 3.87 BM.
MCQ 6Easy RecallPractice
Which statement correctly describes the spin-only magnetic moment formula?
Show answer and why every option is right or wrong
Answer: D. The spin-only formula μ = √(n(n+2)) neglects orbital angular momentum. It works well for first-row transition metals where crystal field effects quench the orbital contribution (NCERT Class 12 Chemistry Chapter 4, page 104).
Why A is wrong: The formula is explicitly 'spin-only' — it ignores orbital angular momentum. This is its defining assumption.
Why B is wrong: Diamagnetic species have n = 0 and μ = 0. The formula is designed for paramagnetic species with n ≥ 1.
Why C is wrong: The formula is used for d-block (transition metal) ions, not lanthanoids. Lanthanoids have significant orbital contribution, making spin-only inadequate.
MCQ 7CalculationPractice
Which pair of ions has the same spin-only magnetic moment? (Z: Ti 22, Cr 24, Fe 26, Co 27, Ni 28)
Show answer and why every option is right or wrong
Answer: B. Step 1, configurations (4s electrons leave first): Fe²⁺ = [Ar] 3d⁶ and Co³⁺ = [Ar] 3d⁶. Step 2, unpaired electrons: a d⁶ ion has 4 (five singly filled, then one pair). Step 3, both give μ = √(4 × 6) = √24 ≈ 4.90 BM, the same value. The spin-only formula μ = √(n(n + 2)) BM, with n the number of unpaired electrons, is in NCERT Class 12 Chemistry, Chapter 4, page 102.
Why A is wrong: A pairs ions by charge rather than by configuration: Cr³⁺ is 3d³ (n = 3, 3.87 BM) but Fe²⁺ is 3d⁶ (n = 4, 4.90 BM).
Why C is wrong: C looks only at the +3 charge: Ti³⁺ is 3d¹ (n = 1, 1.73 BM) while Cr³⁺ is 3d³ (n = 3, 3.87 BM).
Why D is wrong: D treats neighbouring elements as alike: Co²⁺ is 3d⁷ (n = 3, 3.87 BM) and Ni²⁺ is 3d⁸ (n = 2, 2.83 BM).
MCQ 8CalculationPractice
An ion with electronic configuration [Ar] 3d³ is placed in a magnetic field. Its magnetic moment and magnetic behaviour are:
Show answer and why every option is right or wrong
Answer: D. 3d³ has 3 unpaired electrons (one each in three d-orbitals by Hund's rule). μ = √(3×5) = √15 = 3.87 BM. Since n > 0, the ion is paramagnetic, not diamagnetic.
Why A is wrong: 1.73 BM corresponds to n = 1. A 3d³ configuration has 3 unpaired electrons by Hund's rule, not 1.
Why B is wrong: 2.83 BM corresponds to n = 2. In 3d³, all three electrons occupy separate orbitals with parallel spins — three unpaired, not two.
Why C is wrong: The magnetic moment value 3.87 BM is correct for n = 3, but the ion cannot be diamagnetic — diamagnetism requires zero unpaired electrons. Any ion with n > 0 is paramagnetic.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Magnetic Properties: quick recall before you leave
How do you solve a Magnetic Properties question? A worked example
- 1
Given
Co (Z = 27): [Ar] 3d⁷ 4s². Co²⁺ removes 2 electrons (4s² first) → [Ar] 3d⁷.
- 2
Required
Spin-only magnetic moment (μ) and magnetic classification.
- 3
Concept
The spin-only formula relates the number of unpaired electrons to the observed magnetic moment. For high-spin first-row transition metal ions, count unpaired electrons in the d-subshell after removing appropriate electrons.
- 4
Formula
μ = √(n(n+2)) BM
- 5
Substitution
Co²⁺ is 3d⁷. Filling 5 d-orbitals with 7 electrons (Hund's rule, high-spin): ↑↓ ↑↓ ↑ ↑ ↑ → n = 3 unpaired electrons.
μ = √(3 × (3+2)) = √(3 × 5) = √15 - 6
Calculation
μ = √15 = 3.87 BM
- 7
Final answer
μ = 3.87 BM. Since n = 3 > 0, Co²⁺ is paramagnetic.
Note: The integers 3, 5, and 15 in the formula are exact counting numbers and mathematical products — they do not limit significant figures. The result 3.87 BM carries three significant figures by convention of the spin-only formula output. - 8
Common trap
Counting unpaired electrons from the neutral atom (3d⁷ 4s²) instead of the ion (3d⁷). Here both happen to give the same d-configuration for Co²⁺, but for ions like Fe²⁺ (not Fe atom's [Ar] 3d⁶ 4s² → ion is [Ar] 3d⁶), confusing atom vs ion configurations is a frequent source of wrong n values. Always write the ion configuration explicitly.
- 9
Similar NEET-style question
Calculate the magnetic moment of Ni²⁺ and identify whether a complex [Ni(H₂O)₆]²⁺ would be attracted to or repelled by a magnetic field.
What to remember before solving Magnetic Properties questions
Which Magnetic Properties formulas do you need for NEET?
Spin-only magnetic moment
Magnetic moment from n unpaired electrons. 1 unpaired: 1.73 BM; 5: 5.92 BM.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | Bohr magneton |
Valid when
- Spin-only contribution (no orbital contribution)
- Octahedral or tetrahedral complex
Spin-only magnetic moment for transition metal
Predicts paramagnetic moment of d-block ion. n unpaired electrons in d-orbitals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | BM |
Valid when
- Spin-only contribution
- Octahedral or tetrahedral complex
- First-row d-block
More in The d and f Block Elements: 3 exam traps and mistakes · 1 formula · 3 question patterns from its other lessons.
Magnetic Properties questions from past NEET papers
3 questions from NEET 2020, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1Statement I is false but Statement II is true
2Both Statement I and Statement II are true
3Both Statement I and Statement II are false
4Statement I is true but Statement II is false
NTA Answer: Option 4(final)
NEET 2024Revised key
1B and D only
2A and E only
3B and C only
4A and D only
NTA Answer: Option 1(revised_final)
NEET 2020
The calculated spin only magnetic moment of Cr2+ ion is :
12.84 BM
23.87 BM
34.90 BM
45.92 BM
NTA Answer: Option 3(final)
All 15 past-paper questions from The d and f Block Elements →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →