Paramagnetism vs Diamagnetism
When a substance is placed in an external magnetic field, it responds in one of a few characteristic ways. NCERT §4.3.9 identifies the two principal behaviours of interest to a transition-metal chemist: diamagnetism and paramagnetism. Diamagnetic substances are weakly repelled by an applied field, whereas paramagnetic substances are attracted into it. A third, extreme behaviour — ferromagnetism — describes substances such as iron that are attracted very strongly; ferromagnetism is in fact an extreme form of paramagnetism.
The distinction is rooted entirely in electron pairing. A diamagnetic material has all of its electrons paired, so the spins cancel and there is no permanent magnetic moment. A paramagnetic material possesses one or more unpaired electrons, each of which carries an intrinsic magnetic moment. Because so many transition metal ions have partly filled d subshells, paramagnetism is one of the defining family traits of the block.
| Property | Diamagnetic | Paramagnetic |
|---|---|---|
| Unpaired electrons | None (all paired) | One or more |
| Response to field | Weakly repelled | Attracted |
| Net magnetic moment | Zero | Non-zero (∝ √n(n+2)) |
| Typical d-block ions | Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺ | Ti³⁺, V³⁺, Mn²⁺, Fe³⁺ |
Unpaired Electrons as the Source of Magnetism
Paramagnetism arises from the presence of unpaired electrons, each such electron carrying a magnetic moment associated with both its spin angular momentum and its orbital angular momentum. In principle both contributions add to the observed moment of an ion. For the compounds of the first transition series, however, the orbital angular momentum contribution is effectively quenched by the surrounding ligand environment and is therefore of no practical significance.
This quenching is the single physical fact that makes the topic tractable for NEET. Once the orbital term is set aside, the magnetic moment of a first-row ion is governed by one quantity only — the number of unpaired electrons, n. The moment becomes a clean, countable property: more unpaired spins mean a larger moment, and the relationship is given by a simple closed formula.
Paired spins (left) cancel and the ion is repelled by a field; the three unpaired, aligned spins (right) produce a net moment and the ion is drawn into the field.
The Spin-Only Formula
Because only spin survives, the magnetic moment of a first-row transition metal ion is given by the spin-only formula:
$$\mu = \sqrt{n(n+2)}\ \text{BM}$$Here n is the number of unpaired electrons and $\mu$ is the magnetic moment expressed in Bohr magnetons (BM), the natural unit of atomic magnetism. A single unpaired electron ($n = 1$) gives
$$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73\ \text{BM}$$and the moment rises monotonically with each additional unpaired spin. The formula is symmetric and easy to evaluate: in practice you compute the product $n(n+2)$ first and then take its square root. NIOS §21.4.2 gives the same expression and works the $\ce{Ni^2+}$ case ($n = 2$, $\mu = \sqrt{8} = 2.83$ BM) as its illustration.
It is n(n+2), not n²
A frequent error is to compute $\sqrt{n^2}$ or to confuse this with the total spin expression $2\sqrt{S(S+1)}$. The NCERT spin-only formula uses the number of unpaired electrons directly inside $\sqrt{n(n+2)}$ — there is no factor of two and no spin quantum number to track.
Memorise the ladder: n = 1 → 1.73, n = 2 → 2.83, n = 3 → 3.87, n = 4 → 4.90, n = 5 → 5.92 BM.
Counting Unpaired Electrons in Mn+ Ions
Every spin-only problem reduces to one preliminary step: find the d-electron configuration of the ion and count its unpaired electrons. The procedure is mechanical once the order of electron removal is respected.
Transition metals are written with the configuration [noble gas] (n−1)dx ns². When the metal ionises, the ns electrons leave first, before any (n−1)d electron. So for a first-row ion you take the neutral atom, strip the 4s electrons, then strip d electrons to reach the required charge, and finally distribute the remaining d electrons across the five d orbitals following Hund's rule of maximum multiplicity.
Worked through for the common ions, this gives the configurations tabulated below. Note that the maximum number of unpaired electrons, five, occurs at the half-filled $d^5$ configuration.
| Ion | d-configuration | Unpaired e⁻ (n) | μ = √n(n+2) / BM |
|---|---|---|---|
Sc³⁺ | 3d⁰ | 0 | 0 (diamagnetic) |
Ti³⁺ | 3d¹ | 1 | 1.73 |
V³⁺ | 3d² | 2 | 2.83 |
Cr³⁺ / V²⁺ | 3d³ | 3 | 3.87 |
Cr²⁺ / Mn³⁺ | 3d⁴ | 4 | 4.90 |
Mn²⁺ / Fe³⁺ | 3d⁵ | 5 | 5.92 |
Fe²⁺ / Co³⁺ | 3d⁶ | 4 | 4.90 |
Co²⁺ | 3d⁷ | 3 | 3.87 |
Ni²⁺ | 3d⁸ | 2 | 2.83 |
Cu²⁺ | 3d⁹ | 1 | 1.73 |
Zn²⁺ | 3d¹⁰ | 0 | 0 (diamagnetic) |
Hund's rule fills the five d orbitals singly before pairing. Unpaired electrons rise to a maximum of five at d⁵ ($\ce{Mn^2+}$), then fall again; the filled d¹⁰ shell of $\ce{Zn^2+}$ has every electron paired.
Worked Examples: Ti³⁺ to Mn²⁺
Calculate the spin-only magnetic moment of the $\ce{Ti^3+}$ ion.
Titanium (Z = 22) is $\ce{[Ar]}3d^2 4s^2$. Removing three electrons (the two 4s, then one 3d) leaves $\ce{Ti^3+}$ as $3d^1$, so $n = 1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} = \mathbf{1.73\ \text{BM}}$.
Find the spin-only moment of $\ce{V^3+}$.
Vanadium (Z = 23) is $\ce{[Ar]}3d^3 4s^2$. Removing the two 4s and one 3d electron gives $\ce{V^3+}$ as $3d^2$, so $n = 2$.
$\mu = \sqrt{2(2+2)} = \sqrt{8} = \mathbf{2.83\ \text{BM}}$.
Determine the spin-only moment of $\ce{Cr^3+}$.
Chromium (Z = 24) has the anomalous configuration $\ce{[Ar]}3d^5 4s^1$. Removing three electrons (the 4s and two 3d) leaves $\ce{Cr^3+}$ as $3d^3$, so $n = 3$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} = \mathbf{3.87\ \text{BM}}$.
Show that $\ce{Mn^2+}$ and $\ce{Fe^3+}$ carry the largest moment in the series. (This is NCERT Example 4.8 logic for Z = 25.)
Manganese (Z = 25) is $\ce{[Ar]}3d^5 4s^2$; removing the two 4s electrons gives $\ce{Mn^2+}$ as $3d^5$. Iron (Z = 26) is $\ce{[Ar]}3d^6 4s^2$; removing three electrons gives $\ce{Fe^3+}$, also $3d^5$. Both are half-filled with five unpaired electrons, so $n = 5$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} = \mathbf{5.92\ \text{BM}}$ — the maximum value attainable for any 3d ion.
Every magnetic-moment problem starts by fixing the ion's charge. Lock the redox logic in Variable Oxidation States of Transition Elements.
Calculated vs Observed Moments
The spin-only value is a calculated quantity; the laboratory measures an experimental moment, usually for hydrated ions in solution or the solid state. NCERT Table 4.7 compares the two for the first-row divalent ions, and the agreement is striking for the lighter ions but loosens towards the right of the series.
| Ion | Config. | n | Calculated μ / BM | Observed μ / BM |
|---|---|---|---|---|
Sc³⁺ | 3d⁰ | 0 | 0 | 0 |
Ti³⁺ | 3d¹ | 1 | 1.73 | 1.75 |
V²⁺ | 3d³ | 3 | 3.87 | 3.86 |
Cr²⁺ | 3d⁴ | 4 | 4.90 | 4.80 |
Mn²⁺ | 3d⁵ | 5 | 5.92 | 5.96 |
Fe²⁺ | 3d⁶ | 4 | 4.90 | 5.3 – 5.5 |
Co²⁺ | 3d⁷ | 3 | 3.87 | 4.4 – 5.2 |
Ni²⁺ | 3d⁸ | 2 | 2.84 | 2.9 – 3.4 |
Cu²⁺ | 3d⁹ | 1 | 1.73 | 1.8 – 2.2 |
Zn²⁺ | 3d¹⁰ | 0 | 0 | 0 |
For ions from $\ce{Ti^3+}$ to $\ce{Mn^2+}$ the calculated and observed values nearly coincide, confirming that orbital quenching is essentially complete. Beyond $\ce{Fe^2+}$ the observed moments run higher than predicted because the orbital contribution is no longer fully quenched. NEET problems work strictly with the calculated spin-only value, so for the exam the formula governs — but it is worth knowing the agreement is an approximation, not an identity.
Why Sc³⁺ and Zn²⁺ Are Diamagnetic
Two ions sit at the ends of the series with no magnetic moment at all, and they illustrate the two distinct routes to a diamagnetic d-block ion.
$\ce{Sc^3+}$ is $3d^0$. Scandium (Z = 21) is $\ce{[Ar]}3d^1 4s^2$; losing all three valence electrons empties the d subshell entirely. With no d electrons there can be no unpaired spins, so $n = 0$ and $\mu = 0$.
$\ce{Zn^2+}$ is $3d^{10}$. Zinc (Z = 30) is $\ce{[Ar]}3d^{10} 4s^2$; losing the two 4s electrons leaves a completely filled d subshell. Every d electron is paired, so again $n = 0$ and $\mu = 0$.
The same reasoning explains why NIOS §21.4.2 lists $\ce{Ti^4+}$ ($3d^0$), $\ce{V^5+}$ ($3d^0$), $\ce{Cr^6+}$, $\ce{Mn^7+}$ and $\ce{Cu^+}$ ($3d^{10}$) as diamagnetic: each has either an empty or a fully filled d subshell. This is also why the permanganate ion $\ce{MnO4^-}$, with manganese formally $d^0$, is diamagnetic, whereas manganate $\ce{MnO4^2-}$ ($d^1$, one unpaired electron) is paramagnetic.
Diamagnetic does not mean "no d electrons"
A diamagnetic ion needs zero unpaired electrons, which can mean either $d^0$ (Sc³⁺, Ti⁴⁺) or a fully paired $d^{10}$ (Zn²⁺, Cu⁺). Do not assume an ion with ten d electrons is "full of magnetism" — a closed shell is just as non-magnetic as an empty one.
μ = 0 ⇔ n = 0 ⇔ configuration is d⁰ or d¹⁰.
Trend Across the First Series
Reading the table from $\ce{Sc^3+}$ across to $\ce{Zn^2+}$, the number of unpaired electrons — and therefore the spin-only moment — rises to a maximum at the half-filled $d^5$ point and then falls symmetrically. This produces the characteristic inverted-V shape: zero at $d^0$, climbing through 1.73, 2.83, 3.87, 4.90 to the peak of 5.92 BM at $d^5$, then descending back through the same values to zero at $d^{10}$.
The spin-only moment is a concave-rising function of n: each extra unpaired electron adds progressively less, but the count of unpaired electrons itself peaks at d⁵, giving the 5.92 BM maximum.
Reading n From a Measured Moment
The relationship runs both ways. Because the observed moment reflects the number of unpaired electrons, an experimental value can be inverted to deduce an ion's electronic configuration — a favourite NEET twist. Set the measured moment equal to $\sqrt{n(n+2)}$, square both sides, and solve the quadratic $n^2 + 2n - \mu^2 = 0$ for the positive integer root.
A first-row ion shows a magnetic moment of about 3.87 BM. How many unpaired electrons does it carry?
Squaring: $n(n+2) = 3.87^2 \approx 15$. Then $n^2 + 2n - 15 = 0$, which factors as $(n+5)(n-3) = 0$, giving the physical root $n = 3$.
Three unpaired electrons correspond to a $3d^3$ configuration — for example $\ce{Cr^3+}$, $\ce{V^2+}$ or $\ce{Mn^4+}$. This is exactly the reasoning the NEET 2018 matching question rewards.
The same probe distinguishes the green manganate from the purple permanganate (one is paramagnetic, one diamagnetic) and identifies which lanthanoid ions are diamagnetic — those, such as $\ce{Ce^4+}$ ($4f^0$) and $\ce{Yb^2+}$ ($4f^{14}$), with empty or fully filled f subshells. The principle is identical to the d-block case; only the subshell changes. The deeper electronic origin of these spin states, and how ligand fields can force pairing, is developed in Crystal Field Theory.
Magnetic Properties at a Glance
- Paramagnetic = unpaired electrons, attracted to a field; diamagnetic = all paired, repelled.
- Spin-only moment: $\mu = \sqrt{n(n+2)}$ BM, where n is the number of unpaired electrons; valid because orbital momentum is quenched in the first series.
- Memorise the ladder: n = 1 → 1.73, 2 → 2.83, 3 → 3.87, 4 → 4.90, 5 → 5.92 BM.
- Maximum moment 5.92 BM occurs at the half-filled d⁵ configuration — $\ce{Mn^2+}$ and $\ce{Fe^3+}$.
- $\ce{Sc^3+}$ (d⁰) and $\ce{Zn^2+}$ (d¹⁰) are diamagnetic: μ = 0 means n = 0, i.e. an empty or fully filled d subshell.
- The moment can be inverted to find n: solve $n(n+2) = \mu^2$ for the positive integer root.