Magnetic Properties & Formation of Coloured Ions

Magnetic Properties

Place a substance in a magnetic field. Diamagnetic substances (all electrons paired) are weakly repelled; paramagnetic substances (one or more unpaired electrons) are attracted. Most transition-metal ions are paramagnetic because they have unpaired d electrons.

The strength of paramagnetism is measured by the magnetic moment. For transition-metal ions, the orbital contribution is usually quenched, so we use the spin-only formula:

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

where $n$ is the number of unpaired electrons and BM is the Bohr magneton. A single unpaired electron gives $\mu = \sqrt{1\times3} = 1.73$ BM.

Unpaired electrons $n$	$\mu = \sqrt{n(n+2)}$ BM
1	1.73
2	2.83
3	3.87
4	4.90
5	5.92

Using the Spin-Only Formula

The formula works both ways:

• Given the configuration, count unpaired electrons $n$, then compute $\mu$.
• Given the measured $\mu$, solve for $n$ to deduce the number of unpaired electrons (and hence the oxidation state).

Worked example: for Mn$^{2+}$ ($3d^5$), all five d electrons are unpaired, so $n = 5$ and $\mu = \sqrt{5\times7} = \sqrt{35} = 5.92$ BM.

Key Point: To count unpaired electrons, write the d-configuration of the ion (not the atom) and fill the five d orbitals following Hund's rule. For Fe$^{3+}$ ($d^5$): 5 unpaired; for Ni$^{2+}$ ($d^8$): 2 unpaired; for Zn$^{2+}$ ($d^{10}$): 0 unpaired (diamagnetic).

[NEET Important] Spin-only magnetic moment is the single most common numerical in this chapter. Always: (1) find the ion's d-configuration, (2) count $n$, (3) apply $\mu = \sqrt{n(n+2)}$.

Why Transition-Metal Compounds Are Coloured

Most transition-metal ions are coloured — and the reason is their partially filled d subshell.

When a transition-metal ion is surrounded by ligands, the five d orbitals split into two energy levels. A d electron can absorb a photon of visible light and jump from the lower to the higher set — a d-d transition. The light absorbed is removed from white light, and we see the complementary colour.

• Ions with $d^0$ (e.g. Sc$^{3+}$, Ti$^{4+}$) or $d^{10}$ (e.g. Zn$^{2+}$, Cu$^+$) are colourless — there is no partially filled d set for a d-d transition.
• The colour depends on the metal, its oxidation state, and the ligands (which set the size of the d-orbital splitting).

Examples: $[\text{Cu(H}_2\text{O})_6]^{2+}$ is blue; $[\text{Ti(H}_2\text{O})_6]^{3+}$ ($d^1$) is purple; Mn$^{2+}$ salts are pale pink. Sc$^{3+}$ and Zn$^{2+}$ salts are white.

[JEE Tip] "Why is Zn$^{2+}$ (or Sc$^{3+}$) colourless?" — because it has a $d^{10}$ (or $d^0$) configuration with no possible d-d transition. This is a guaranteed exam question.

Solved Examples

Example 1: Spin-only moment of Mn2+

Calculate the spin-only magnetic moment of Mn$^{2+}$ ($Z = 25$).

Solution: Mn$^{2+}$ is $3d^5$ with 5 unpaired electrons. $\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$ BM.

Example 2: Moment of a d2 ion

Calculate the spin-only moment of an ion with $d^2$ configuration.

Solution: $d^2$ has 2 unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83$ BM.

Example 3: Number of unpaired electrons from moment

A transition-metal ion has a spin-only moment of 3.87 BM. How many unpaired electrons does it have?

Solution: $3.87 = \sqrt{n(n+2)}$, so $n(n+2) = 15$, giving $n = 3$. Three unpaired electrons (a $d^3$ ion such as Cr$^{3+}$).

Example 4: Moment of M2+ with Z = 27

Calculate the spin-only magnetic moment of the M$^{2+}$ ion with $Z = 27$ (cobalt).

Solution: Co is $[\text{Ar}]\,3d^7 4s^2$; Co$^{2+}$ is $3d^7$ with 3 unpaired electrons. $\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87$ BM.

Example 5: Diamagnetic ion

Which of Zn$^{2+}$, Cu$^{2+}$, Fe$^{3+}$ is diamagnetic, and why?

Solution: Zn$^{2+}$ ($3d^{10}$) is diamagnetic — all d electrons are paired ($n = 0$, $\mu = 0$). Cu$^{2+}$ ($d^9$, 1 unpaired) and Fe$^{3+}$ ($d^5$, 5 unpaired) are paramagnetic.

Example 6: Why Sc3+ is colourless

Why are Sc$^{3+}$ salts colourless?

Solution: Sc$^{3+}$ has a $3d^0$ configuration — no d electrons for a d-d transition — so it cannot absorb visible light and appears colourless.

Example 7: Colour of Cu+ vs Cu2+

Why is Cu$^+$ colourless but Cu$^{2+}$ coloured?

Solution: Cu$^+$ is $3d^{10}$ (fully filled) — no d-d transition possible, so colourless. Cu$^{2+}$ is $3d^9$ with a vacancy in the d set, allowing a d-d transition that absorbs visible light — so it is coloured (blue).

Example 8: Moment of Ni2+

Calculate the spin-only moment of Ni$^{2+}$ ($Z = 28$).

Solution: Ni is $[\text{Ar}]\,3d^8 4s^2$; Ni$^{2+}$ is $3d^8$ with 2 unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83$ BM.

Example 9: Origin of colour

What causes the colour of transition-metal compounds?

Solution: d-d electronic transitions. In a ligand field the d orbitals split; a d electron absorbs visible light to jump between the split levels. The colour seen is the complement of the light absorbed.

Example 10: Identify maximum paramagnetism

Among Ti$^{3+}$ ($d^1$), V$^{3+}$ ($d^2$), Cr$^{3+}$ ($d^3$) and Mn$^{2+}$ ($d^5$), which has the largest spin-only moment?

Solution: Mn$^{2+}$ ($d^5$) has the most unpaired electrons (5), giving the largest moment, $\mu = 5.92$ BM.