CBSE Class 12 d-Block — Colour, Magnetism and Catalysis

Three questions from the d-block chapter come back again and again: why transition metal compounds are coloured, how to calculate a magnetic moment, and why these metals are such good catalysts. All three have the same root cause — a partly filled d subshell. Understand that one idea properly and the chapter becomes short.

The magnetic-moment question is the most reliably scoring of the three because it is pure calculation. We will do it with full arithmetic below.

First, what counts as a transition element

A transition element is one that has an incompletely filled d subshell either in its ground state or in one of its common oxidation states. That definition is worth quoting exactly, because it decides the standard follow-up question.

Zinc, cadmium and mercury are in the d block but are not typical transition elements: their configuration is d¹⁰ as the atom and d¹⁰ as the common M²⁺ ion, so there is no partly filled d subshell anywhere in their ordinary chemistry. That single fact explains why their compounds are white or colourless and diamagnetic.

One configuration rule you must apply automatically: when a transition metal forms a cation, the 4s electrons leave before the 3d electrons. Iron is [Ar]3d⁶4s², so Fe²⁺ is [Ar]3d⁶ (not 3d⁴4s²) and Fe³⁺ is [Ar]3d⁵.

Magnetism — the spin-only formula

A substance with unpaired electrons is attracted into a magnetic field and is paramagnetic. With every electron paired it is weakly repelled and is diamagnetic. For 3d ions the magnetic moment is estimated from the number of unpaired electrons alone:

Learn the five values rather than recomputing a square root every time. Here they are, with the arithmetic done:

Notice that the moment rises to a maximum at d⁵ and then falls again, because after d⁵ the electrons must start pairing up.

Worked example 1 — Fe²⁺ and Fe³⁺

Worked example 2 — working backwards from a measured value

Worked example 3 — the diamagnetic cases

Colour — and the exception that catches everyone

In a complex, the five d orbitals are no longer at the same energy: the ligands split them into two sets separated by a small energy gap. Visible light supplies exactly the right amount of energy to promote an electron across that gap — a d–d transition. The colour you see is the complement of the colour absorbed.

Now the exception. In MnO₄⁻ (permanganate, deep purple) manganese is in the +7 state and in Cr₂O₇²⁻ (dichromate, orange) chromium is +6 — both are d⁰. By the rule above they should be colourless, yet they are among the most intensely coloured ions in the lab. Their colour comes from a charge-transfer transition: an electron jumps from an oxygen-based orbital to a metal-based orbital. It is not a d–d transition. Charge transfer is also why these colours are so much more intense than, say, the pale pink of Mn²⁺.

Catalysis — why d-block metals are so good at it

Three properties combine:

• Variable oxidation states. A catalyst must be able to accept electrons at one step and give them back at another. A metal that moves easily between, say, Fe²⁺ and Fe³⁺ can do that repeatedly without being consumed.
• Formation of unstable intermediates. Partly filled d orbitals let the metal bind a reactant just firmly enough to weaken its bonds, then release the product. This provides an alternative path of lower activation energy.
• Surface adsorption. In heterogeneous catalysis, reactant molecules are adsorbed on the metal surface, which brings them close together in the right orientation and stretches their bonds.

Remember the limit of what a catalyst does: it lowers the activation energy and speeds up both the forward and backward reactions equally. It never shifts the position of equilibrium and never changes ΔG for the reaction.

Mistakes that cost marks

Where this appears in your exam

These are dependable, self-contained questions — but confirm the current syllabus and any chapter changes from the official CBSE notification and the latest NCERT textbook.