JEE Mains Chemistry · The d- and f-Block Elements

Electronic Configuration and General Properties

The d-block fills the (n−1)d subshell after ns, its ions lose the ns electrons first, and the half-filled and filled d shells explain the odd configurations, the kinks in ionisation enthalpy and the soft, low-melting zinc group.

All The d- and f-Block Elements notes JEE Mains Chemistry strategy

Why this matters

Twenty-four PYQs, twenty-one of them multiple choice, and six from 2026. Seven write a configuration for an atom or an ion, including the 4d exceptions; seven compare ionisation enthalpies across chromium, manganese and iron; ten test melting points, atomisation enthalpy, density, catalysts and interstitial compounds. Everything on the page starts from one skill: counting d electrons correctly.

Concept 1 of 3: Configurations of d-block atoms and ions

The 4s subshell fills before 3d, but once the 3d electrons are in place they sit lower in energy. So an atom is written 3dⁿ4s², and an ion loses its 4s electrons first. Two atoms break the pattern in the 3d series: chromium and copper move one 4s electron into 3d to reach a half-filled 3d⁵ or a filled 3d¹⁰.

Definition

3d atoms: $[\mathrm{Ar}]\,3d^{n}4s^{2}$ , except $\mathrm{Cr}\ [\mathrm{Ar}]\,3d^{5}4s^{1}$ and $\mathrm{Cu}\ [\mathrm{Ar}]\,3d^{10}4s^{1}$ .
4d exceptions: $\mathrm{Nb}\ 4d^{4}5s^{1}$ , $\mathrm{Mo}\ 4d^{5}5s^{1}$ , $\mathrm{Ru}\ 4d^{7}5s^{1}$ , $\mathrm{Rh}\ 4d^{8}5s^{1}$ , $\mathrm{Pd}\ 4d^{10}5s^{0}$ , $\mathrm{Ag}\ 4d^{10}5s^{1}$ .
5d: $\mathrm{Pt}\ [\mathrm{Xe}]\,4f^{14}5d^{9}6s^{1}$ , $\mathrm{Au}\ [\mathrm{Xe}]\,4f^{14}5d^{10}6s^{1}$ .
Ions: remove the 4s electrons first, one at a time, then 3d. So $\mathrm{Mn^{+}} = 3d^{5}4s^{1}$ but $\mathrm{Cr^{+}} = 3d^{5}$ .
For a 3d ion with charge 2+ or more, the d count is $Z - 18 - n$ .
Unpaired electrons in the atoms: Sc 1, Ti 2, V 3, Cr 6, Mn 5, Fe 4, Co 3, Ni 2, Cu 1, Zn 0.
A full d subshell (d¹⁰) in the atom: Cu, Zn, Pd, Ag, Cd, Au, Hg.

d electrons in a 3d ion

n_d(\mathrm{M^{n+}}) = Z - 18 - n \qquad (n \ge 2)

Worked example

Write the configurations of

\mathrm{Co^{2+}}

(Z = 27) and

\mathrm{Cu^{+}}

(Z = 29), and say which has more unpaired electrons.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q122Moderate

Example 1 · The d- and f-Block Elements · Electronic Configuration and General Properties

Niobium ( Nb ) and ruthenium ( Ru ) have " x " and "

y

" number of electrons in their respective

4d

orbitals. The value of

x + y

\_\_\_\_\_\_

Ions lose 4s before 3d

\mathrm{Fe^{2+}}

[\mathrm{Ar}]\,3d^{6}

, not

[\mathrm{Ar}]\,3d^{4}4s^{2}

. The 4s electrons fill first but leave first too. Writing the ion as the atom minus 3d electrons gives the wrong count of unpaired electrons and the wrong magnetic moment.

The 4d series has more exceptions than the 3d

In 3d only Cr and Cu take a single s electron. In 4d, Nb, Mo, Ru, Rh and Ag all do, and Pd has none at all. Do not copy the 3d pattern down the group: Nb is

4d^{4}5s^{1}

although V above it is

3d^{3}4s^{2}

Drill 6 more on configurations of d-block atoms and ions

Concept 2 of 3: Ionisation enthalpies across the 3d series

Ionisation enthalpy rises slowly across the series, because each extra proton is only partly shielded by the new 3d electron. The kinks come from stable shells. Removing an electron is hard when it breaks a 3d⁵ or 3d¹⁰ shell, and easy when it leaves one behind.

Definition

First IE: Cr (653) is lower than Mn (717). Cr loses its lone 4s electron; Mn must break a paired 4s².
Second IE: Cr is the highest from Sc to Fe (1592), because $\mathrm{Cr^{+}}$ is 3d⁵. Cu is higher still (1958), because $\mathrm{Cu^{+}}$ is 3d¹⁰.
Third IE: Mn is very high, because $\mathrm{Mn^{2+}}$ is 3d⁵. Fe is low, because $\mathrm{Fe^{2+}}$ (3d⁶) reaches 3d⁵ by losing one electron.
So Mn²⁺ is hard to oxidise and Fe²⁺ is easy: this is why $\mathrm{Fe^{3+}}$ is common and $\mathrm{Mn^{3+}}$ is an oxidant.
Zinc has the highest first IE of the series (906), because it loses an electron from a filled 4s² above a filled 3d¹⁰.

Metal	First IE (kJ/mol)	Second IE (kJ/mol)	Third IE (kJ/mol)	What it shows
Sc	631	1235	2389	Sc³⁺ is d⁰, so +3 is easy and is its only state
Ti	656	1309	2652	A steady rise with the nuclear charge
V	650	1414	2828	A steady rise with the nuclear charge
Cr	653	1592	2987	Low first IE (lone 4s); high second IE (breaks 3d⁵) Highest second IE from Sc to Fe, but its third IE is below Mn's.
Mn	717	1509	3248	High third IE: Mn²⁺ is 3d⁵
Fe	762	1561	2957	Low third IE: Fe²⁺ loses one electron to reach 3d⁵
Co	758	1644	3232	Rises again after the dip at Fe
Ni	736	1752	3393	Rises again after the dip at Fe
Cu	745	1958	3554	Highest second IE of the series: Cu⁺ is 3d¹⁰
Zn	906	1734	3833	Highest first IE: a filled 4s² over a filled 3d¹⁰

Values rounded to the nearest kJ/mol. The kinks, not the exact numbers, decide the questions.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q29Moderate

Example 2 · The d- and f-Block Elements · Electronic Configuration and General Properties

Given below are two statements : Statement-I : The first ionization enthalpy of Cr is lower than that of Mn. Statement-II : The second and third ionization enthalpies of Cr are higher than those of Mn . In the light of the above statements, choose the correct answer from the options given below :

Cr beats Mn on the second IE only

Cr's second IE is higher than Mn's, but its third IE is lower. A statement that 'the second and third IEs of Cr are both higher than those of Mn' is false, because

\mathrm{Mn^{2+}}

is the 3d⁵ ion at the third step.

Cr is not the highest second IE of the whole series

Cr has the highest second IE only up to Fe. Copper's second IE (1958) is higher, because it breaks a filled 3d¹⁰. Read which metals the question lists before answering.

Drill 6 more on ionisation enthalpies across the 3d series

Concept 3 of 3: Melting points, atomisation, density, catalysts and interstitial compounds

Transition metals are strong, dense and high-melting because both their ns and their unpaired (n−1)d electrons join the metallic bonding. The more unpaired d electrons, the stronger the bonding, so the enthalpy of atomisation peaks near the middle of each series. Manganese and zinc are the weak points: Mn's 3d⁵ holds its electrons back, and Zn's 3d¹⁰ gives none.

Definition

Atomisation enthalpy peaks at V (515 kJ/mol) and dips at Mn (281). Zn (126) is the lowest, so Zn, Cd and Hg are soft and low-melting.
4d and 5d metals have higher atomisation enthalpies than 3d metals, so they form more metal–metal bonds.
Melting points (°C): Mn 1246 < Fe 1538; Tc 2157 < Ru 2334; but Re 3186 > Os 3033. W (3422) is the highest of all.
Density rises across the series: Zn 7.14 < Cr 7.19 < Fe 7.87 < Co 8.90 < Cu 8.96 g/cm³.
Catalysts: $\mathrm{V_2O_5}$ (contact process), Fe (Haber process), Ni (hydrogenation), $\mathrm{TiCl_4}$ with $\mathrm{Al(C_2H_5)_3}$ (Ziegler–Natta), $\mathrm{PdCl_2}$ (Wacker process, ethene to ethanal).
A catalyst surface bonds reactants using BOTH 3d and 4s electrons. This raises their concentration at the surface and WEAKENS their bonds, lowering the activation energy.
Interstitial compounds (TiC, $\mathrm{Mn_4N}$ , $\mathrm{Fe_3H}$ , $\mathrm{TiH_{1.7}}$ ): small atoms trapped in the metal lattice. They are non-stoichiometric, very hard, higher-melting than the metal, still conduct, and are chemically inert.
Zn, Cd, Hg: full d subshell, so they are not typical transition metals. Zn and Cd show only +2; Hg shows +1 (as $\mathrm{Hg_2^{2+}}$ ) and +2. Their compounds are white and diamagnetic.

Metal	Atomisation enthalpy (kJ/mol)	Metallic radius (pm)	Density (g/cm³)	Point tested
Sc	326	164	2.99	Largest atom of the series
Ti	473	147	4.51	Ti⁴⁺ in TiCl₄ is d⁰: the Ziegler–Natta catalyst is diamagnetic
V	515	135	6.11	Highest atomisation enthalpy of the 3d series
Cr	397	129	7.19	Smallest radius among Sc, Ti, V, Cr, Mn and Zn
Mn	281	137	7.21	A dip: 3d⁵ holds its d electrons out of the bonding
Fe	416	126	7.87	Catalyst of the Haber process
Co	425	125	8.90	Dense, high-melting
Ni	430	125	8.91	Catalyst for hydrogenating oils
Cu	339	128	8.96	Densest of the listed 3d metals
Zn	126	137	7.14	Lowest atomisation enthalpy: soft, low-melting Zn, Cd and Hg have filled d subshells; they are the soft end of each series.

The atomisation enthalpy tracks the number of unpaired d electrons that join the metallic bond.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q36Moderate

Example 3 · The d- and f-Block Elements · Electronic Configuration and General Properties

The correct option with order of melting points of the pairs

(Mn,Fe),(Tc,Ru)

and

(Re,Os)

is :

A catalyst weakens bonds and uses 4s electrons too

Two false statements recur: that first-row catalysts use only their 3d electrons, and that adsorption strengthens the reactant bonds. The surface bonds through 3d AND 4s electrons, and adsorption weakens the reactant bonds, which is why the activation energy falls.

The group-7/group-8 order flips in the 5d series

Mn melts below Fe and Tc below Ru, but Re melts above Os. Do not extend the 3d and 4d pattern to the 5d pair.

Drill 9 more on melting points, atomisation, density, catalysts and interstitial compounds

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Configurations of d-block atoms and ions
d electrons in a 3d ion
$n_d(\mathrm{M^{n+}}) = Z - 18 - n \qquad (n \ge 2)$

Reference tables (2)

Ionisation enthalpies across the 3d series10 rows

Metal	First IE (kJ/mol)	Second IE (kJ/mol)	Third IE (kJ/mol)	What it shows
Sc	631	1235	2389	Sc³⁺ is d⁰, so +3 is easy and is its only state
Ti	656	1309	2652	A steady rise with the nuclear charge
V	650	1414	2828	A steady rise with the nuclear charge
Cr	653	1592	2987	Low first IE (lone 4s); high second IE (breaks 3d⁵) Highest second IE from Sc to Fe, but its third IE is below Mn's.
Mn	717	1509	3248	High third IE: Mn²⁺ is 3d⁵
Fe	762	1561	2957	Low third IE: Fe²⁺ loses one electron to reach 3d⁵
Co	758	1644	3232	Rises again after the dip at Fe
Ni	736	1752	3393	Rises again after the dip at Fe
Cu	745	1958	3554	Highest second IE of the series: Cu⁺ is 3d¹⁰
Zn	906	1734	3833	Highest first IE: a filled 4s² over a filled 3d¹⁰

Values rounded to the nearest kJ/mol. The kinks, not the exact numbers, decide the questions.

Melting points, atomisation, density, catalysts and interstitial compounds10 rows

Metal	Atomisation enthalpy (kJ/mol)	Metallic radius (pm)	Density (g/cm³)	Point tested
Sc	326	164	2.99	Largest atom of the series
Ti	473	147	4.51	Ti⁴⁺ in TiCl₄ is d⁰: the Ziegler–Natta catalyst is diamagnetic
V	515	135	6.11	Highest atomisation enthalpy of the 3d series
Cr	397	129	7.19	Smallest radius among Sc, Ti, V, Cr, Mn and Zn
Mn	281	137	7.21	A dip: 3d⁵ holds its d electrons out of the bonding
Fe	416	126	7.87	Catalyst of the Haber process
Co	425	125	8.90	Dense, high-melting
Ni	430	125	8.91	Catalyst for hydrogenating oils
Cu	339	128	8.96	Densest of the listed 3d metals
Zn	126	137	7.14	Lowest atomisation enthalpy: soft, low-melting Zn, Cd and Hg have filled d subshells; they are the soft end of each series.

The atomisation enthalpy tracks the number of unpaired d electrons that join the metallic bond.

Watch out for (6)

Ions lose 4s before 3d
→ Configurations of d-block atoms and ions
The 4d series has more exceptions than the 3d
→ Configurations of d-block atoms and ions
Cr beats Mn on the second IE only
→ Ionisation enthalpies across the 3d series
Cr is not the highest second IE of the whole series
→ Ionisation enthalpies across the 3d series
A catalyst weakens bonds and uses 4s electrons too
→ Melting points, atomisation, density, catalysts and interstitial compounds
The group-7/group-8 order flips in the 5d series
→ Melting points, atomisation, density, catalysts and interstitial compounds

Test yourself on The d- and f-Block Elements

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the The d- and f-Block Elements chapter test Drill all 24 questions

Related notes

Magnetic Moment and Colour — from a d count to a value in BM →