JEE Mains Chemistry · Classification of Elements and Periodicity

Ionization Enthalpy

Ionization enthalpy is the energy needed to remove the most loosely held electron from an isolated gaseous atom; it rises across a period with two dips, falls down a group with a few exceptions, and jumps sharply once the valence electrons are gone.

All Classification of Elements and Periodicity notes JEE Mains Chemistry strategy

Why this matters

Twenty-seven PYQs, the largest page, twenty-six of them multiple choice, and nine from 2026. Fifteen rank the first ionization enthalpies of a period, where the marks go on the two dips; four compare elements down groups 13 and 14; eight use second and later ionization enthalpies, to name a group from a jump or to find the energy for a mass of atoms.

Concept 1 of 3: First ionization enthalpy across a period

Across a period the nuclear charge rises on the same shell, so electrons get harder to pull off. The rise is broken twice. Group 13 dips below group 2, because its first p electron is held less tightly than an s electron. Group 16 dips below group 15, because its fourth p electron shares an orbital and is pushed out by the partner.

Definition

$\Delta_i H$ is always positive: removing an electron always takes energy.
Across a period: rises overall, from the alkali metal (lowest) to the noble gas (highest).
Dip 1, group 2 > group 13: Be > B and Mg > Al; a $2p$ electron penetrates less than a $2s$ electron.
Dip 2, group 15 > group 16: N > O and P > S; the half-filled $np^3$ is stable and the paired electron in $np^4$ is repelled.
Period 2 in order: Li < B < Be < C < O < N < F < Ne.
Period 3 in order: Na < Al < Mg < Si < S < P < Cl < Ar.

Group	Period 2 (kJ mol⁻¹)	Period 3 (kJ mol⁻¹)	Why
1	Li 520	Na 496	One s electron outside a noble-gas core: lowest in the period
2	Be 899	Mg 737	Filled $ns^2$ subshell
13	B 801	Al 577	Dip: the lone $np$ electron is less penetrating Group 13 sits BELOW group 2. The option with a smooth rise (Be < B) is the trap.
14	C 1086	Si 786	Rises again
15	N 1402	P 1012	Half-filled $np^3$ : extra stable
16	O 1314	S 1000	Dip: pairing in $np^4$ adds repulsion Group 16 sits BELOW group 15: N > O and P > S.
17	F 1681	Cl 1256	Rises again
18	Ne 2080	Ar 1520	Filled shell: highest in the period

Read down each column for the period order; the two amber rows are the exceptions every question tests.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q28Moderate

Example 1 · Classification of Elements and Periodicity · Ionization Enthalpy

The correct trend in the first ionization enthalpies of the elements in the

3^{\text{rd~}}

period of periodic table is:

The smooth order is the wrong option

Li < Be < B < C < N < O < F looks right and is always offered. It misses both dips. The true order swaps two pairs: Li < B < Be < C < O < N < F.

Drill 14 more on first ionization enthalpy across a period

Concept 2 of 3: Down a group, and where it fails

Down a group the outer electron sits in a bigger shell, so it is easier to remove. After the d and f subshells fill, though, those inner electrons shield the nucleus badly. The heavier element then holds its outer electrons more tightly than expected, and the trend stalls or reverses.

Definition

Groups 1, 2 and 18 fall steadily: Li > Na > K; Rn is the lowest noble gas.
Group 13: Ga (579) is just above Al (577) because of ten poorly shielding $3d$ electrons; Tl (589) is above In (558) because of the $4f$ electrons.
Second ionization enthalpy in group 13 follows B > Ga > Al.
Group 14: Pb (715) is above Sn (708); the order is C > Si > Ge > Pb > Sn.
Across period 4, Zn (906, filled $3d^{10}4s^2$ ) is far above Ga (579).

Group	First ionization enthalpy (kJ mol⁻¹)	Order and exception
1	Li 520, Na 496, K 419, Rb 403, Cs 376	Steady fall
2	Be 899, Mg 737, Ca 590, Sr 549, Ba 503	Steady fall
13	B 801, Al 577, Ga 579, In 558, Tl 589	B > Tl > Ga > Al > In Ga is not below Al, and Tl is above both.
13, second	B 2427, Al 1816, Ga 1979, In 1820, Tl 1971	B > Ga > Tl > In > Al
14	C 1086, Si 786, Ge 761, Sn 708, Pb 715	C > Si > Ge > Pb > Sn Pb is above Sn.
18	He 2372, Ne 2080, Ar 1520, Kr 1351, Xe 1170, Rn 1037	Steady fall; Rn lowest

The exceptions appear only after a filled d or f subshell: from Ga, Tl and Pb onwards.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q115Moderate

Example 2 · Classification of Elements and Periodicity · Ionization Enthalpy

Given below are two statements : Statement (I): The first ionization energy of Pb is greater than that of Sn Statement(II) : The first ionization energy of Ge is greater than that of Si . In the light of the above statements, choose the correct answer from the options given below :

A simple fall down group 13 or 14 is wrong

An order like Al > Ga or Sn > Pb applies the group rule blindly. Poor shielding by d and f electrons lifts Ga and Pb. Check these two pairs before choosing.

Drill 3 more on down a group, and where it fails

Concept 3 of 3: Successive ionization enthalpies and the energy for a mass

Each electron removed leaves a more positive ion, so the next one is harder to pull off: IE₂ is always larger than IE₁. Once all the valence electrons are gone, the next one comes from a noble-gas core, and the value jumps several times over. The position of that jump counts the valence electrons.

Definition

$\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \dots$ , all positive.
A large jump between $\Delta_i H_k$ and $\Delta_i H_{k+1}$ means $k$ valence electrons: group $k$ for s-block, group $10 + k$ for p-block.
Second ionization enthalpies compare the $\mathrm{M^+}$ ions: Na > Mg, because $\mathrm{Na^+}$ has a neon core; for C, N, O, F the order is C < N < F < O, because $\mathrm{O^+}$ is a half-filled $2p^3$ .
Ca has a very high third ionization enthalpy: $\mathrm{Ca^{2+}}$ has the argon configuration.
Energy for a given mass = moles × the sum of the enthalpies for the electrons removed.

Energy to ionize a mass of gaseous atoms

E = \frac{m}{M}\,\left(\Delta_i H_1 + \Delta_i H_2 + \dots\right)

mmass of the gaseous atoms, in g
Mmolar mass, in g mol⁻¹

Worked example

An element's first four ionization enthalpies are 577, 1816, 2744 and 11577 kJ mol⁻¹. Find its group and the ion it forms.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 6 · Q34Moderate

Example 3 · Classification of Elements and Periodicity · Ionization Enthalpy

Identify the elements

X

and

Y

using the ionisation energy values (kJ/mol) given below:

Element	1st IE	2nd IE
X	495	4563
Y	731	1450

Negative or smaller second values

Options such as −856 kJ mol⁻¹ or 590 kJ mol⁻¹ for the second ionization enthalpy of Mg are impossible. The second value must be positive and larger than 737.

Second ionization compares the cations

For IE₂, look at the ion that loses the electron.

\mathrm{O^+}

2p^3

and

\mathrm{N^+}

2p^2

, so the dip moves one place: O is above F for IE₂, the reverse of the first ionization order.

Drill 7 more on successive ionization enthalpies and the energy for a mass

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)

Successive ionization enthalpies and the energy for a mass
Energy to ionize a mass of gaseous atoms
$E = \frac{m}{M}\,\left(\Delta_i H_1 + \Delta_i H_2 + \dots\right)$

Reference tables (2)

First ionization enthalpy across a period8 rows

Group	Period 2 (kJ mol⁻¹)	Period 3 (kJ mol⁻¹)	Why
1	Li 520	Na 496	One s electron outside a noble-gas core: lowest in the period
2	Be 899	Mg 737	Filled $ns^2$ subshell
13	B 801	Al 577	Dip: the lone $np$ electron is less penetrating Group 13 sits BELOW group 2. The option with a smooth rise (Be < B) is the trap.
14	C 1086	Si 786	Rises again
15	N 1402	P 1012	Half-filled $np^3$ : extra stable
16	O 1314	S 1000	Dip: pairing in $np^4$ adds repulsion Group 16 sits BELOW group 15: N > O and P > S.
17	F 1681	Cl 1256	Rises again
18	Ne 2080	Ar 1520	Filled shell: highest in the period

Read down each column for the period order; the two amber rows are the exceptions every question tests.

Down a group, and where it fails6 rows

Group	First ionization enthalpy (kJ mol⁻¹)	Order and exception
1	Li 520, Na 496, K 419, Rb 403, Cs 376	Steady fall
2	Be 899, Mg 737, Ca 590, Sr 549, Ba 503	Steady fall
13	B 801, Al 577, Ga 579, In 558, Tl 589	B > Tl > Ga > Al > In Ga is not below Al, and Tl is above both.
13, second	B 2427, Al 1816, Ga 1979, In 1820, Tl 1971	B > Ga > Tl > In > Al
14	C 1086, Si 786, Ge 761, Sn 708, Pb 715	C > Si > Ge > Pb > Sn Pb is above Sn.
18	He 2372, Ne 2080, Ar 1520, Kr 1351, Xe 1170, Rn 1037	Steady fall; Rn lowest

The exceptions appear only after a filled d or f subshell: from Ga, Tl and Pb onwards.

Watch out for (4)

The smooth order is the wrong option
→ First ionization enthalpy across a period
A simple fall down group 13 or 14 is wrong
→ Down a group, and where it fails
Negative or smaller second values
→ Successive ionization enthalpies and the energy for a mass
Second ionization compares the cations
→ Successive ionization enthalpies and the energy for a mass

Test yourself on Classification of Elements and Periodicity

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 Classification of Elements and Periodicity chapter test Drill all 27 questions