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A favourite question in Class 11 chemistry, JEE and NEET gives you a list such as 578, 1817, 2745, 11578 kJ/mol and asks which group the element belongs to. You do not need to memorise any values. You need one idea and a habit of dividing rather than subtracting. This post explains the idea with real data and shows the traps.
What are successive ionisation enthalpies?
The NCERT Class 11 chapter on Classification of Elements defines the first ionisation enthalpy as the enthalpy change for X(g) → X⁺(g) + e⁻, and the second as the energy to remove the next electron, X⁺(g) → X²⁺(g) + e⁻. It notes that each one is higher than the one before, because it is harder to pull an electron from an ion that is already positive. Ionisation enthalpy is listed under Unit 9, Classification of Elements and Periodicity in Properties, in the JEE (Main) 2026 syllabus.
So a steady rise is normal. What you look for is a jump far bigger than the rest.
Why is there one very large jump?
Valence electrons sit in the outermost shell, far from the nucleus and shielded by the inner electrons. Once they are all gone, the next electron comes from the shell below. That electron is closer to the nucleus and much less shielded, so removing it costs far more energy. The position of the big jump therefore counts the valence electrons.
What do real values look like?
The NIST Atomic Spectra Database lists ionisation energies in electron volts. We converted them using 1 eV per particle = 96.485 kJ/mol, which follows from the CODATA value of the electron volt and the Avogadro constant. Rounded to the nearest kJ/mol:
Magnesium: 738, 1451, 7733, 10543
Aluminium: 578, 1817, 2745, 11578, 14842
Silicon: 787, 1577, 3232, 4356, 16091
Magnesium's 738 and 1451 match element VI in NCERT's Exercise 3.31, which lists 738 and 1451 kJ/mol as its first two values.
Worked example: find the group of aluminium from its data
Take the five aluminium values above and divide each by the one before it.
1817 ÷ 578 = 3.14
2745 ÷ 1817 = 1.51
11578 ÷ 2745 = 4.22
14842 ÷ 11578 = 1.28
The largest ratio, 4.22, comes after the third electron. So the element has three valence electrons. In the modern 1 to 18 numbering, a p-block element in period 3 with three valence electrons is in group 13. That is aluminium.
Now magnesium: 1451 ÷ 738 = 1.97, then 7733 ÷ 1451 = 5.33, then 10543 ÷ 7733 = 1.36. The jump comes after the second electron, so two valence electrons and group 2.
Silicon: the ratios are 2.00, 2.05, 1.35 and then 16091 ÷ 4356 = 3.69. The jump comes after the fourth electron, so group 14.
Why does aluminium have a second, smaller jump?
Look again at aluminium's first ratio, 3.14. It is large, and students sometimes stop there and say "group 1". The first electron in aluminium is a 3p electron, and the next two come from 3s, which is held a little more tightly. That gives a noticeable step, but nothing like the jump to the inner shell. This is the same effect behind boron's low first ionisation enthalpy, which our post on periodic trends and their exceptions explains. Always compare all the ratios before deciding.
Which mistakes cost marks?
Reading the first large-looking step. Check every ratio, not just the first.
Subtracting. In silicon, 4356 − 3232 = 1124 looks small and 16091 − 4356 = 11735 looks huge, which happens to work here. But differences mislead when the values are already large.
Off-by-one counting. If the jump is between the third and fourth values, three electrons were easy to remove. The answer is three, not four.
Mixing old and new group numbers. Three valence electrons is group 13 (old IIIA). Say which system you are using if the question does not.
Too little data. With only two values, a large ratio such as 520 to 7300 (NCERT's element I, ratio about 14) tells you group 1. A small ratio tells you only that the element is not in group 1.
A practice routine
Do this over two sessions.
From the NIST database, copy the first five ionisation energies of any four elements from periods 2 and 3, and convert them to kJ/mol.
Without labels, swap lists with a friend and find each other's groups by the ratio method.
Solve NCERT Exercise 3.31 and explain each choice in one sentence.
Write a short explanation of why the big jump happens, using the words shell, shielding and nuclear charge.
If shells and shielding are the shaky part, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise effective nuclear charge before you return to the data.
Start now: find the group of an element whose values are 496, 4562 and 6910 kJ/mol.
