ScienceCBSE Class 9 · Journey Inside the Atom

Atomic Number, Electronic Configuration, Valency, Isotopes and Isobars (Class 9)

Updated 17 October 2026 · Fact-checked

The atomic number Z is the number of protons (and, in a neutral atom, of electrons); the mass number A is protons plus neutrons, so neutrons = A − Z. Electrons fill shells K, L, M in order with at most 2n² electrons in a shell and at most 8 in the outermost shell (for the first 18 elements); valency is the number of electrons an atom loses, gains or shares to complete its outer shell. Isotopes have the same Z but different A, isobars have the same A but different Z, and the average atomic mass is the abundance-weighted mean of the isotope masses.

Sodium has configuration 2, 8, 1; its single outer electron is lost, so valency is 1, and neutrons equal A minus Z.

Illustration drawn for QuestionBuddy.

Understand Atomic Number, Electronic Configuration, Valency and Isotopes

Particles in the atom. Electron: charge −1, mass about 1/1840 of a proton, in shells. Proton: +1, relative mass 1, in the nucleus. Neutron: charge 0, relative mass 1, in the nucleus (Chadwick, 1932); every nucleus has neutrons except ordinary hydrogen. A neutral atom has equal numbers of protons and electrons.

Z, A and notation. Atomic number Z = number of protons; it fixes the identity of the element. Mass number A = protons + neutrons (nucleons); electrons are ignored because they are so light. Number of neutrons = A − Z. The notation is ᴬ_Z X: for ²³₁₁Na, A = 23 and Z = 11, so there are 11 protons, 12 neutrons and 11 electrons.

Distribution of electrons (first 18 elements). The Bohr-Bury rules: a shell can hold at most 2n² electrons (K 2, L 8, M 18); the outermost shell holds at most 8 electrons (K alone holds 2); shells fill in order K, L, M. Configurations: H 1; He 2; Li 2,1; Be 2,2; B 2,3; C 2,4; N 2,5; O 2,6; F 2,7; Ne 2,8; Na 2,8,1; Mg 2,8,2; Al 2,8,3; Si 2,8,4; P 2,8,5; S 2,8,6; Cl 2,8,7; Ar 2,8,8. A Bohr diagram has a nucleus labelled with protons and neutrons and concentric circles for the shells.

Valence electrons and valency. The outermost occupied shell is the valence shell and its electrons are valence electrons. Atoms with 8 (an octet) or, for helium, 2 are stable and unreactive (noble gases). Valency is the combining capacity: with 1 to 3 valence electrons the atom usually loses them and valency = number of valence electrons (Na 1, Mg 2, Al 3); with 5 to 7 it usually gains electrons and valency = 8 − valence electrons (N 3, O 2, F and Cl 1); with 4 it shares, so valency 4 (carbon); with 8 the valency is zero. Combining capacity can also be read from compounds: in H₂O oxygen joins 2 H (valency 2), in NH₃ nitrogen joins 3 H (valency 3), in MgCl₂ magnesium joins 2 Cl (valency 2).

Isotopes. Isotopes are atoms of the same element with the same atomic number but different mass numbers (different numbers of neutrons). Hydrogen has protium ¹₁H (0 neutrons), deuterium ²₁H (1 neutron) and tritium ³₁H (2 neutrons); carbon has ¹²₆C, ¹³₆C and ¹⁴₆C. Isotopes have the same chemical properties (same electrons) and different physical properties (mass, density, melting and boiling points). Uses: uranium-235 as fuel in nuclear reactors, cobalt-60 in cancer radiotherapy, iodine-131 for thyroid problems, carbon-14 for dating fossils.

Average atomic mass. Average = sum of (isotope mass × % abundance ÷ 100). For chlorine, 75% ³⁵Cl and 25% ³⁷Cl give 35 × 0.75 + 37 × 0.25 = 35.5 u; no single atom has a mass of 35.5 u, and the simple average (36) is wrong because it ignores abundance.

Isobars. Isobars are atoms of different elements with the same mass number but different atomic numbers, for example ⁴⁰₁₈Ar, ⁴⁰₁₉K and ⁴⁰₂₀Ca (40 nucleons each, but 18, 19 and 20 protons). Isotopes: same element, same Z, different A, same chemical properties. Isobars: different elements, different Z, same A, different chemical properties.

Atomic Number, Electronic Configuration, Valency and Isotopes: formulas and key facts

Neutral atom: $P r o t o n s = e l ec t r o n s = a t o mi c n u mb er Z$
Mass number and neutrons: $A = p r o t o n s + n e u t r o n s ∣ n e u t r o n s = A - Z$
Shell capacity: 2n²: K 2, L 8, M 18 | outermost shell at most 8
Valency: $1 t o 3 v a l e n cee l ec t r o n s : v a l e n cy = t ha t n u mb er ∣5 t o 7 : v a l e n cy = 8 - n u mb er ∣4 : s ha r es 4∣8 : z er o$
Average atomic mass: Sum of (isotope mass × % abundance ÷ 100)
Isotopes versus isobars: $I so t o p es : s am e Z, d i f f er e n t A ∣ i so ba r s : s am e A, d i f f er e n tZ$

How to solve Atomic Number, Electronic Configuration, Valency and Isotopes questions

Use this method for particle counts, configurations, valency and average mass.

1Read Z and A from the symbol; protons = electrons = Z, neutrons = A − Z.
2Fill the shells in order with 2, 8, 8 and stop when the electrons are used up; do not exceed 8 in the outermost shell.
3Count the valence electrons; valency is that number (1 to 3) or 8 minus it (5 to 7).
4For average mass, convert each percentage to a fraction, multiply by the isotope mass and add.
5To find an unknown abundance, let it be x and solve (mass₁ × x) + (mass₂ × (1 − x)) = average.
6To classify atoms as isotopes or isobars, compare Z and A.

Atomic Number, Electronic Configuration, Valency and Isotopes in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Counting particles in a sodium atom

Suppose a school chart gives the symbol ²³₁₁Na for a sodium atom and asks students to complete a table of particles.

Find the protons, neutrons and electrons, the electronic configuration and the valency.

Try it first, then show the solutionSolution

Z = 11, so protons = 11 and electrons = 11 in a neutral atom.
Neutrons = A − Z = 23 − 11 = 12.
Configuration: 2, 8, 1. One valence electron, which is easily lost, so valency = 1.

Answer: 11 protons, 12 neutrons, 11 electrons; 2, 8, 1; valency 1.

Memory hook: Z counts protons, A minus Z counts neutrons.

Why an atomic mass need not be a whole number

Suppose a student sees 6.9 u printed for lithium on a chart and asks whether some lithium atoms have a fractional number of neutrons. Assume a sample in which 10% of the atoms are ⁶Li and 90% are ⁷Li.

Explain the value 6.9 u using the abundances given.

Try it first, then show the solutionSolution

Average = 6 × 0.10 + 7 × 0.90 = 0.6 + 6.3 = 6.9 u.
Every lithium atom has a whole-number mass number (6 or 7); 6.9 u is the average over many atoms.
The simple mean (6 + 7) ÷ 2 = 6.5 is wrong because it ignores that ⁷Li is nine times more abundant.

Answer: 6.9 u is the abundance-weighted average of the two isotopes.

Memory hook: Average mass is weighted, not simple.

Tricks to remember Atomic Number, Electronic Configuration, Valency and Isotopes

Short memory aids. Say each one aloud once, then use it on a question the same day.

Z is for Zero in on the element
Z (protons) decides who the element is. Change Z and the element changes; change only A and you get an isotope.
Use it when: Isotopes versus different elements.
2, 8, 8
For the first 18 elements fill 2, then 8, then up to 8 again; the outer shell never exceeds 8.
Use it when: Writing configurations quickly.
Lose up to 3, gain 5 to 7
1 to 3 valence electrons: lose them, valency equals the number. 5 to 7: gain, valency is 8 minus the number.
Use it when: Finding valency.
Iso-tope, iso-bar
Isotopes have the same place (topos) in the periodic table, same Z. Isobars have the same weight (baros), same A.
Use it when: Distinguishing the two terms.

Atomic Number, Electronic Configuration, Valency and Isotopes: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Counting protons, neutrons and electrons · 3 marks · Short answerBoard level

Give the number of protons, neutrons and electrons in (a) ²³₁₁Na, (b) ³⁷₁₇Cl, (c) ⁴⁰₂₀Ca.

Try it first, then show the solutionSolution

(a) Na: p = 11, n = 23 − 11 = 12, e = 11. (1)
(b) Cl: p = 17, n = 37 − 17 = 20, e = 17. (1)
(c) Ca: p = 20, n = 40 − 20 = 20, e = 20. (1)

Answer: (a) 11, 12, 11; (b) 17, 20, 17; (c) 20, 20, 20.

Example 2: Configuration and valency · 3 marks · Short answerBoard level

Write the electronic configuration and the valency of (a) beryllium (Z = 4), (b) phosphorus (Z = 15), (c) fluorine (Z = 9).

Try it first, then show the solutionSolution

(a) Be: 2, 2. Two valence electrons are lost, so valency = 2. (1)
(b) P: 2, 8, 5. Five valence electrons, gains 3 to make 8, valency = 8 − 5 = 3. (1)
(c) F: 2, 7. Seven valence electrons, gains 1 to make 8, valency = 8 − 7 = 1. (1)

Answer: (a) 2, 2 and 2; (b) 2, 8, 5 and 3; (c) 2, 7 and 1.

Example 3: Average atomic mass and unknown abundance · 5 marks · Long answerHarder board level

(a) Magnesium has isotopes ²⁴Mg (79%), ²⁵Mg (10%) and ²⁶Mg (11%). Find the average atomic mass. (2)
(b) An element has two isotopes of masses 63 u and 65 u, and its average atomic mass is 63.5 u. Find the percentage of each isotope. (2)
(c) Why is the simple average of 24, 25 and 26 not used for magnesium? (1)

Try it first, then show the solutionSolution

(a) 24 × 0.79 = 18.96; 25 × 0.10 = 2.50; 26 × 0.11 = 2.86. Sum = 18.96 + 2.50 + 2.86 = 24.32 u. (2)
(b) Let the fraction of 63 u be x. 63x + 65(1 − x) = 63.5 gives 65 − 2x = 63.5, so x = 0.75. The 63 u isotope is 75% and the 65 u isotope is 25%. Check: 63 × 0.75 + 65 × 0.25 = 47.25 + 16.25 = 63.5. (2)
(c) The simple average (24 + 25 + 26) ÷ 3 = 25 u ignores abundance; ²⁴Mg is far more abundant, so the true average is nearer 24. (1)

Answer: (a) 24.32 u; (b) 75% and 25%; (c) the simple mean ignores abundance.

Example 4: Isotopes, isobars and neutron counts · 5 marks · Long answerHarder board level

Consider ¹²₆C, ¹³₆C, ¹⁴₆C, ¹⁴₇N, ⁴⁰₁₈Ar, ⁴⁰₁₉K and ⁴⁰₂₀Ca.
(a) Name the isotopes among them. (1)
(b) Name the sets of isobars. (1)
(c) Find the number of neutrons in ¹⁴₆C and ¹⁴₇N. (1)
(d) Explain why the isotopes of carbon behave alike chemically. (1)
(e) Give one use of carbon-14. (1)

Try it first, then show the solutionSolution

(a) ¹²C, ¹³C and ¹⁴C (same Z = 6, different A). (1)
(b) ¹⁴₆C and ¹⁴₇N (A = 14); and ⁴⁰₁₈Ar, ⁴⁰₁₉K and ⁴⁰₂₀Ca (A = 40). (1)
(c) ¹⁴C: 14 − 6 = 8 neutrons. ¹⁴N: 14 − 7 = 7 neutrons. (1)
(d) They have the same number of electrons and the same configuration (2, 4), and chemical behaviour depends on the electrons. (1)
(e) Carbon-14 is used to date fossils and ancient objects. (1)

Answer: (a) the three carbons; (b) C-14 with N-14, and Ar-40, K-40, Ca-40; (c) 8 and 7; (d) same electrons; (e) dating fossils.

Example 5: Identify the element from clues · 4 marks · Case-basedHOTS / competency

An atom has mass number 27 and 14 neutrons.
(a) Find its atomic number and the number of electrons. (1)
(b) Write its electronic configuration and its valency. (1)
(c) The ion formed has the same configuration as neon (2, 8). Find the charge on the ion and its protons and electrons. (1)
(d) Using oxygen (2, 6), write the formula of the compound formed and find its formula unit mass (Al = 27, O = 16). (1)

Try it first, then show the solutionSolution

(a) Z = A − neutrons = 27 − 14 = 13, so electrons = 13 (this is aluminium). (1)
(b) 2, 8, 3. Three valence electrons are lost, so valency = 3. (1)
(c) It loses 3 electrons to reach 2, 8, giving Al³⁺: 13 protons and 10 electrons. (1)
(d) Al³⁺ and O²⁻: criss-cross gives Al₂O₃. Mass = 2 × 27 + 3 × 16 = 54 + 48 = 102 u. (1)

Answer: (a) 13; (b) 2, 8, 3 and valency 3; (c) Al³⁺ with 13 protons and 10 electrons; (d) Al₂O₃, 102 u.

Where students lose marks in Atomic Number, Electronic Configuration, Valency and Isotopes

The slips examiners see most often, and what to write instead.

Mistake: Counting electrons in the mass number.
Confusion about what A includes.
Wrong working
A = protons + neutrons + electrons
Right working
A = protons + neutrons
Do this: Electrons are too light to count.
Usually costs: 1 mark.
Mistake: Putting more than 8 electrons in the outermost shell, or filling M before L is full.
The 2n² rule alone is applied without the octet limit.
Wrong working
Ar = 2, 8, 8 written as 2, 6, 10
Right working
2, 8, 8
Do this: Fill K 2, L 8, then M up to 8 for the first 18 elements.
Usually costs: 1 mark each.
Mistake: Using the simple mean for average atomic mass.
Abundance is ignored.
Wrong working
Cl = (35 + 37) ÷ 2 = 36 u
Right working
Cl = 35 × 0.75 + 37 × 0.25 = 35.5 u
Do this: Weight each isotope by its percentage.
Usually costs: 2 marks.
Mistake: Calling isotopes and isobars the same.
The two words sound alike.
Wrong working
¹⁴C and ¹⁴N are isotopes.
Right working
They are isobars (same A, different Z).
Do this: Isotopes: same Z. Isobars: same A.
Usually costs: 1 mark.
Mistake: Giving the valency of an atom with 5 to 7 valence electrons as the number of valence electrons.
The rule for losing is applied to gaining.
Wrong working
Oxygen (2, 6) has valency 6.
Right working
Oxygen gains 2 to make 8, so valency = 8 − 6 = 2.
Do this: For 5 to 7, valency is 8 minus the number.
Usually costs: 1 mark.

Exam tips

Write the symbol with A and Z and then list p, n and e in a mini-table.
Draw Bohr diagrams with the nucleus labelled and each shell as a circle with the right number of dots.
For average atomic mass show each product and the sum; for unknown percentage define x clearly.
Use the phrase 'same atomic number, different mass number' for isotopes and 'same mass number, different atomic number' for isobars.
Link valency to the valence electrons when asked 'why'.

Practice questions on Atomic Number, Electronic Configuration, Valency and Isotopes

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Define isotopes and write the notation for the three isotopes of hydrogen, stating the neutrons in each.
Hint
Same Z, different A.
Answer
Isotopes are atoms of the same element with the same atomic number but different mass numbers (1 mark). Protium ¹₁H has 0 neutrons, deuterium ²₁H has 1 neutron, tritium ³₁H has 2 neutrons (1 mark).
Q2 · 3 marks · Short answerHarder board level
Neon has two isotopes, ²⁰Ne (90%) and ²²Ne (10%). (a) Find its average atomic mass. (b) A different sample has an average atomic mass of 20.4 u; find the percentage of ²²Ne in it. (c) State how many neutrons the neon-22 atom has (Z = 10).
Hint
Let the fraction of ²²Ne be x.
Answer
(a) 20 × 0.90 + 22 × 0.10 = 18 + 2.2 = 20.2 u (1 mark). (b) 20(1 − x) + 22x = 20.4 gives 20 + 2x = 20.4, x = 0.2, so 20% (1 mark). (c) 22 − 10 = 12 neutrons (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): Carbon-14 (¹⁴₆C) and nitrogen-14 (¹⁴₇N) are isobars. Reason (R): They have the same mass number but different atomic numbers. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Compare Z and A of the two atoms.
Answer
(a). Both have A = 14 but Z = 6 and 7, which is exactly the definition of isobars (different elements, same mass number).
Q4 · 3 marks · Short answerHarder board level
Write the electronic configuration of sulphur (Z = 16) and silicon (Z = 14), find the valency of each, and explain why neon has zero valency.
Hint
Compare valence electrons with 8.
Answer
Sulphur: 2, 8, 6, so it gains 2 electrons and has valency 8 − 6 = 2 (1 mark). Silicon: 2, 8, 4, so it shares 4 and has valency 4 (1 mark). Neon (2, 8) has a complete outer shell of 8 electrons, so it neither gains, loses nor shares electrons and its valency is zero (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
Suppose natural uranium in a sample is 1% ²³⁵U and 99% ²³⁸U (both Z = 92). (a) Find the neutrons in each isotope. (b) Find the average atomic mass of the sample. (c) After enrichment the sample is 5% ²³⁵U; find the new average. (d) State one use of ²³⁵U and say why the average mass changes so little.
Hint
Weight each mass by its fraction.
Answer
(a) ²³⁵U: 235 − 92 = 143 neutrons; ²³⁸U: 238 − 92 = 146 neutrons (1 mark). (b) 235 × 0.01 + 238 × 0.99 = 2.35 + 235.62 = 237.97 u (1 mark). (c) 235 × 0.05 + 238 × 0.95 = 11.75 + 226.1 = 237.85 u (1 mark). (d) ²³⁵U is a fuel in nuclear reactors; the two isotopes differ in mass by only 3 u out of about 237, so changing their proportions alters the average only slightly (1 mark).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the maximum number of electrons in the M shell?Show answer
18 (2n² with n = 3).
2.What is the electronic configuration of chlorine?Show answer
2, 8, 7.
3.How many neutrons are in ¹⁴₆C?Show answer
8.
4.What is the valency of an element with 2, 8, 3?Show answer
3 (aluminium).
5.Name an isotope used to treat cancer.Show answer
Cobalt-60.
6.What are isobars?Show answer
Atoms of different elements with the same mass number.

Atomic Number, Electronic Configuration, Valency and Isotopes: frequently asked questions

Why do isotopes have different physical properties?

They have different numbers of neutrons, so their masses differ, which affects properties such as density, melting point and boiling point.

Why does the atomic mass of chlorine come out as 35.5 u?

It is the average of ³⁵Cl (75%) and ³⁷Cl (25%); no individual atom has that mass.

Can the outermost shell of an atom have more than 8 electrons?

Not for the first 18 elements. The outermost shell holds at most 8 (or 2 for K).

Do all atoms have equal numbers of protons and neutrons?

No. Light atoms often do (carbon 6 and 6), but heavier atoms have more neutrons than protons (uranium 92 and 146).

Why does helium have valency zero although it has only 2 electrons?

Its K shell is complete with 2 electrons, so it is stable and does not gain, lose or share electrons.