Atomic Structure – NDA Chemistry PYQs

Solve NDA Chemistry previous-year questions on Atomic Structure with detailed solutions and explanations.

Chapter-wise PYQs • Concept-based explanations • Exam insights

NDA 2026-I

Q. 1. If x is the number of electrons and y is the number of nucleons present in ¹⁶₃₂S²⁻, then which one of the following is correct?

(a) x = 14 and y = 32

(b) x = 16 and y = 30

(d) x = 18 and y = 30

Answer: (c) x = 18 and y = 32

Explanation: S has atomic number 16, mass number 32. S²⁻ has gained 2 electrons: electrons = 16+2 = 18. Nucleons = protons + neutrons = mass number = 32 (charge does not affect nucleon count). So x = 18, y = 32. Concept Tested: Electron and nucleon count in anions – sulphide ion

NDA 2025-I

Q. 2. Which one of the following statements is correct?

(a) A neutron is formed by combination of an electron and a proton. Therefore, it is neutral

(b) The mass of an electron is about 1/2000 times that of a proton

(d) J. J. Thomson proposed that the nucleus of an atom contains only neutrons

Answer: (b) The mass of an electron is about 1/2000 times that of a proton

Explanation: The mass of an electron is approximately 1/1836 times the mass of a proton – often approximated as 1/2000 in school-level texts, making (b) the closest correct statement. (a) is incorrect – a neutron is a fundamental particle, not a proton-electron combination. (c) is incorrect – cobalt-60 is used in cancer treatment, not goitre treatment (iodine isotopes treat goitre). (d) is incorrect – J.J. Thomson proposed the plum-pudding model, not a neutron-only nucleus. Concept Tested: Properties of subatomic particles – mass of electron relative to proton

Q. 3. Which one of the following particles in the nucleus of an atom was discovered by J. Chadwick?

(a) Electron

(b) Proton

(d) Neutron

Answer: (d) Neutron

Explanation: James Chadwick discovered the neutron in 1932. The electron was discovered by J.J. Thomson. The proton was associated with Rutherford and Goldstein. The positron was discovered by Carl Anderson. Concept Tested: Discovery of the neutron – J. Chadwick

Q. 4. Which one of the following findings is NOT observed in Rutherford’s Nuclear Model of Atom?

(a) There is a neutral centre in an atom called nucleus

(b) Nearly all the mass of an atom resides in the nucleus

(d) The size of a nucleus is very small as compared to the size of atom

Answer: (a) There is a neutral centre in an atom called nucleus

Explanation: Rutherford’s model proposed a positively charged nucleus, not a neutral one. The nucleus contains protons and is positively charged. Options (b), (c) and (d) are all findings from Rutherford’s model. (a) incorrectly describes the nucleus as neutral – it is positively charged.
Concept Tested: Rutherford’s nuclear model – characteristics of the nucleus

NDA 2024-II

Q. 5. What is the maximum number of electrons that can occupy the first shell of an atom?

(a) 2

(b) 4

(d) 16

Answer: (a) 2

Explanation: The first shell (K-shell, n=1) can hold a maximum of 2n² = 2×1² = 2 electrons.
Concept Tested: Electron capacity of the first shell

Q. 6. For Aluminium (Al) (atomic number: 13, mass number: 27), which one among the following statements is NOT correct?

(a) Number of electrons present in Al is 13

(b) Number of protons present in Al is 13

(d) The valency of Al is 2

Answer: (d) The valency of Al is 2

Explanation: Al has atomic number 13, so electrons = 13 (a correct), protons = 13 (b correct). Neutrons = 27−13 = 14 (c correct). Aluminium has electronic configuration 2,8,3 – it has 3 valence electrons and a valency of 3, not 2. So (d) is NOT correct.
Concept Tested: Properties of aluminium – valency and subatomic particles

NDA 2023-II

Q. 7. As proposed by Dalton, which of the following symbol represents phosphorus ?

(a) [Circle with P]

(b) [Circle with L]

(d) [Circle with three-spoke]

Answer: [Circle with P]

Explanation: In Dalton’s atomic notation, phosphorus was represented by a circle with a cross/plus sign inside.
Concept Tested: Dalton’s atomic symbols – phosphorus

NDA 2022-I

Q. 8. The mass number of argon is 40. Which one of the following statements is correct?

(a) The number of protons in argon is 22.

(b) The number of neutrons in argon is 18.

(d) The sum of numbers of protons and electrons in argon is 40.

Answer: (c) The number of electrons in argon is 18.

Explanation: Argon has atomic number 18. In a neutral atom, electrons = protons = 18. (a) is wrong – protons = 18, not 22. (b) is wrong – neutrons = 40−18 = 22, not 18. (c) is correct – electrons = 18. (d) is wrong – protons + electrons = 18+18 = 36, not 40 (the 40 is the mass number = protons + neutrons).
Concept Tested: Subatomic particle counts from atomic number and mass number – argon

Q. 9. For an element with atomic number 35, which one of the following will be the correct number of electrons in its valence shell based on Bohr’s model of an atom?

(a) 1

(b) 3

(d) 7

Answer: (d) 7

Explanation: Atomic number 35 is Bromine (Br). Electronic configuration: 2, 8, 18, 7. The outermost (valence) shell has 7 electrons. Bromine is a halogen, like chlorine, with 7 valence electrons.
Concept Tested: Valence electrons from electronic configuration – Bromine

NDA 2021-II

Q. 10. Rutherford’s alpha-particle (α) scattering experiment was responsible for the discovery of which one of the following?

(a) Electron

(b) Proton

(d) Neutron

Answer: (c) Atomic Nucleus

Explanation: Rutherford’s gold foil experiment showed that most alpha particles passed through, but some deflected at large angles, indicating a small dense positive centre – the atomic nucleus.
Concept Tested: Rutherford’s alpha-scattering experiment – nucleus discovery

Q. 11. What is the maximum number of electrons in the M-Shell?

(a) 6

(b) 8

(d) 32

Answer: (c) 18

Explanation: The formula for maximum electrons in a shell is 2n², where n is the shell number.
M-shell is n=3: 2×3² = 2×9 = 18. The M-shell holds a maximum of 18 electrons.
Concept Tested: Electron capacity of atomic shells – M-shell

★ JOVIK Exam Insight: Shell capacities: K(n=1)=2, L(n=2)=8, M(n=3)=18, N(n=4)=32. Formula: 2n².

Q. 12. Chlorine occurs in nature in two isotopic forms of masses 35 u and 37 u in the ratio of 3 : 1 respectively. What is the average atomic mass of the Chlorine atom?

(a) 36·1 u

(b) 35·5 u

(d) 35·1 u

Answer: (b) 35·5 u

Explanation: Average = (35 × 3/4) + (37 × 1/4) = 26.25 + 9.25 = 35.5 u.
This matches the standard atomic weight of chlorine (35.5).
Concept Tested: Average atomic mass calculation – chlorine isotopes

★ JOVIK Exam Insight: Chlorine’s 35.5 u average atomic mass is a classic NDA numerical. The 3:1 ratio of Cl-35 to Cl-37 is the standard data given.

Q. 13. Which one of the following ions is not iso-electronic with F⁻?

(a) O²⁻

(b) Na⁺

(d) N⁻

Answer: (d) N⁻

Explanation: F⁻ has 9+1 = 10 electrons. O²⁻ has 8+2 = 10 electrons (isoelectronic). Na⁺ has 11−1 = 10 electrons (isoelectronic). Ne has 10 electrons (isoelectronic). N⁻ has 7+1 = 8 electrons – NOT isoelectronic with F⁻. So N⁻ is the answer.
Concept Tested: Isoelectronic species – identifying the non-member

NDA 2021-I

Q. 14. Which one of the following conclusions could not be derived from Rutherford’s α-particle scattering experiment?

(a) Most of the space in the atom is empty.

(b) The radius of the atom is about 10⁵ times the radius of the nucleus.

(d) Nearly all the mass of the atom resides in the nucleus.

Answer: (c) Electrons move in a circular path of fixed energy called orbits.

Explanation: Rutherford’s scattering experiment established: (a) most atomic space is empty (most particles passed through), (b) the nucleus is very small compared to the atom, and (d) most mass is concentrated in the nucleus. It did NOT establish circular electron orbits of fixed energy – that concept came from Bohr’s model published later. Concept Tested: Limitations of Rutherford’s atomic model – Bohr’s contribution

★ JOVIK Exam Insight: A common NDA trap: Rutherford’s model did not explain electron orbits. Fixed energy orbits are Bohr’s contribution. Rutherford showed the nucleus exists.

NDA 2020

Q. 15. Which one of the following is the most fundamental characteristic of an element?

(a) Melting point

(b) Atomic number

(d) Atomic weight

Answer: (b) Atomic number

Explanation: The atomic number (number of protons) uniquely identifies every element. No two elements have the same atomic number. Melting point, colour and atomic weight can vary with isotopes or conditions. The atomic number is the defining property of an element. Concept Tested: Atomic number as the fundamental identifier of an element

Q. 16. Neutrons were discovered by

(a) James Chadwick

(b) Ernest Rutherford

(d) John Dalton

Answer: (a) James Chadwick

Explanation: James Chadwick discovered the neutron in 1932 by bombarding beryllium with alpha particles. Rutherford discovered the nucleus. J.J. Thomson discovered the electron. John Dalton proposed the atomic theory. Concept Tested: Discovery of the neutron – James Chadwick

Q. 17. Atomic mass of an element is equal to the sum of number of

(a) electrons and protons only

(b) protons and neutrons only

(d) electrons, protons and neutrons

Answer: (b) protons and neutrons only

Explanation: Atomic mass (mass number) = protons + neutrons. Electrons have negligible mass compared to protons and neutrons and are not included in the mass number calculation. Concept Tested: Atomic mass – composition from protons and neutrons

NDA 2019-I

Q. 18. In which of the following pairs are the ions isoelectronic?

(a) Mg²⁺, Ar

(b) Na⁺, O²⁻

(d) K⁺, Ne

Answer: (b) Na⁺, O²⁻

Explanation: Na⁺: Na has 11 electrons; Na⁺ loses 1 → 10 electrons. O²⁻: O has 8 electrons; O²⁻ gains 2 → 10 electrons. Both have 10 electrons – isoelectronic. Check others: Mg²⁺ = 10e, Ar = 18e (not equal). Al³⁺ = 10e, Cl⁻ = 18e (not equal). K⁺ = 18e, Ne = 10e (not equal). Concept Tested: Isoelectronic ion pairs – electron counting

NDA 2018-II

Q. 19. A sample of oxygen contains two isotopes of oxygen with masses 16 u and 18 u respectively. The proportion of these isotopes in the sample is 3 : 1. What will be the average atomic mass of oxygen in this sample?

(a) 17·5 u

(b) 17 u

(d) 16·5 u

Answer: (d) 16·5 u

Explanation: Average atomic mass = (mass₁ × fraction₁) + (mass₂ × fraction₂). Ratio 3:1 means fractions are 3/4 and 1/4. Average = (16 × 3/4) + (18 × 1/4) = 12 + 4.5 = 16.5 u. Concept Tested: Calculation of average atomic mass from isotopic proportions

★ JOVIK Exam Insight: Show your working in NDA numericals. Average atomic mass = Σ(mass × fraction). Always convert ratio to fraction first.

NDA 2017-II

Q. 20. The ionization energy of hydrogen atom in the ground state is

(a) 13·6 MeV

(b) 13·6 eV

(d) Zero

Answer: (b) 13·6 eV

Explanation: The ionization energy of the hydrogen atom in its ground state (n=1) is 13.6 eV (electron-volts). MeV is a unit used for nuclear energies – much larger scale. Joule is correct SI but the numerical value 13.6 J would be enormously larger than the actual value. Concept Tested: Ionization energy of hydrogen – value and units

Q. 21. The species that has the same number of electrons as ³⁵₁₇Cl is

(a) ³²₁₆S

(b) ³⁴₁₆S⁺

(d) ³⁵₁₆S²⁻

Answer: (c) ⁴⁰₁₈Ar⁺

Explanation: ³⁵₁₇Cl is a neutral chlorine atom with 17 electrons. Check each option: (a) ³²S: atomic number 16 → 16 electrons (not 17). (b) ³⁴S⁺: 16−1 = 15 electrons (not 17). (c) ⁴⁰Ar⁺: atomic number 18 minus 1 lost electron = 17 electrons – isoelectronic with Cl. (d) ³⁵S²⁻: 16+2 = 18 electrons (not 17). The correct answer is (c). Concept Tested: Isoelectronic species – electron counting in ions

NDA 2017-I

Q. 22. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of

(a) Electron

(b) Proton

(d) Helium

Answer: (c) Nucleus

Explanation: Rutherford fired alpha particles at a thin gold foil. Most passed through, but some were deflected at large angles and a few bounced straight back. This showed that the atom has a small, dense, positively charged centre – the nucleus. This was not the discovery of the electron (Thomson) or proton (Goldstein/Rutherford later). Concept Tested: Rutherford’s alpha-scattering experiment – discovery of the nucleus

★ JOVIK Exam Insight: NDA asks this repeatedly. Rutherford = nucleus. Thomson = electron. Chadwick = neutron. Goldstein = proton. Do not confuse them.

NDA 2016-II

Q. 23. There are six electrons, six protons and six neutrons in an atom of an element. What is the atomic number of the element?

(a) 6

(b) 12

(d) 24

Answer: (a) 6

Explanation: Atomic number = number of protons = 6. The number of electrons equals protons in a neutral atom, which is consistent here. Mass number = protons + neutrons = 6 + 6 = 12. The element is Carbon (C). Concept Tested: Atomic number from subatomic particle count

NDA 2016-I

Q. 24. An atom of carbon has 6 protons. Its mass number is 12. How many neutrons are present in an atom of carbon ?

(a) 12

(b) 6

(d) 14

Answer: (b) 6

Explanation: Neutrons = mass number − number of protons = 12 − 6 = 6. Concept Tested: Calculation of neutrons – carbon atom

NDA 2015-II

Q. 25. Which of the following are the properties of an electron ?

1. Electron is a constituent of cathode ray

2. Electron is a negatively charged particle

3. The mass of the electron is equal to the mass of the proton

4. Electron is deflected by the electric field but not by magnetic field

Select the correct answer using the code given below:

(a) 1 and 2 only

(b) 1, 2 and 3

(d) 1 and 4

Answer: (a) 1 and 2 only

Explanation: Statement 1 is correct: cathode rays are streams of electrons. Statement 2 is correct: electrons carry negative charge. Statement 3 is wrong: the mass of an electron is approximately 1/1836 times the mass of a proton – much smaller. Statement 4 is wrong: electrons are deflected by both electric and magnetic fields. Concept Tested: Properties of the electron

Q. 26. Which one of the following statements is correct?

(a) Rutherford’s alpha-particle scattering experiment led to the discovery of electron

(b) J J Thomson suggested that the nucleus of an atom contains protons

(d) The mass number of an atom is equal to the number of electrons in its shells

Answer: (c) The atomic number of an element is the same as the number of protons in the nucleus of its atom

Explanation: The atomic number equals the number of protons. (a) is wrong – Rutherford’s experiment led to discovery of the nucleus, not the electron. (b) is wrong – J.J. Thomson did not propose the nuclear proton model. (d) is wrong – mass number = protons + neutrons, not electrons. Concept Tested: Atomic number – definition and identification

NDA 2014-II

Q. 27. A monatomic species that has 18 electrons and a net charge of 2⁻ has

(a) the same number of electrons as a neutral argon atom

(b) more protons than electrons

(d) 20 protons

Answer: (a) the same number of electrons as a neutral argon atom

Explanation: A neutral argon atom has atomic number 18, so it has 18 electrons. This species also has 18 electrons. So option (a) is correct. Since the charge is 2⁻, it has gained 2 electrons. Protons = 18 − 2 = 16 (which is sulphur, S²⁻). Protons (16) < electrons (18), so (b) is wrong. S²⁻ has a full outer shell, no unpaired electrons, so (c) is wrong. Protons = 16, not 20, so (d) is wrong. Concept Tested: Electron count of ions and isoelectronic relationship with noble gases

Q. 28. Which one among the following contains the most neutrons? (Note: Atomic numbers – Fe: 26, Cu: 29, Zn: 30)

(a) ⁵⁹₂₆Fe

(b) ⁶¹₂₉Cu

(d) ⁶⁰₃₀Zn²⁺

Answer: (a) ⁵⁹₂₆Fe

Explanation: Neutrons = mass number − atomic number. (a) ⁵⁹Fe: 59−26 = 33 neutrons. (b) ⁶¹Cu: 61−29 = 32 neutrons. (c) ⁶¹Zn: 61−30 = 31 neutrons. (d) ⁶⁰Zn²⁺: 60−30 = 30 neutrons (ionic charge does not affect neutrons). ⁵⁹Fe has the most neutrons (33). Concept Tested: Calculation of neutrons – comparison across isotopes

NDA 2014-I

Q. 29. The number of valence electrons in the O²⁻ ion is

(a) 4

(b) 6

(d) 10

Answer: (c) 8

Explanation: Oxygen has atomic number 8 with electron configuration 2,6. The O²⁻ ion has gained 2 extra electrons, giving 8 total electrons: configuration 2,6+2 = 2,8. The valence shell (outermost) now holds 8 electrons. So the answer is 8. Concept Tested: Electron configuration of anions – O²⁻

Q. 30. Which one of the following is the correct electronic configuration of chlorine?

(a) 2, 7, 8

(b) 2, 8, 7

(d) 7, 8, 2

Answer: (b) 2, 8, 7

Explanation: Chlorine has atomic number 17. Electrons fill shells as: shell 1 = 2, shell 2 = 8, shell 3 = 7. Total = 17. The correct configuration is 2, 8, 7. Concept Tested: Electronic configuration of chlorine

Q. 31. Which of the following pairs represents isoelectronic ions?

(a) Na⁺, K⁺

(b) K⁺, Mg²⁺

(d) Ca²⁺, S²⁻

Answer: (d) Ca²⁺, S²⁻

Explanation: Ca has atomic number 20; Ca²⁺ loses 2 electrons → 18 electrons. S has atomic number 16; S²⁻ gains 2 electrons → 18 electrons. Both have 18 electrons – they are isoelectronic. Na⁺ = 10, K⁺ = 18 (not equal). K⁺ = 18, Mg²⁺ = 10 (not equal). Mg²⁺ = 10, Ca²⁺ = 18 (not equal). Concept Tested: Isoelectronic ions – identification using electron count

Q. 32. The mass number of an atom is determined by

(a) adding the number of neutrons and number of electrons

(b) adding the number of protons and number of electrons

(d) adding the number of neutrons and number of protons

Answer: (d) adding the number of neutrons and number of protons

Explanation: Mass number = number of protons + number of neutrons. Electrons have negligible mass and are not included in the mass number. The number of protons alone gives the atomic number. Concept Tested: Definition of mass number

NDA 2013-I

Q. 33. Two atoms are said to be isotopes if

(a) they have the same atomic number but different mass numbers

(b) they have the same number of neutrons but different mass numbers

(d) they have the same number of neutrons but different modes of radioactive decay

Answer: (a) they have the same atomic number but different mass numbers

Explanation: Isotopes are atoms of the same element – they have the same atomic number (same number of protons) but different mass numbers due to different numbers of neutrons. For example, ¹H and ²H (deuterium) are isotopes of hydrogen. Concept Tested: Definition of isotopes

NDA 2012-II

Q. 34. Match List I with List II and select the correct answer using the code given below the Lists:

List I (Scientist)

A. Goldstein

B. Chadwick

C. JJ Thomson

D. John Dalton

List II (Discovery)

1. Atomic theory

2. Proton

3. Neutron

4. Electron

Code:

A B C D

(a) 2 3 4 1

(b) 2 4 3 1

(d) 1 3 4 2

(a) 2 3 4 1

(b) 2 4 3 1

(d) 1 3 4 2

Answer: (a) A-2, B-3, C-4, D-1

Explanation: Goldstein discovered the proton (positive rays). Chadwick discovered the neutron. J.J. Thomson discovered the electron. John Dalton proposed the atomic theory. So A→2, B→3, C→4, D→1 matches code (a). Concept Tested: Discoverers of subatomic particles and atomic theory

★ JOVIK Exam Insight: NDA regularly tests scientist-discovery matching. The four key pairs: Dalton = atomic theory, Thomson = electron, Rutherford = nucleus, Chadwick = neutron, Goldstein = proton.

NDA 2012-I

Q. 35. Which one among the following is correct regarding ²⁰Ne, ²³Na⁺, ¹⁹F⁻ and ²⁴Mg²⁺?

(a) They are isomers of each other

(b) They are isotopes of each other

(d) All of the above

Answer: (c) They are isoelectronic with each other

Explanation: Isoelectronic species have the same number of electrons. Ne has 10 electrons. Na⁺ has lost 1 from 11 = 10 electrons. F⁻ has gained 1 to 9+1 = 10 electrons. Mg²⁺ has lost 2 from 12 = 10 electrons. All four have 10 electrons – they are isoelectronic. Concept Tested: Isoelectronic species – identification

★ JOVIK Exam Insight: NDA regularly tests isoelectronic species. The 10-electron set (Ne, Na⁺, F⁻, Mg²⁺, O²⁻, Al³⁺) is the most frequently tested group.

Q. 36. Which of the following pairs is/are correctly matched?

1. Isotopes : Atoms with same atomic number but different atomic mass

2. Isobars : Atoms with same number of neutrons but different atomic number

3. Isotones : Atoms with same mass number but different atomic number

Select the correct answer using the code given below:

(a) 1, 2 and 3

(b) 1 only

(d) 2 only

Answer: (b) 1 only

Explanation: Statement 1 is correct: isotopes have the same atomic number but different mass numbers. Statement 2 is wrong: isobars have the same mass number but different atomic numbers. Statement 3 is wrong: isotones have the same number of neutrons but different atomic numbers (not the same mass number). Concept Tested: Definitions of isotopes, isobars and isotones

★ JOVIK Exam Insight: Isobars and isotones are commonly confused in NDA. Isobars = same mass number, different atomic number. Isotones = same neutron number, different atomic number.

Q. 37. Which one among the following transitions of electron of hydrogen atom emits radiation of the shortest wavelength?

(a) n = 2 to n = 1

(b) n = 3 to n = 2

(d) n = 5 to n = 4

Answer: (a) n = 2 to n = 1

Explanation: Shorter wavelength corresponds to higher energy emission. The energy of emitted radiation is proportional to the difference in energy levels. The gap between n=2 and n=1 is the largest among these options because lower levels have greater energy differences. So the transition from n=2 to n=1 releases the most energy and produces the shortest wavelength. Concept Tested: Electron transitions in hydrogen – energy and wavelength

Q. 38. Which one among the following is the most appropriate statement with respect to the atomic weight of an element?

(a) The atomic weight of an element is the sum total of the number of protons and neutrons present in the atom of the element

(b) Unlike mass number, the atomic weight of an element can be a fraction

(d) The atomic weight of all the atoms in an element is the same

Answer: (b) Unlike mass number, the atomic weight of an element can be a fraction

Explanation: Mass number is always a whole number (count of protons + neutrons in one specific isotope). Atomic weight (relative atomic mass) is the weighted average of all naturally occurring isotopes – so it can be a decimal, for example chlorine is 35.5. Option (a) describes mass number, not atomic weight. Option (c) and (d) are incorrect for the same reason. Concept Tested: Distinction between mass number and atomic weight

NDA 2011-II

Q. 39. The number of protons in a negatively charged atom (anion) is

(a) more than the atomic number of the element

(b) less than the atomic number of the element

(d) less than the number of electrons in the atom

Answer: (d) less than the number of electrons in the atom

Explanation: An anion is formed when an atom gains extra electrons. The number of protons stays fixed and equals the atomic number. Because extra electrons are added, electrons > protons. So the proton count is less than the electron count. Concept Tested: Anion formation – relationship between protons and electrons

NDA 2010-I

Q. 40. The number of neutrons in ₁₃Al²⁷ is

(a) 40

(b) 27

(d) 13

Answer: (c) 14

Explanation: Aluminium-27 has atomic number 13 (protons = 13) and mass number 27. Neutrons = mass number − atomic number = 27 − 13 = 14. Concept Tested: Calculation of neutrons from atomic number and mass number

★ JOVIK Exam Insight: A quick NDA formula: neutrons = mass number − atomic number. This is one of the most straightforward calculation questions in atomic structure.