Structure of Atom NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Structure of Atom question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Structure of Atom previous year questions
34 questions. Tap an option to check your answer.
Q1NEET 2021
A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light, c = 3.0 × 10⁸ ms⁻¹]
Not quite — the answer is A.
λ = c/ν = (3.0 × 10⁸)/(1368 × 10³) = 219.30 m. 2192 m results from a decimal slip; the value 219.2 m is a rounding error.
Q2NEET 2020 · Q124
The number of protons, neutrons and electrons in ¹⁷⁵₇₁Lu, respectively, are :
Not quite — the answer is A.
Atomic number 71 gives 71 protons and, for the neutral atom, 71 electrons; neutrons = 175 − 71 = 104. Using the mass number 175 as the proton count is the common slip.
Q3NEET 2019 · Q49
4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is:
Not quite — the answer is A.
(n + l) values: 5f = 8, 6p = 7, 5p = 6, 4d = 6; for equal n + l the higher n is higher in energy, so 5p > 4d. Order: 5f > 6p > 5p > 4d. Putting 4d above 5p ignores the tie-break on n.
Q4NEET 2019 · Q87
Which of the following series of transitions in the spectrum of hydrogen atom falls in visible region?
Not quite — the answer is B.
Balmer lines (transitions ending at n = 2) lie in the visible region. Lyman lines (ending at n = 1) are ultraviolet; Paschen and Brackett are infrared.
Q5NEET 2018 · Q90
Which one is a wrong statement?
Not quite — the answer is A.
By Hund's rule, the three unpaired 2p electrons of nitrogen must have parallel spins; the box diagram with one 2p electron spin-down is wrong. An s electron has l = 0 so zero orbital angular momentum, and d_z² corresponds to m = 0.
Q6NEET 2017 · Q10
Which one is the wrong statement?
Not quite — the answer is D.
In hydrogen-like (one-electron) species the energy depends only on n, so 2s and 2p are degenerate; the 2s-below-2p order holds only for multi-electron atoms. The other statements are correct: λ = h/mv, ΔE·Δt ≥ h/4π, and half/fully filled subshells are extra stable.
Q7NEET 2017 (Gujarati paper) · Q27
The total number of orbitals present for principle quantum number, n = 4 is:
Not quite — the answer is C.
The number of orbitals in a shell is n², so n = 4 has 16 orbitals (one 4s, three 4p, five 4d, seven 4f). 32 would be the number of electrons, 2n².
Q8NEET 2016 Phase 2 · Q70
Which of the following pairs of d-orbitals will have electron density along the axes?
Not quite — the answer is C.
The lobes of d_x²−y² lie along the x and y axes and the main lobes of d_z² lie along the z axis. d_xy, d_yz and d_xz have their lobes between the axes, so any pair containing them is wrong.
Q9NEET 2016 Phase 1 · Q61
Two electrons occupying the same orbital are distinguished by:
Not quite — the answer is D.
Electrons in the same orbital share n, l and m; by the Pauli exclusion principle they must differ in spin quantum number (+½ and −½). The other three quantum numbers define the orbital itself and are therefore identical.
Q10AIPMT 2015
The angular momentum of electron in 'd' orbital is equal to:
Not quite — the answer is B.
Orbital angular momentum = √(l(l + 1)) ħ. For a d orbital l = 2, giving √6 ħ. √2 ħ is the value for a p orbital.
Q11AIPMT 2015
The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following?
Not quite — the answer is C.
Fe²⁺ is [Ar]3d⁶, so it has 6 d-electrons. Ne has 6 p-electrons, Mg has 6 s-electrons and Fe has 6 d-electrons, but Cl (1s²2s²2p⁶3s²3p⁵) has 11 p-electrons.
Q12AIPMT 2014
Calculate the energy in joule corresponding to light of wavelength 45 nm: (Planck's constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 10⁸ ms⁻¹)
Not quite — the answer is C.
E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(45 × 10⁻⁹) = 4.42 × 10⁻¹⁸ J.
Q13AIPMT 2014
Be²⁺ is isoelectronic with which of the following ions?
Not quite — the answer is A.
Be²⁺ has 4 − 2 = 2 electrons, the same as Li⁺ (3 − 1 = 2). Na⁺ and Mg²⁺ have 10 electrons and H⁺ has none.
Q14AIPMT 2014
What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l = 1, mₗ = 0
Not quite — the answer is D.
n = 3 and l = 1 specify the 3p subshell, and mₗ = 0 picks out exactly one of its three orbitals. Three would be the answer if mₗ were not given.
Q15NEET 2013
The value of Planck's constant is 6.63 × 10⁻³⁴ J s. The speed of light is 3 × 10¹⁷ nm s⁻¹. Which value is closest to the wavelength in nanometre of a quantum of light with frequency of 6 × 10¹⁵ s⁻¹?
Not quite — the answer is C.
λ = c/ν = (3 × 10¹⁷ nm s⁻¹)/(6 × 10¹⁵ s⁻¹) = 50 nm. Planck's constant is not needed to find the wavelength.
Q16NEET 2013
What is the maximum number of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = −1.
Not quite — the answer is D.
n = 3, l = 1, m = −1 specifies one particular 3p orbital. One orbital holds at most two electrons with opposite spins. 6 is the capacity of the whole 3p subshell.
Q17NEET 2013
Based on equation E = −2.178 × 10⁻¹⁸ J (Z²/n²), certain conclusions are written. Which of them is not correct?
Not quite — the answer is D.
For n = 1 the energy is the most negative, which means the electron is most tightly bound in the smallest orbit, not more loosely bound, so that conclusion is wrong. The other statements correctly describe the negative sign, the growth of radius with n and use of the equation for energy changes.
Q18AIPMT 2012 Mains
The orbital angular momentum of a p-electron is given as :
Not quite — the answer is A.
Orbital angular momentum = √(l(l + 1))·h/2π. For a p-electron l = 1, giving √2·h/2π = h/(√2 π). √6·h/2π would be for a d-electron (l = 2).
Q19AIPMT 2012 Prelims
Maximum number of electrons in a subshell with l = 3 and n = 4 is:
Not quite — the answer is B.
l = 3 is an f subshell, with 2l + 1 = 7 orbitals, each holding 2 electrons: 14 electrons. The value n = 4 only fixes the shell (4f) and does not change the subshell capacity; 10 is the capacity of a d subshell (l = 2).
Q20AIPMT 2012 Prelims
The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is
Not quite — the answer is D.
Rb (Z = 37) is [Kr] 5s¹, so the valence electron has n = 5, l = 0 (s), m = 0 and s = +½. Sets with l = 1 describe a 5p electron, and n = 6 is the wrong shell.
Q21AIPMT 2011 Mains
According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
Not quite — the answer is A.
ΔE ∝ (1/n₁² − 1/n₂²): for 6 → 5 it is 1/25 − 1/36 = 0.0122, for 5 → 4 it is 1/16 − 1/25 = 0.0225, and the others are larger. So the 6 → 5 transition gives the least energetic photon.
Q22AIPMT 2011 Prelims
The total number of atomic orbitals in fourth energy level of an atom is
Not quite — the answer is C.
The number of orbitals in shell n is n². For n = 4 (4s, 4p, 4d, 4f: 1 + 3 + 5 + 7) this is 16 orbitals. 32 is the maximum number of electrons (2n²), not orbitals.
Q23AIPMT 2011 Prelims
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. λ₁ and λ₂ will be
Not quite — the answer is C.
E = hc/λ, so λ ∝ 1/E. E₂ is twice E₁, so λ₂ = λ₁/2, i.e. λ₁ = 2λ₂. The higher-energy radiation has the shorter wavelength, so λ₁ = λ₂/2 is the reverse.
Q24AIPMT 2011 Prelims
If n = 6, the correct sequence of filling of electrons will be
Not quite — the answer is B.
For n = 6 the Aufbau order is 6s → 4f → 5d → 6p (as in Cs, Ba, La/Ce onwards, Hf, then Tl), i.e. ns → (n−2)f → (n−1)d → np. Filling np before the inner d and f orbitals is the trap.
Q25AIPMT 2010 Mains · Q100
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be (h = 6.6 × 10⁻³⁴ J s)
Not quite — the answer is C.
λ = h/(mv) = 6.6 × 10⁻³⁴/(0.66 × 100) = 6.6 × 10⁻³⁴/66 = 1.0 × 10⁻³⁵ m. 6.6 × 10⁻³⁴ m comes from forgetting to divide by the momentum.
Q26AIPMT 2009 Prelims · Q161
The energy absorbed by each molecule (A₂) of a substance is 4.4 × 10⁻¹⁹ J and bond energy per molecule is 4.0 × 10⁻¹⁹ J. The kinetic energy of the molecule per atom will be:
Not quite — the answer is D.
Energy left after breaking the bond = 4.4 × 10⁻¹⁹ − 4.0 × 10⁻¹⁹ = 0.4 × 10⁻¹⁹ J per molecule. It is shared by the two atoms: 2.0 × 10⁻²⁰ J per atom. 4.0 × 10⁻²⁰ J forgets to divide between the two atoms.
Q27AIPMT 2009 Prelims · Q168
Maximum number of electrons in a subshell of an atom is determined by the following:
Not quite — the answer is D.
A subshell with quantum number l has (2l + 1) orbitals, each holding 2 electrons, so the maximum is 2(2l + 1) = 4l + 2. 2l + 1 counts orbitals, not electrons, and 2n² is for a whole shell.
Q28AIPMT 2009 Prelims · Q173
Which of the following is not a permissible arrangement of electrons in an atom?
Not quite — the answer is B.
For l = 2 the magnetic quantum number can only take values from −2 to +2, so m = −3 is not allowed. The other sets obey all the rules (l < n, |m| ≤ l, s = ±1/2).
Q29AIPMT 2008 Prelims · Q51
If uncertainty in position and momentum are equal, then uncertainty in velocity is
Not quite — the answer is B.
Δx·Δp = h/4π with Δx = Δp gives Δp² = h/4π, so Δp = (1/2)√(h/π). Then Δv = Δp/m = (1/2m)√(h/π). The forms without m are momenta, not velocities.
Q30AIPMT 2008 Prelims · Q56
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 × 10⁻¹⁸ g cm s⁻¹. The uncertainty in electron velocity is, (mass of an electron is 9 × 10⁻²⁸ g)
Not quite — the answer is B.
Δv = Δp/m = (1 × 10⁻¹⁸ g cm s⁻¹)/(9 × 10⁻²⁸ g) ≈ 1.1 × 10⁹ cm s⁻¹. Both quantities are already in CGS units, so no conversion is needed, and converting only one of them gives the wrong power of ten.
Q31AIPMT 2007 Prelims · Q90
Consider the following sets of quantum numbers (n, l, m, s): (i) 3, 0, 0, +1/2 (ii) 2, 2, 1, +1/2 (iii) 4, 3, −2, −1/2 (iv) 1, 0, −1, −1/2 (v) 3, 2, 3, +1/2 Which of the following sets of quantum number is not possible?
Not quite — the answer is B.
(ii) is impossible because l must be less than n (l = 2 with n = 2), (iv) because m must lie between −l and +l (m = −1 with l = 0) and (v) because m = 3 exceeds l = 2. (i) and (iii) are valid, so the impossible sets are (ii), (iv) and (v).
Q32AIPMT 2006 Prelims · Q60
Given : The mass of electron is 9.11 × 10⁻³¹ kg, Planck constant is 6.626 × 10⁻³⁴ Js, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is :
Not quite — the answer is A.
Heisenberg: Δx·mΔv ≥ h/4π, so Δv = 6.626 × 10⁻³⁴/(4 × 3.1416 × 9.11 × 10⁻³¹ × 1 × 10⁻¹¹) = 6.626 × 10⁻³⁴/1.145 × 10⁻⁴⁰ ≈ 5.79 × 10⁶ m s⁻¹. Remember 0.1 Å = 10⁻¹¹ m; using 10⁻¹⁰ m would change the power of ten.
Q33AIPMT 2006 Prelims · Q95
The orientation of an atomic orbital is governed by :
Not quite — the answer is C.
The magnetic quantum number mₗ fixes the orientation of an orbital in space, e.g. the three p orbitals along x, y and z. The azimuthal quantum number fixes the shape, the principal quantum number the size and energy, and the spin quantum number refers to the electron, not the orbital.
Q34AIPMT 2005 Prelims · Q92
The energy of second Bohr orbit of the hydrogen atom is −328 kJ mol⁻¹ ; hence the energy of fourth Bohr orbit would be :
Not quite — the answer is D.
Eₙ ∝ 1/n², so E₄ = E₂ × (2/4)² = −328/4 = −82 kJ mol⁻¹. Dividing by 8 (−41 kJ mol⁻¹) wrongly uses Eₙ ∝ 1/n³.
How often Structure of Atom came up
Questions from this chapter in each paper on this site.
- NEET 2021: 1
- NEET 2020: 1
- NEET 2019: 2
- NEET 2018: 1
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 1
- AIPMT 2015: 2
- AIPMT 2014: 3
- NEET 2013: 3
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 2
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 3
- AIPMT 2010 Mains: 1
- AIPMT 2009 Prelims: 3
- AIPMT 2008 Prelims: 2
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 2
- AIPMT 2005 Prelims: 1
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