40 practice MCQs on sub-atomic particles, atomic models, quantum numbers, orbitals and electronic configuration. Every question has the correct answer and an explanation.
Start free — 20 questionsAtomic Structure is part of the MDCAT Chemistry syllabus. This page collects Atomic Structure MCQs in the style of past MDCAT papers — below are 12 of them with answers and explanations, and the full set of 40 questions is in the HighYield MDCAT QBank.
This chapter covers: sub-atomic particles, atomic models, quantum numbers, orbitals and electronic configuration.
Correct answer: C. 20.18 amu
The average atomic mass of neon is calculated by summing each isotope's mass multiplied by its fractional abundance: (20×0.9092)+(21×0.0026)+(22×0.0882) ≈ 20.18 amu.
Correct answer: B. Increase
As the principal quantum number n increases, s-orbitals retain their spherical shape but increase in size (radius), since the electron is found farther from the nucleus on average.
Correct answer: D. Na⁺, F⁻
Na+ (11 protons, 10 electrons) and F- (9 protons, 10 electrons) both have 10 electrons, giving them the same electronic configuration as neutral neon.
Correct answer: C. 2n²
The maximum number of electrons that can occupy a given principal energy level n is given by the formula 2n².
Correct answer: D. Increase in number of shells
Across a period, atomic radius decreases mainly because the increasing number of protons in the nucleus exerts a stronger pull on the electrons in the same principal shell.
Correct answer: D. Nickel
Iron, in the 3d series, has the electronic configuration [Ar] 3d6 4s2, though this question's answer key indicates zinc among these options.
Correct answer: D. 26
The number of unpaired electrons in an element's ground state corresponds to the element's atomic number, which is the number of protons in its nucleus. In this case, an element with four unpaired electrons would have an atomic number of 26. Iron has an atomic number of 26; it has four unpaired electrons in its ground state electronic configuration:1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
Correct answer: A. 2.19 x 106 ms-1
According to Bohr's model of the hydrogen atom, the speed of an electron in the first Bohr orbit can be calculated using the formula: v = Ze² / 2εοh Where, Atomic number = Z = 1 Charge = e = 1.602 × 10-19 C Permittivity of free space = εο =8.854×10-12 C²/Nm² Planck's constant = h = 6.626 × 10-34 J's Putting the values in the above formula, we get: v = 1 × (1.602 × 10-19)² / 2 × (8.854×10-12) × (6.626 × 10-34) v = 2.19 × 106 ms-1
Correct answer: B. Alkali metals
The correct answer is indeed B, alkali metals. Among the options provided, the alkali metals, specifically cesium (Cs), have the highest atomic radius in their respective periods. Cesium has the largest atomic radius among all the alkali metals.
Correct answer: C. 1s2, 2s2, 2p2
This option is correct. This is the correct configuration for carbon.
Correct answer: B. S
S will have the smallest atomic radius because it lies at the end among the mentioned elements. Across the period, the atomic radius decreases.
Correct answer: B. Balmer series
The Balmer series is a series of spectral lines in the hydrogen atom spectrum that lies in the visible region of the electromagnetic spectrum, with wavelengths between 400-700 nanometers (nm). This series corresponds to electron transitions from higher energy levels (n ≥ 3) to the second energy level (n = 2).
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