Free Atomic Structure MCQs with Answers

1. The atomic number of an element is equal to the number of:

• A. Neutrons in the nucleus
• B. Protons in the nucleus
• C. Protons and neutrons combined
• D. Electron shells

Explanation: The atomic number equals the number of protons in the nucleus and determines the identity of the element. The mass number is the sum of protons and neutrons; isotopes of an element share an atomic number but differ in mass number.

2. The principal quantum number n of an electron determines its

• A. shape of orbital
• B. main energy level and distance from the nucleus
• C. orientation of the orbital in space
• D. direction of spin

Explanation: The principal quantum number sets the shell, and therefore the main energy level and the average distance of the electron from the nucleus. Orbital shape is given by the azimuthal quantum number l, orientation by the magnetic quantum number m, and spin by s, which takes the values plus or minus one half.

3. How many orbitals are present in the d subshell?

• A. Three
• B. Five
• C. Seven
• D. Nine

Explanation: For the d subshell l equals 2, so m runs from minus 2 to plus 2, giving five values and therefore five orbitals. With two electrons per orbital the d subshell holds a maximum of ten electrons, which is exactly the width of the d block in the periodic table. The p subshell has three orbitals and the f subshell has seven.

4. Which set of quantum numbers is not possible?

• A. n = 2, l = 1, m = 0
• B. n = 3, l = 2, m = -2
• C. n = 2, l = 2, m = 1
• D. n = 4, l = 0, m = 0

Explanation: The azimuthal quantum number l is restricted to values from 0 up to n minus 1, so with n equal to 2 the only allowed values of l are 0 and 1. A set with n equal to 2 and l equal to 2 would describe a 2d orbital, which does not exist. Every other set listed satisfies both that rule and the requirement that m lies between minus l and plus l.

5. The maximum number of electrons that can occupy the shell with n = 3 is

• A. 8
• B. 10
• C. 18
• D. 32

Explanation: The capacity of a shell is 2n squared, so for n equal to 3 that is 2 times 9, which is 18. Those 18 are distributed as 2 in the 3s, 6 in the 3p and 10 in the 3d subshells. The answer 8 is the common trap, taken from the octet rule rather than from the shell capacity.

6. The shape of a p orbital is

• A. spherical
• B. dumbbell shaped
• C. doughnut shaped
• D. tetrahedral

Explanation: A p orbital has two lobes on opposite sides of the nucleus with a node at the nucleus itself, giving the familiar dumbbell shape, and the three p orbitals lie along the x, y and z axes. Spherical shape belongs to the s orbital. Tetrahedral describes a molecular geometry rather than the shape of a single orbital.

7. According to Hund's rule, electrons entering a set of degenerate orbitals will

• A. pair up in the first orbital before entering the next
• B. occupy separate orbitals with parallel spins before pairing
• C. occupy the orbital of highest energy first
• D. always pair with opposite spins immediately

Explanation: Electrons repel one another, so they spread across orbitals of equal energy singly and with parallel spins before any orbital takes a second electron. This is why nitrogen, with three 2p electrons, has three unpaired electrons rather than one pair and one single. Filling lowest energy orbitals first is the Aufbau principle, a separate rule.

8. The correct electronic configuration of chromium (Z = 24) is

• A. [Ar] 3d4 4s2
• B. [Ar] 3d5 4s1
• C. [Ar] 3d6 4s0
• D. [Ar] 3d3 4s3

Explanation: Chromium promotes one 4s electron into the 3d subshell because a half filled d5 configuration alongside a half filled 4s is more stable than d4 with a full 4s. Copper does the same thing to reach a full d10. The 4s subshell can never hold three electrons, which rules out the last option outright.

9. Planck's quantum theory states that energy is emitted or absorbed

• A. continuously in any amount
• B. in discrete packets whose energy is proportional to frequency
• C. only as heat
• D. in packets whose energy is proportional to wavelength

Explanation: Planck proposed that radiation is emitted in quanta of energy E equal to h times the frequency, where h is Planck's constant. Since frequency and wavelength are inversely related, energy is inversely proportional to wavelength, which is why the last option is wrong rather than merely imprecise. The quantum idea was what resolved the black body radiation problem that classical physics could not.

10. The lines of the Balmer series in the hydrogen spectrum are produced when an electron falls to which energy level?

• A. n = 1
• B. n = 2
• C. n = 3
• D. n = 4

Explanation: The Balmer series arises from transitions ending at n equal to 2 and it is the only hydrogen series that falls in the visible region, which is why it was discovered first. Transitions ending at n equal to 1 give the Lyman series in the ultraviolet, and those ending at n equal to 3 give the Paschen series in the infrared.