Atomic Structure & Periodicity

An atom of an element is represented as ⁵⁶Fe³⁺ (atomic number 26). How many protons, neutrons and electrons does this ion contain?

• A26 protons, 30 neutrons, 23 electrons✓
• B26 protons, 26 neutrons, 26 electrons
• C26 protons, 30 neutrons, 26 electrons
• D30 protons, 26 neutrons, 23 electrons

Answer: Protons = atomic number = 26. Neutrons = mass number − protons = 56 − 26 = 30. A 3+ ion has lost 3 electrons, so electrons = 26 − 3 = 23.

Chlorine has two isotopes, ³⁵Cl and ³⁷Cl, with relative isotopic masses of 34.97 and 36.97. The relative atomic mass of chlorine is 35.5. What is the approximate percentage abundance of the ³⁵Cl isotope?

• A25%
• B50%
• C75%✓
• D90%

Answer: Let abundance of ³⁵Cl = x%, so ³⁷Cl = (100−x)%. Ar = [34.97x + 36.97(100−x)]/100 = 35.5. Solving: 34.97x + 3697 − 36.97x = 3550, so −2.00x = −147, x ≈ 73.5% ≈ 75%.

In a time-of-flight (TOF) mass spectrometer, which statement correctly describes why ions of the same charge but different mass reach the detector at different times?

• AHeavier ions are accelerated to a higher velocity by the electric field
• BAll ions gain the same kinetic energy, so heavier ions travel more slowly through the drift region✓
• CHeavier ions are deflected more strongly by the magnetic field
• DLighter ions gain more kinetic energy and therefore arrive last

Answer: During acceleration all ions of equal charge gain the same kinetic energy (KE = ½mv²). Because KE is fixed, a larger mass m gives a smaller velocity v, so heavier ions take longer to cross the field-free drift region and arrive later.

What is the full electron configuration of a chromium atom (Z = 24) in its ground state?

• A1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
• B1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
• C1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵✓
• D1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p⁴

Answer: Chromium is an exception to the Aufbau filling order. A half-filled 3d sub-shell is more stable, so one 4s electron is promoted, giving [Ar] 4s¹ 3d⁵, i.e. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵.

The successive ionisation energies (kJ mol⁻¹) of an element are: 738, 1451, 7733, 10540, 13630. In which group of the periodic table is this element?

• AGroup 1
• BGroup 2✓
• CGroup 13
• DGroup 14

Answer: There is a large jump between the 2nd (1451) and 3rd (7733) ionisation energies. This shows two electrons are easily removed before reaching a far more stable, closer-to-nucleus shell, so the element has 2 outer electrons and is in Group 2.

Which equation correctly represents the SECOND ionisation energy of magnesium?

• AMg(g) → Mg⁺(g) + e⁻
• BMg⁺(aq) → Mg²⁺(aq) + e⁻
• CMg(g) → Mg²⁺(g) + 2e⁻
• DMg⁺(g) → Mg²⁺(g) + e⁻✓

Answer: The nth ionisation energy refers to removing one mole of electrons from one mole of gaseous +(n−1) ions. The second ionisation energy starts from the gaseous 1+ ion: Mg⁺(g) → Mg²⁺(g) + e⁻.

The first ionisation energy of aluminium is slightly LOWER than that of magnesium. Which statement best explains this anomaly?

• AAluminium has a larger nuclear charge, so its outer electron is held more loosely
• BThe outer electron of aluminium is removed from a 3p sub-shell, which is higher in energy and better shielded than the 3s sub-shell of magnesium✓
• CMagnesium has a half-filled 3p sub-shell, giving it extra stability
• DAluminium atoms are smaller than magnesium atoms

Answer: In Mg the outer electron is removed from the 3s sub-shell, whereas in Al it is removed from the higher-energy 3p sub-shell, which is shielded by the 3s electrons. The 3p electron is therefore easier to remove, lowering the first ionisation energy despite Al's greater nuclear charge.

Going across Period 3 from sodium to argon, what is the general trend in atomic radius and the main reason for it?

• ADecreases, because increasing nuclear charge pulls the same shell of electrons closer✓
• BIncreases, because more electron shells are added
• CStays constant, because electrons are added to the same shell
• DIncreases, because shielding increases across the period

Answer: Across a period the number of protons (nuclear charge) increases while electrons are added to the same outer shell, so shielding is roughly constant. The stronger nuclear attraction pulls the outer shell in, so atomic radius decreases.

Which of the following correctly describes the trend in melting point across Period 3 from Na to Ar?

• AIt rises to a maximum at silicon, drops sharply at phosphorus, and is lowest at argon✓
• BIt rises steadily from Na to Ar
• CIt peaks at sodium then falls steadily to argon
• DIt is highest at chlorine because of strong covalent bonds

Answer: Metallic bonding strengthens Na→Al; silicon is a giant covalent (macromolecular) solid with the highest melting point. From phosphorus onwards the elements are simple molecules (P₄, S₈, Cl₂) held by weak London forces, so melting points fall sharply, and argon (monatomic) has the lowest.

Which species has the smallest radius?

• ANa⁺
• BO²⁻
• CF⁻
• DMg²⁺✓

Answer: Na⁺, Mg²⁺, F⁻ and O²⁻ are all isoelectronic (10 electrons). The species with the most protons exerts the greatest pull on the electron cloud. Mg²⁺ has 12 protons (more than Na⁺ 11, F⁻ 9, O²⁻ 8), so it has the smallest radius.