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IBDP Chemistry - Structure 3.1 The periodic table: Classification of elements- IB Style Questions For SL Paper 1A - FA 2025

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Question

Germanium can exist in multiple stable oxidation states. Which row lists these species in order of increasing atomic radius?
(A) \(Ge^{4+}, Ge^{2+}, Ge, Ge^{4-}\)
(B) \(Ge^{4-}, Ge^{4+}, Ge^{2+}, Ge\)
(C) \(Ge, Ge^{4-}, Ge^{4+}, Ge^{2+}\)
(D) \(Ge^{2+}, Ge, Ge^{4-}, Ge^{4+}\)
▶️ Answer/Explanation
Detailed solution

1. General Trends in Ionic Radius:

  • Cations (positive ions) are smaller than their neutral atoms because they lose valence electrons (and sometimes entire shells), increasing effective nuclear charge per electron. Higher positive charge = Smaller radius. (\(Ge^{4+} < Ge^{2+} < Ge\)).
  • Anions (negative ions) are larger than their neutral atoms because added electrons increase electron-electron repulsion, expanding the cloud. (\(Ge < Ge^{4-}\)).

2. Arrange in Increasing Order:
Smallest (most positive) \(\rightarrow\) Largest (most negative).
Order: \(Ge^{4+} < Ge^{2+} < Ge < Ge^{4-}\).
This matches row A.
✅ Answer: (A)

Question

Which statement best explains the trend in first ionisation energy from sodium (Na) to chlorine (Cl)?
(A) Nuclear charge increases.
(B) Electronegativity decreases.
(C) Atomic radius increases.
(D) Shielding decreases.
▶️ Answer/Explanation
Detailed solution

1. Analyze the Trend Across a Period:
Moving from Na to Cl (left to right across Period 3), protons are added to the nucleus one by one. This increases the nuclear charge.

2. Analyze Shielding:
Electrons are added to the same principal energy shell (n=3). The inner shell electrons (n=1, n=2) remain constant, so the shielding effect is approximately constant, not decreasing.

3. Relate to Ionization Energy:
Since nuclear charge increases while shielding stays roughly the same, the effective nuclear charge increases. This pulls the valence electrons closer (decreasing atomic radius) and holds them more tightly. Therefore, more energy is required to remove an electron (higher first ionization energy). The primary driver is the increasing nuclear charge.
✅ Answer: (A)

Question

Which correctly states the strongest intermolecular forces in the compounds below?
 \(\text{CH}_4\)\(\text{CH}_3\text{Cl}\)\(\text{CH}_3\text{NH}_2\)
Adipole–dipoleLondon forceshydrogen bonding
BLondon forcesdipole–dipolehydrogen bonding
Chydrogen bondingLondon forcesdipole–dipole
DLondon forceshydrogen bondingdipole–dipole
▶️ Answer/Explanation
Detailed solution

\(\text{CH}_4\) is a non-polar molecule (tetrahedral with symmetric \(\text{C–H}\) bonds), so its strongest intermolecular forces are London forces.

\(\text{CH}_3\text{Cl}\) is polar due to the electronegative \(\text{Cl}\) atom, so it has dipole–dipole forces as its strongest intermolecular forces (plus London forces).

\(\text{CH}_3\text{NH}_2\) contains an \(\text{N–H}\) bond and a lone pair on nitrogen, so molecules can form hydrogen bonds with each other. Hydrogen bonding is stronger than simple dipole–dipole interactions.

Therefore, the correct combination is London forces for \(\text{CH}_4\), dipole–dipole for \(\text{CH}_3\text{Cl}\), and hydrogen bonding for \(\text{CH}_3\text{NH}_2\), which corresponds to option B.

✅ Answer: (B)

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