IB DP Chemistry Mock Exam SL Paper 1A Set 1 - 2025 Syllabus

IB DP Chemistry Mock Exam SL Paper 1A Set 1

Prepare for the IB DP Chemistry Exam with our comprehensive IB DP Chemistry Exam Mock Exam SL Paper 1A Set 1. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam

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Question

Which pair of the following species has the same number of outer-shell (valence) electrons? The letters do not correspond to real element symbols.

\({}^{24}_{12}W^{2+} \quad {}^{31}_{16}X^{-} \quad {}^{45}_{21}Y^{2+} \quad {}^{19}_{9}Z^{-}\)

A. \(X\) and \(Y\)

B. \(X\) and \(W\)

C. \(Z\) and \(Y\)

D. \(Z\) and \(W\)

▶️ Answer/Explanation

Detailed solution

1. Find the total number of electrons for each species:
\(W^{2+}:\) atomic number \(12 \Rightarrow\) electrons \(= 12 – 2 = 10\).
\(X^{-}:\) atomic number \(16 \Rightarrow\) electrons \(= 16 + 1 = 17\).
\(Y^{2+}:\) atomic number \(21 \Rightarrow\) electrons \(= 21 – 2 = 19\).
\(Z^{-}:\) atomic number \(9 \Rightarrow\) electrons \(= 9 + 1 = 10\).

2. Compare electron arrangements and outer electrons:
Species with \(10\) electrons have the same configuration as neon: \(2, 8\), so they each have \(8\) outer (valence) electrons.
Thus \(W^{2+}\) and \(Z^{-}\) both possess \(8\) valence electrons.
The species with \(17\) and \(19\) electrons have different outer-electron counts.

✅ Answer: (D)

Question

Which row corresponds to a gas that would show the smallest deviation from ideal gas behaviour?

	Gas	Pressure	Temperature
A	Phosphine, \(PH_{3}\)	Low	High
B	Ammonia, \(NH_{3}\)	Low	High
C	Phosphine, \(PH_{3}\)	High	Low
D	Ammonia, \(NH_{3}\)	High	Low

▶️ Answer/Explanation

Detailed solution

1. Ideal Gas Conditions:
Gases behave most ideally at low pressure (negligible volume of particles) and high temperature (negligible intermolecular forces). This eliminates options C and D.

2. Intermolecular Forces:
To deviate the least, the gas should have the weakest intermolecular forces.

Ammonia (\(NH_{3}\)): Exhibits hydrogen bonding (strong).
Phosphine (\(PH_{3}\)): Exhibits dipole-dipole forces and London dispersion forces, but no hydrogen bonding (P is not electronegative enough).

Since \(PH_{3}\) has weaker intermolecular forces than \(NH_{3}\), it will behave more ideally.

3. Conclusion:
\(PH_{3}\) at Low Pressure and High Temperature.
✅ Answer: (A)

Question

Which row provides the correct formula and electron configuration of the ions present in calcium nitride?

	Formula	Electron config of Ca species	Electron config of N species
A	\(Ca_{2}N_{3}\)	[Ar]	[Ne]
B	\(Ca_{2}N_{3}\)	\(4s^2\) [Ar]	[He] \(2s^2 2p^3\)
C	\(Ca_{3}N_{2}\)	[Ar] \(4s^2\)	[He] \(2s^2 2p^3\)
D	\(Ca_{3}N_{2}\)	[Ar]	[Ne]

▶️ Answer/Explanation

Detailed solution

1. Determine Ionic Charges:

Calcium (Group 2) forms \(Ca^{2+}\) ions.
Nitrogen (Group 15) forms \(N^{3-}\) ions.

2. Determine Formula:
To balance charges: \(3 \times (+2) + 2 \times (-3) = 0\). The formula is \(Ca_{3}N_{2}\). This eliminates A and B.

3. Determine Electron Configurations of Ions:

\(Ca^{2+}\): Calcium (\([Ar]4s^2\)) loses 2 electrons \(\rightarrow\) \([Ar]\).
\(N^{3-}\): Nitrogen (\([He]2s^22p^3\)) gains 3 electrons \(\rightarrow\) \([He]2s^22p^6\), which is isoelectronic with Neon \([Ne]\).

4. Match to Options:
Formula \(Ca_{3}N_{2}\), Ca species \([Ar]\), N species \([Ne]\).
✅ Answer: (D)

IB DP Chemistry Mock Exam SL Paper 1A Set 1 - 2025 Syllabus