Atomic Structure and Properties: Unit Diagnostic (Chemistry Unit 1)

Answer: (A). Find moles of molecules (n = m / M), then multiply by atoms per molecule. CH₄: 16.0 ÷ 16.04 = 0.997 mol × 5 = 4.99 mol of atoms. NH₃: 0.998 × 4 = 3.99. H₂O: 27.0 ÷ 18.02 = 1.50 × 3 = 4.50. CO₂: 1.00 × 3 = 3.00.

• (B) also has about 1 mol of molecules, but only 4 atoms in each.
• (C) has the most molecules (1.50 mol). The question asks about atoms.
• (D) has the largest mass, but a larger mass does not mean more particles.

If you missed this: read the Topic 1.1 study guide, especially Worked example 1.

Answer: (A). The midpoint of 50.0 and 53.0 is 51.5. The average, 51.6, is just above the midpoint, so the heavier isotope must be slightly more common. Only (A) has more than 50% of mass 53.0. Check: (50.0 × 0.467) + (53.0 × 0.533) = 51.6 amu.

• (B) swaps the two abundances. It gives 51.4 amu, just below the midpoint.
• (C) is the simple mean (51.5 amu). It ignores the fact that the average is not exactly in the middle.
• (D) gives 51.2 amu. Being lighter does not make an isotope more common.

If you missed this: read the Topic 1.2 study guide, including the two quick checks and Worked example 3.

(a)

1. Mass of oxygen: 1.000 − 0.4364 = 0.5636 g.
2. Moles: P 0.4364 ÷ 30.97 = 0.01409 mol; O 0.5636 ÷ 16.00 = 0.03523 mol.
3. Divide by the smaller: P 1.000; O 0.03523 ÷ 0.01409 = 2.500.
4. A ratio ending in .50 means multiply both numbers by 2: P 2, O 5.

Empirical formula: P₂O₅.

(b) By the law of definite proportions, every pure sample of the compound has the same mass fraction of phosphorus: 0.4364 g per 1.000 g, or 43.64%. In 3.50 g: 3.50 × 0.4364 = 1.53 g of phosphorus. (The formula gives the same value: P₂O₅ is 61.94 ÷ 141.94 = 43.64% phosphorus.)

Common slip: rounding 2.50 to 2 or 3. PO₂ (49.18% P) and PO₃ (39.22% P) do not fit the data.

If you missed this: read the Topic 1.3 study guide, especially the table for turning a ratio into whole numbers.

Answer: (B). All the chlorine is in the MgCl₂.

1. M(MgCl₂) = 24.31 + 2(35.45) = 95.21 g mol⁻¹. Mass fraction of Cl = 70.90 ÷ 95.21 = 0.7447.
2. Mass of MgCl₂ = 1.06 g ÷ 0.7447 = 1.423 g.
3. Purity = 1.423 ÷ 2.00 × 100 = 71.2%.

• (A) is 1.06 ÷ 2.00 × 100, the percent of chlorine in the sample, not the percent of the compound.
• (C) is the percent of chlorine in pure MgCl₂. It ignores the measured data.
• (D) uses only one Cl per formula unit (35.45 ÷ 95.21). A purity above 100% is impossible, which is the warning sign.

If you missed this: read the Topic 1.4 study guide, especially Worked example 1.

Answer: (B). [Ar]3d⁸ holds 18 + 8 = 26 electrons. A nickel atom (Z = 28) is [Ar]4s²3d⁸. Ni²⁺ has 26 electrons, and the two electrons removed come from 4s, the highest shell: [Ar]3d⁸.

• (A) is tempting because nickel has eight 3d electrons. But the atom also has two 4s electrons (28 in total).
• (C) has the right count (26 electrons), but an iron atom is [Ar]4s²3d⁶: in a neutral atom 4s fills before 3d.
• (D) Zn²⁺ has 28 electrons: [Ar]3d¹⁰.

If you missed this: read “Configurations of ions” in the Topic 1.5 study guide.

(a) Ga: [Ar]4s²3d¹⁰4p¹ (18 + 2 + 10 + 1 = 31). Ga³⁺ has 28 electrons. Remove from the highest shell (n = 4) first: the 4p electron, then both 4s electrons. Ga³⁺ = [Ar]3d¹⁰.

(b) Valence electrons are in the highest shell, 4s²4p¹: 3 valence electrons. The other 28 are core electrons; the full 3d subshell belongs to the core.

(c) The 3d electron. It is in shell 3, closer to the nucleus than the 4p electron, and it is shielded by fewer inner electrons, so it feels a larger effective nuclear charge. By Coulomb’s law (F ∝ q₁q₂ / r²), a larger charge at a smaller distance means a stronger attraction.

Common slips: Ga³⁺ as [Ar]4s²3d⁸ (removing the “last in” electrons); counting 3d electrons as valence; naming only one factor in (c).

If you missed this: read “Core and valence electrons” and “How hard is an electron to remove?” in the Topic 1.5 study guide.

Answer: (A). S is 1s²2s²2p⁶3s²3p⁴ and Cl is 1s²2s²2p⁶3s²3p⁵: five subshells each, so five peaks, with a taller 3p peak for chlorine (5 against 4). Chlorine has 17 protons against 16 for electrons in the same subshells, so by Coulomb’s law every peak sits at a higher binding energy.

• (B) The extra electron joins the occupied 3p subshell. One peak per subshell, not per electron.
• (C) An electron in the same shell adds very little shielding. The extra proton wins.
• (D) Heights count electrons, and the 3p subshells differ (4 against 5).

If you missed this: read “How to read a spectrum” and “Explaining peak positions with Coulomb’s law” in the Topic 1.6 study guide.

Answer: (A). The values are Be 899, B 801, Mg 738 and Al 578 kJ mol⁻¹. Beryllium’s outer electrons are in a shell closer to the nucleus than magnesium’s. Boron’s outer electron is in 2p, higher in energy and partly shielded by the 2s electrons, so it is easier to remove than beryllium’s 2s electron.

• (B) is the “further right means higher” trap: the group 13 dip breaks the trend.
• (C) Magnesium’s 3s electrons are farther from the nucleus.
• (D) Aluminium is lower in the group and past the dip: the lowest of the four.

If you missed this: read “Ionization energy” and Figure 2 in the Topic 1.7 study guide.

Answer: (C). Na⁺ and Mg²⁺ both have 10 electrons (1s²2s²2p⁶), so they are isoelectronic. Mg²⁺ has 12 protons against 11 for Na⁺, so it pulls the same electron cloud in more tightly. Both cations are much smaller than a sodium atom, which still has its n = 3 shell.

• (A) A sodium atom still has its third shell, so it is larger than both cations.
• (B) Na⁺ is small, but it has one proton fewer than Mg²⁺ for the same 10 electrons.
• (D) Cl⁻ has 18 electrons in three shells, so it is much larger than either cation.

If you missed this: read “Ionic radius” and Worked example 1 in the Topic 1.7 study guide.

Answer: (A). Caesium’s single 6s valence electron is far from the nucleus and heavily shielded, so it is very weakly held. Fluorine’s nucleus attracts an extra electron into its small n = 2 shell very strongly. The electron transfers, giving Cs⁺ and F⁻, which form the ionic compound CsF.

• (B) Both nonmetals hold their valence electrons strongly, so they share electrons (CO₂ is covalent).
• (C) Neon holds its full valence shell very tightly and has no room for another electron in it, so it forms no stable compound with chlorine.
• (D) Two nonmetals again: they would share electrons, not transfer them.

If you missed this: read “Why valence electrons control chemistry” in the Topic 1.8 study guide.