NCEA Level 3 Chemistry 91390 Particles, Substances and Thermochemistry: past papers, practice and marking
- NCEA Level 3
- Chemistry
- AS91390
- 5 credits
- External exam
About 91390
AS91390, Demonstrate understanding of thermochemical principles and the properties of particles and substances. Electron configuration, periodic trends, Lewis structures, shape, polarity, intermolecular forces, enthalpy, Hess’s law and entropy, where saying exactly the right thing is most of the mark.
What 91390 covers
- Lewis structuresCount regions of electron density (up to six), then name the shape from the atoms; bond angles
- PolarityBond dipoles from electronegativity difference, and whether the shape makes them cancel
- Intermolecular forcesIdentify temporary dipole, permanent dipole and hydrogen bonding; justify boiling points, ΔvapH and heating curves
- Periodic trendsElectron configurations (s, p, d), atomic and ionic radii, ionisation energy and electronegativity, with reasons
- EnthalpyΔfH, Hess’s law, calorimetry (q = mcΔT) and phase-change enthalpies
- EntropyEntropy changes of the system and the surroundings, and why a reaction is spontaneous
91390 past papers
Open a 91390 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 3 Chemistry past papers
Practice questions
Question 1: Lewis structures
Predict and justify the shapes of PCl₅, BF₃ and I₃⁻.
- For each species, state the number of bonding and non-bonding regions and name the shape.(up to Merit)
- PCl₅ and I₃⁻ both have five regions of electron density around the central atom. Explain why they have different shapes, and give the bond angles in each.(up to Excellence)
Check your answer
Part a: what gets the marks
- P uses all 5 valence electrons in five P–Cl bonds, so it has five bonding regions and no lone pairs.
- PCl₅ is trigonal bipyramidal, with angles of 90° (axial to equatorial) and 120° (equatorial to equatorial).
- B uses all 3 valence electrons in three B–F bonds, so it has three bonding regions and no lone pairs (boron does not complete an octet).
- BF₃ is trigonal planar, with angles of 120°.
- The central I in I₃⁻ has 7 + 1 = 8 valence electrons, including one for the charge; 2 are used in bonds and 6 form three lone pairs, so it has two bonding regions and three lone pairs.
- Its five regions form a trigonal bipyramidal arrangement with the three lone pairs equatorial, so I₃⁻ is linear at 180°.
A full answer: PCl₅ (phosphorus pentachloride): 5 bonding, 0 lone pairs, trigonal bipyramidal. BF₃ (boron trifluoride): 3 bonding, 0 lone pairs, trigonal planar, 120°. I₃⁻ (triiodide ion): 2 bonding, 3 lone pairs, linear, 180°.
Part b: what gets the marks
- Five regions repel to maximise separation, so both have a trigonal bipyramidal arrangement of electron density.
- PCl₅ has five bonding regions and no lone pairs, so its shape is trigonal bipyramidal, with angles of 90°, 120° and 180°.
- The three lone pairs take the equatorial positions, leaving the two I atoms axial.
- With the two outer I atoms axial, I₃⁻ is linear with an I–I–I angle of 180°, because the shape is named from the atoms only.
A full answer: Same electron geometry, different shapes. Both species have five regions of electron density, which repel to be as far apart as possible: a trigonal bipyramidal electron geometry.
PCl₅ (phosphorus pentachloride): all five regions are bonding pairs, so an atom sits in every position and the shape is trigonal bipyramidal, with angles of 90°, 120° and 180°.
I₃⁻ (triiodide ion): only two regions are bonding; the three lone pairs occupy the equatorial positions. The shape is named from the atoms, and the two outer I atoms are axial, so I₃⁻ is linear, with an angle of 180°.
What gets each grade
Question 2: Lewis structures
Arsenic pentafluoride, AsF₅, and selenium hexafluoride, SeF₆, are both gases in which every region of electron density around the central atom is a bonding pair.
- Name the shape of each molecule.(up to Achieved)
- Justify why the F–As–F bond angles in AsF₅ are different from the F–Se–F bond angles in SeF₆.(up to Excellence)
Check your answer
Part a: what gets the marks
- As has five bonding pairs and no lone pairs, so AsF₅ is trigonal bipyramidal.
- Se has six bonding pairs and no lone pairs, so SeF₆ is octahedral.
A full answer: AsF₅ (arsenic pentafluoride): trigonal bipyramidal. SeF₆ (selenium hexafluoride): octahedral.
Part b: what gets the marks
- Regions of electron density around a central atom repel each other and are arranged as far apart as possible to minimise repulsion.
- As has five regions, which repel to a trigonal bipyramid with F–As–F angles of 120° (equatorial), 90° (axial to equatorial) and 180° (axial to axial).
- Se has six regions, which repel to an octahedron with F–Se–F angles of 90° between adjacent bonds and 180° between opposite bonds.
- The different number of regions of electron density (five compared with six) gives a different electron geometry, so the angles differ: AsF₅ (arsenic pentafluoride) has 120° angles, SeF₆ (selenium hexafluoride) does not.
A full answer: Five versus six regions of electron density. In both molecules the regions of electron density around the central atom repel each other and take up the arrangement that maximises separation and minimises repulsion.
As has five regions (all bonding), giving a trigonal bipyramidal geometry with bond angles of 120° in the equatorial plane, 90° between axial and equatorial bonds, and 180° between the two axial bonds.
Se has six regions (all bonding), giving an octahedral geometry in which every adjacent pair of bonds is at 90° (and opposite bonds at 180°).
The difference in the number of regions of electron density is therefore why AsF₅ (arsenic pentafluoride) has 120° angles that SeF₆ (selenium hexafluoride) does not.
What gets each grade
Question 3: Lewis structures
Sulfur trioxide, SO₃, is trigonal planar, but the sulfite ion, SO₃²⁻, is trigonal pyramidal. Each has three O atoms bonded to S.
- Explain why SO₃ and SO₃²⁻ have different shapes, and compare their O–S–O bond angles.(up to Excellence)
- Explain why SO₃ is non-polar even though its S=O bonds are polar.(up to Merit)
Check your answer
Part a: what gets the marks
- SO₃ has three regions of electron density on S and no lone pairs, while SO₃²⁻ has three bonding regions and one lone pair on S.
- In SO₃ the three regions (each double bond counts as one region) repel as far apart as possible, so SO₃ is trigonal planar with O–S–O angles of 120°.
- In SO₃²⁻ the four regions repel to a tetrahedral arrangement; one is a lone pair, so the ion is trigonal pyramidal with O–S–O angles a little less than 109.5°.
- The two extra electrons from the 2− charge give S a lone pair, so S has four regions instead of three, and SO₃²⁻ has smaller bond angles than SO₃ (about 107° against 120°).
A full answer: SO₃: 24 valence electrons. S has three regions of electron density (a double bond counts as one region) and no lone pairs. Three regions repel as far apart as possible: trigonal planar, O–S–O 120°.
SO₃²⁻: 26 valence electrons (two added for the 2− charge), so S also has one lone pair: four regions, which repel to a tetrahedral arrangement. Only three positions hold O atoms, so the ion is trigonal pyramidal, with O–S–O angles a little less than 109.5° (about 107°).
The extra pair of electrons from the charge adds a fourth region on S, which changes the shape and makes the bond angles smaller.
Part b: what gets the marks
- The S=O bonds are polar, because O is more electronegative than S.
- SO₃ is trigonal planar, so its three equal S=O dipoles are symmetrically arranged at 120° and cancel, leaving no overall dipole.
A full answer: The S=O bonds are polar, because O is more electronegative than S.
SO₃ is trigonal planar, so the three equal dipoles are symmetrically arranged at 120° and cancel.
There is no overall dipole, so SO₃ is non-polar.