Unit notes and practice questions
Everything here is for unit 2 of the Grade 11 Chemistry textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.
The cards filed under this chapter on the Grade 11 Chemistry page. Open one at a time.
When you use it
Use this card when determining whether a molecule with polar bonds is overall polar or non-polar.
Watch out
Polar bonds do not always produce a polar molecule. In symmetric shapes like carbon dioxide or boron trichloride, individual bond dipoles cancel out to give zero net dipole moment.
Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up
When you use it
Use this formula to calculate the dipole moment of a polar covalent bond or diatomic molecule.
Watch out
In non-SI units, 1 Debye equals 3.33564 multiplied by 10^-30 coulomb metres.
Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up
| Category | Electrons on Central Atom | Examples |
|---|---|---|
| Less than octet | Fewer than 8 electrons | , , |
| Expanded octet | More than 8 electrons | , , |
| Odd-electron molecules | Odd number of electrons | , , |
When you use it
Use this reference when identifying molecules that do not obey the standard eight-electron octet rule.
Watch out
Third period and lower non-metals expand their octet using available empty d orbitals, but second period elements cannot.
Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up
When you use it
Use this card when interpreting resonance structures and Lewis representations of molecules like ozone.
Watch out
Resonance does not mean a molecule flips back and forth between structures. The real molecule is a single, permanent hybrid structure with intermediate bonds.
Drafted from Grade 11 Chemistry, pages 67-144, then checked twice before it went up
| Solid Type | Particles | Attractive Forces | Key Properties | Examples |
|---|---|---|---|---|
| Ionic | Cations and anions | Ionic bonds | High melting point, brittle, conducts in molten state | |
| Molecular | Molecules or atoms | Intermolecular forces | Low melting point, soft, non-conductors | |
| Covalent network | Atoms | Covalent bonds | Extremely high melting point, very hard, non-conductors | |
| Metallic | Metal cations in electron sea | Metallic bonds | Variable melting point, malleable, conducts electricity |
When you use it
Use to classify crystalline solids based on particle types, bonding forces, and physical properties.
Watch out
Ionic solids do not conduct electricity in the solid state, they only conduct when molten or dissolved in water.
Drafted from Grade 11 Chemistry, pages 145-192, then checked twice before it went up
When you use it
Use when a question asks how a covalent bond actually forms, or why the two bonds of a C=C are not alike. A bond appears where two half filled valence orbitals overlap and their electrons pair with opposite spins: head on overlap along the line joining the nuclei gives a sigma bond, while sideways overlap of two parallel p orbitals gives a pi bond whose density sits above and below that line.
Watch out
Students count a double bond as two of the same thing and draw the second bond of C=C as another head on overlap squeezed between the nuclei. Only the first bond between two atoms is a sigma bond, and every extra one is a pi bond, so C=C is one sigma plus one pi and a pi bond never stands alone. Its pair sits above and below the axis rather than between the nuclei, which is why it is the weaker bond and why the two ends of a double bond cannot rotate. The + and - marked on the p lobes are wave signs, not electric charges.
When you use it
Use when a question asks how many sigma and how many pi bonds a molecule holds. Count every line in the structure as one sigma bond, then add one pi for each extra line inside a double or a triple bond.
Watch out
Students count only the bonds inside the multiple bond and forget that every other line in the structure is a sigma bond too. Asked about ethene, C2H4, they answer 1 sigma and 1 pi. The real count is four C-H sigma bonds plus the C=C, which is itself one sigma and one pi, so 5 sigma and 1 pi. Ethyne, C2H2, is 3 sigma and 2 pi. The other habit to break is calling a double bond 2 sigma or a triple bond 3 sigma: there is never more than one sigma between the same two atoms.
| Hybrid set | Orbitals mixed | Electron geometry | Bond angle | Examples |
|---|---|---|---|---|
| linear | 180° | , | ||
| trigonal planar | 120° | , | ||
| tetrahedral | 109.5° | , , | ||
| trigonal bipyramidal | 120°, 90° and 180° | |||
| octahedral | 90° and 180° |
When you use it
Use when you have to name the shape around a central atom, or read its hybridization off a Lewis structure. Count the electron domains on that atom with the lone pairs included, and the count picks the row: the number of hybrid orbitals always equals the number of atomic orbitals mixed.
Watch out
Students read the geometry column as the shape of the molecule and write NH3 as tetrahedral at 109.5 degrees. The two answers come from different counts. Hybridization counts every electron domain, lone pairs included, so NH3 and H2O are both sp3 with tetrahedral electron geometry. The shape is named from the atoms alone, so NH3 is trigonal pyramidal at about 107 degrees and H2O is bent at about 104.5 degrees. The mirror image of the mistake is dropping the lone pairs and calling H2O sp2 because it has two bonds.
| Bond | Bond order | Bond energy (kJ/mol) | Bond length (pm) |
|---|---|---|---|
| H-H | 1 | 436 | 74 |
| Cl-Cl | 1 | 242 | 199 |
| H-Cl | 1 | 431 | 127 |
| C-H | 1 | 413 | 109 |
| C-C | 1 | 348 | 154 |
| C=C | 2 | 612 | 134 |
| C≡C | 3 | 837 | 120 |
When you use it
Use when a question compares two bonds for strength or for length, or asks how much energy breaking one mole of a bond costs. The first three rows are exact values for those diatomic molecules; the carbon rows are averages taken over many compounds, so another book may print a few units different.
Watch out
Students memorise that a shorter bond is a stronger bond and then use it across different pairs of atoms. Asked which is stronger, C-H at 109 pm or C=C at 134 pm, most answer C-H because it is shorter, and it is the weaker of the two at 413 against 612. The rule is only safe between the same two atoms, where C-C, C=C and C≡C get shorter and stronger together. H-H at 74 pm and C-C at 154 pm are different atoms, so the length tells you nothing about which is harder to break.
When you use it
Use when a question asks whether a diatomic molecule or ion can exist, how strong its bond is, or whether it is paramagnetic. Fill the molecular orbitals in energy order first, count the bonding and the antibonding electrons, then halve the difference: a bond order of 0 means the molecule does not form.
Watch out
Students count only the electrons they can see in the printed configuration and then subtract with the wrong totals. Decide once whether the 1s core is in or out and stay with it: N2 is half of 10 minus 4 with the core and half of 8 minus 2 without it, giving 3 either way, while half of 10 minus 2 mixes the two and gives 4. The second habit is pairing both pi 2p electrons into one orbital. B2 puts one in pi 2px and one in pi 2py with parallel spins, which is the whole reason B2 is paramagnetic. And do not carry the pi 2p below sigma 2p diagram past N2: O2 and F2 use the flipped order.
10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.
Question 1
Which statement is the definition of a chemical bond in this unit?
Answer: B The attractive force which holds atoms, ions, and molecules together.
Why
The unit says that the attractive force which holds atoms, ions, and molecules together is called a chemical bond. The other options describe repulsion, energy, or drawing, which are not the definition.
Question 2
According to the octet rule, how many electrons should an atom try to have in its outer shell?
Answer: C Eight
Why
The octet rule says atoms tend to gain or lose electrons until their outer shell contains an octet, which means eight electrons. Helium is the exception with two electrons.
Question 3
What is the bond length of the H2 molecule?
Answer: A 74 pm
Why
The unit states that in the H2 molecule the bond length is 74 pm. This is the H-H distance at minimum energy, where the molecule is most stable. The values 102 pm and 138 pm are ionic radii of sodium and potassium ions, and 1.87 D is the dipole moment of water.
Question 4
Which of these diatomic molecules is non-polar?
Answer: D H<sub>2</sub>
Why
Diatomic molecules made of atoms of the same element, such as H2, O2 and F2, are non-polar because they have no dipole moment. HCl, CO and NO contain atoms of different elements, so they are polar.
Question 5
The ions in LiF and in MgO are of similar size. Which compound has the stronger ionic bonding?
Answer: A MgO
Why
The lattice energy depends directly on the product of the ionic charges (q1 × q2) and inversely on the distance between them. MgO has ions with higher charges, so the attraction is stronger when the sizes are similar.
Question 6
The radii of the sodium and potassium ions are 102 pm and 138 pm. Which compound has stronger ionic attractions, NaCl or KCl?
Answer: B NaCl
Why
The sodium ion is smaller (102 pm), so the ions in NaCl sit closer together. Lattice energy is inversely related to the distance between the charges, so NaCl has the stronger ionic attraction.
Question 7
Using the VSEPR model, what is the shape of an ammonia (NH3) molecule?
Answer: B Trigonal pyramidal
Why
NH3 has three bonding pairs and one lone pair on the nitrogen. The overall arrangement of the four electron pairs is tetrahedral, but because one pair is a lone pair, the geometry of the molecule is trigonal pyramidal.
Question 8
How does the ammonium ion, NH4+, form?
Answer: D The lone-pair electrons from the nitrogen atom of NH<sub>3</sub> bond to H<sup>+</sup>.
Why
The ammonium ion forms when the lone-pair electrons from the nitrogen atom of ammonia bond to H+. This kind of bond is called a coordinate covalent bond, or dative bond. One atom donates both electrons.
Question 9
For NaCl, the enthalpy of formation is ΔfH° = -411 kJ mol-1. The Born-Haber steps are: sublimation of Na = +108 kJ mol-1; half the bond energy of Cl2 = +120 kJ mol-1; ionization energy of Na = +496 kJ mol-1; electron affinity of Cl = -349 kJ mol-1. Using Hess's law, what is the lattice energy U of NaCl?
Answer: A -786 kJ mol<sup>-1</sup>
Why
Use U = ΔfH° - (ΔHstep1 + ΔHstep2 + ΔHstep3 + ΔHstep4). The sum of the steps is 108 + 120 + 496 + (-349) = 375. So U = -411 - 375 = -786 kJ mol-1. The -1050 kJ value belongs to LiF, not NaCl.
Question 10
Water has a resultant dipole moment of 6.23 × 10-30 C m (1.87 D). What does this fact prove about the shape of the H2O molecule?
Answer: B The molecule is bent, because its O-H bond polarities do not cancel each other.
Why
In CO2, the linear shape makes the bond dipoles cancel, so the dipole moment is zero. Water has a non-zero dipole moment, so its bond polarities do not cancel. This only happens in a bent shape, and the O-H bonds are polar because oxygen is more electronegative.
The app holds 12 questions on this unit, marked as you go. These are the first 10.
Unit 2 runs from page 67 to page 144 of the Ministry of Education edition.
Ministry of Education edition, English, 342 pages
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