IB Chemistry HL 2.2 Models of Bonding Question Bank

Practise IB Chemistry HL 2.2 with evidence-led questions on bonding models, molecular geometry, polarity and structure.

Covered syllabus points

2.2.1—Covalent bonding 2.2.2—Bond types and strength 2.2.3—Coordination bonds 2.2.4—VSEPR model 2.2.5—Bond polarity 2.2.6—Molecular polarity 2.2.7—Covalent network structures 2.2.8—Intermolecular forces (IMF)2.2.9—IMF strength and properties 2.2.10—Chromatography 2.2.11 (HL)—Resonance structures 2.2.12 (HL)—Benzene (C₆H₆)2.2.13 (HL)—Expanded octets 2.2.14 (HL)—Formal charge 2.2.15 (HL)—Sigma (σ) and pi (π) bonds 2.2.16 (HL)—Hybridization

[Maximum number: 4]

Phosphine (IUPAC name phosphane) is a hydride of phosphorus, with the formula $\mathrm{PH}_{3}$ .

(a)

Draw a Lewis (electron dot) structure of phosphine.

[ 1 ]

(b)

State the hybridization of the phosphorus atom in phosphine.

[ 1 ]

(c)

Outline whether you expect the bonds in phosphine to be polar or non-polar, giving a brief reason.

Figure for Question (c) — IB Chemistry HL

Phosphine has a much greater molar mass than ammonia. Explain why phosphine has a significantly lower boiling point than ammonia.

[ 2 ]

[Maximum number: 12]

(a)

An organic compound, X, with a molar mass of approximately $88 \mathrm{~g} \mathrm{~mol}^{-1}$ contains 54.5 % carbon, 36.3 % oxygen and 9.2 % hydrogen by mass.

[ 5 ]

(i)

The carboxylic acid contains two different carbon-oxygen bonds. Identify which bond is stronger and which bond is longer.

Stronger bond:

Longer bond:

[ 2 ]

(ii)

Predict and explain the bond lengths and bond strengths of the carbon-oxygen bonds in $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}$ .

[ 3 ]

(b)

State the meaning of the term hybridization.

[ 1 ]

(c)

Describe the hybridization of the carbon atom in methane and explain how the concept of hybridization can be used to explain the shape of the methane molecule.

[ 3 ]

$\mathrm{sp}^{3}$ ;
isolated C atom electron configuration $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2} /$ excited state C electron configuration is $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{1} 2 \mathrm{p}^{3}$ ;
$2 \mathrm{~s}^{1} 2 \mathrm{p}^{3}$ electrons blend to form four identical hybrid orbitals; hybrid orbitals lower in total energy than atomic orbitals; repulsion of (identical hybrid) orbitals creates a tetrahedral shape; Accept suitably annotated diagram for M2, M3 and M4.

(d)

Identify the hybridization of the carbon atoms in diamond and graphite and explain why graphite is an electrical conductor.

[ 3 ]

diamond:
$\mathrm{sp}^{3}$ ;
graphite:
$\mathrm{sp}^{2}$ ;
(p) electrons delocalized (around layer);

[Maximum number: 2]

Solid ionic compounds form crystal lattices.

Cobalt also forms chlorides with the formula $\mathrm{CoCl}_{2}$ .

Hydrated cobalt (II) ions, $\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{2+}$ , are pink. Describe the interaction between the cobalt ion and a water molecule in terms of the type of bond and how this bond is formed.

Type of bond:
How the bond forms:
$\mathrm{CoCl}_{4}{ }^{2-}$ ions are blue.

Type of bond: coordinate/dative/covalent

How the bond forms: oxygen/water molecule /ligand donates $\mathrm{e}^{-}$ pair to cobalt«(II) ion»

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Covered syllabus points

2.2 The covalent model question 1

Question (a)

Question (b)

Question (c)

2.2 The covalent model question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)

2.2 The covalent model question 3