IB Chemistry Quiz: Apply The Covalent Model

What this quiz covers

This quiz focuses on Apply The Covalent Model, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. C-F
2. N-F
3. O-F
4. B-F (correct answer)

Question 2

1. The triple bond is more than five times as strong as the single bond. (correct answer)
2. The energy required to break one mole of N₂ molecules into N atoms is 160 kJ.
3. The average energy of a π bond in N₂ is less than the energy of the σ bond.
4. The N≡N bond is approximately one-third the length of the N-N bond.

Question 3

1. Water (H₂O)
2. Iodine (I₂) (correct answer)
3. Sodium chloride (NaCl)
4. Ammonia (NH₃)

Question 4

1. Pentane is a polar molecule due to its chain structure, whereas 2,2-dimethylpropane is nonpolar.
2. The straight-chain structure of pentane allows for a larger surface area of contact between molecules, leading to stronger London dispersion forces. (correct answer)
3. The branched structure of 2,2-dimethylpropane has more C-H bonds, resulting in stronger overall intermolecular attractions.
4. Pentane and 2,2-dimethylpropane have different molar masses, which accounts for the difference in boiling points.

Question 5

1. X is the most polar and Z is the least polar.
2. Y is the most polar and X is the least polar. (correct answer)
3. Z is the most polar and Y is the least polar.
4. X is the most polar and Y is the least polar.

Question 6

1. The prediction is correct; all three molecules have trigonal planar geometry with bond angles of approximately 120° due to identical bonding patterns with fluorine
2. The prediction is incorrect; $BF_3$ has 120° bond angles due to $sp^2$ hybridization, while $NF_3$ and $PF_3$ have smaller bond angles due to lone pair repulsion in $sp^3$ hybridization (correct answer)
3. The prediction is incorrect; $BF_3$ and $PF_3$ have 120° bond angles due to $sp^2$ hybridization, while $NF_3$ has 107° bond angles due to $sp^3$ hybridization and lone pair repulsion
4. The prediction is correct; all three molecules have pyramidal geometry with bond angles of approximately 107° due to lone pairs on each central atom affecting bond angles equally

Question 7

1. Electron domain geometry: trigonal bipyramidal; Molecular geometry: seesaw with bond angles of 120° and 90° due to one lone pair occupying an equatorial position
2. Electron domain geometry: octahedral; Molecular geometry: square planar with bond angles of 90° due to two lone pairs occupying opposite axial positions for minimum repulsion (correct answer)
3. Electron domain geometry: tetrahedral; Molecular geometry: square planar with bond angles of 90° due to expanded octet allowing four fluorine atoms around xenon in a planar arrangement
4. Electron domain geometry: octahedral; Molecular geometry: tetrahedral with bond angles of 109.5° due to four bonding pairs arranged to minimize electron-electron repulsion around the central atom

Question 8

1. Phosphorus uses $sp^3$ hybridization in all three compounds, but bond angles decrease as more chlorine atoms are added due to increased steric repulsion between larger numbers of substituents
2. Phosphorus undergoes promotion of electrons to higher energy levels, enabling $sp^2$ hybridization in $PCl_3$, $sp^3$ hybridization in $PCl_5$, and $sp^3d$ hybridization in $PCl_6^-$
3. Phosphorus can expand its octet using available 3d orbitals, allowing $sp^3$ hybridization in $PCl_3$, $sp^3d$ hybridization in $PCl_5$, and $sp^3d^2$ hybridization in $PCl_6^-$ (correct answer)
4. Phosphorus forms ionic bonds with chlorine atoms in all compounds, with the number of bonds determined by the charge transfer capability rather than orbital hybridization patterns

Question 9

1. $CO$ has the shortest bond due to triple bond character from molecular orbital overlap, $CO_2$ has intermediate length due to double bonds, and $CO_3^{2-}$ has the longest due to resonance delocalization reducing effective bond order (correct answer)
2. $CO$ has the shortest bond due to maximum s-character in $sp$ hybridization, $CO_2$ has intermediate length due to $sp^2$ hybridization, and $CO_3^{2-}$ has the longest due to $sp^3$ hybridization
3. $CO$ has the shortest bond due to electronegativity differences creating stronger ionic character, $CO_2$ has intermediate length due to partial ionic character, and $CO_3^{2-}$ has the longest due to purely covalent bonding
4. $CO$ has the shortest bond due to small molecular size allowing closer approach, $CO_2$ has intermediate length due to linear geometry constraints, and $CO_3^{2-}$ has the longest due to trigonal planar geometry spreading electron density

Question 10

1. The N-O bond length should be equal to a typical N-O single bond length because the negative charge on the ion weakens all bonds equally throughout the structure
2. The N-O bond length should be equal to a typical N=O double bond length because resonance stabilization strengthens all bonds to their maximum possible strength
3. The N-O bond length should be intermediate between typical N-O single and N=O double bond lengths, approximately 1.33 times a single bond due to partial double bond character
4. The N-O bond length should be intermediate between typical N-O single and N=O double bond lengths, being shorter than single bonds but longer than double bonds due to resonance averaging (correct answer)

Question 11

1. Structure A is preferred because single bonds minimize formal charges on sulfur, with sulfur having a formal charge of +2 and each oxygen having a formal charge of -1
2. Structure A is preferred because it avoids violating the octet rule for sulfur, even though it results in larger formal charges on the oxygen atoms compared to Structure B
3. Structure B is preferred because it places the smallest formal charges on the most electronegative atoms, with sulfur having a formal charge of +1 and each oxygen having a formal charge of 0 (correct answer)
4. Structure B is preferred because double bonds provide better formal charge distribution, with sulfur having a formal charge of 0 and each oxygen having a formal charge of -1 through resonance

Question 12

1. Both molecules have $sp^3d$ hybridization; $SF_4$ has seesaw geometry due to one lone pair, while $ClF_3$ has T-shaped geometry due to two lone pairs (correct answer)
2. Both molecules have $sp^3d$ hybridization; $SF_4$ has tetrahedral geometry due to four bonding pairs, while $ClF_3$ has trigonal pyramidal geometry due to one lone pair
3. Both molecules have $sp^3$ hybridization; $SF_4$ has tetrahedral geometry while $ClF_3$ has trigonal planar geometry due to different numbers of fluorine atoms
4. Both molecules have $sp^3d^2$ hybridization; $SF_4$ has square planar geometry due to two lone pairs, while $ClF_3$ has linear geometry due to three lone pairs

Question 13

1. $BeCl_2$ is linear due to ionic bonding character eliminating directional constraints, while $H_2O$ is bent due to covalent bonding requiring specific orbital overlap angles
2. $BeCl_2$ is linear due to $sp^2$ hybridization with no lone pairs, while $H_2O$ is bent due to $sp^2$ hybridization with 1 lone pair on oxygen creating asymmetric electron distribution
3. $BeCl_2$ is bent due to lone pair repulsion on beryllium, while $H_2O$ is linear due to the small size of hydrogen atoms allowing closer approach to oxygen
4. $BeCl_2$ is linear due to $sp$ hybridization with 2 electron domains, while $H_2O$ is bent due to $sp^3$ hybridization with 4 electron domains including 2 lone pairs on oxygen (correct answer)

Question 14

1. SO₂
2. BF₃
3. PCl₃ (correct answer)
4. CCl₄

Question 15

1. The Si=O double bonds in the SiO₂ lattice are significantly stronger than the C=O double bonds in CO₂ molecules.
2. Silicon dioxide is a polar substance with strong dipole-dipole interactions, while carbon dioxide is nonpolar.
3. Silicon dioxide exists as a covalent network structure, while carbon dioxide is a simple molecular substance with weak intermolecular forces. (correct answer)
4. Silicon has a larger atomic radius than carbon, which allows for more effective orbital overlap and stronger covalent bonds.

Question 16

1. CO₂
2. CH₄
3. H₂O (correct answer)
4. BF₃

Question 17

1. Graphite is extremely hard because it has a three-dimensional network of strong covalent bonds throughout its structure.
2. Graphite conducts electricity because it contains delocalized electrons within its layered structure. (correct answer)
3. Graphite has a low melting point because the layers are held together by weak covalent bonds.
4. Graphite is transparent because all valence electrons are tightly held in localized sigma bonds between carbon atoms.

Question 18

1. Trigonal pyramidal, <109.5°
2. Trigonal planar, 120° (correct answer)
3. Bent, <120°
4. Tetrahedral, 109.5°

Question 19

1. SeF₂, because it has a significantly greater number of electrons, leading to stronger London dispersion forces. (correct answer)
2. OF₂, because it has a smaller molar mass, which allows the molecules to pack more closely and interact more strongly.
3. OF₂, because the O-F bond is more polar than the Se-F bond, leading to stronger dipole-dipole interactions.
4. SeF₂, because its molecular geometry is less symmetrical than that of OF₂, resulting in a larger net dipole moment.

Question 20

1. H₂S
2. CH₃Cl
3. SO₃ (correct answer)
4. NCl₃