Chemical Bonding — Science

  1. Grade 11 — Ionic Bonding
  2. Grade 11 — Covalent Bonding
  3. Grade 11 — Metallic Bonding
  4. Grade 11 — Molecular Geometry & Polarity
  5. Grade 11 — Intermolecular Forces

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Problem 1

Two atoms have electronegativity values of 0.8 and 3.0. Based on their electronegativity difference (about 2.2), what type of bond is most likely to form between them?
  1. Nonpolar covalent — the electrons are shared equally.
  2. Polar covalent — the electrons are shared, but unequally.
  3. Ionic — the difference is large enough that an electron is essentially transferred.
  4. Metallic — the electrons pool into a shared 'sea'.
Hints

A rough rule of thumb: ΔEN below ~0.5 is nonpolar covalent, ~0.5–1.7 is polar covalent, and above ~1.7 is ionic.

Compute the difference: 3.0 − 0.8.

A ΔEN of about 2.2 is well above the ionic threshold.

Answer and explanation

Ionic — the difference is large enough that an electron is essentially transferred.

Electronegativity difference ΔEN = 3.0 − 0.8 = 2.2.

By the usual rule of thumb, ΔEN greater than about 1.7 means the more electronegative atom effectively takes the electron: an ionic bond.

So the bond is ionic (choice C).

Problem 2

Two nonmetal atoms bond with an electronegativity difference of about 0.9 (for example, the H–Cl bond). How is this bond best described?
  1. Nonpolar covalent — the electrons sit exactly in the middle.
  2. Polar covalent — the electrons are shared but pulled toward the more electronegative atom, giving partial charges.
  3. Ionic — one atom completely removes the electron from the other.
  4. Metallic — delocalized electrons flow freely between the atoms.
Hints

Two nonmetals share electrons, so the bond is covalent — the only question is whether the sharing is equal.

A ΔEN around 0.9 falls in the polar-covalent range (roughly 0.5–1.7).

Unequal sharing creates a slightly negative end and a slightly positive end.

Answer and explanation

Polar covalent — the electrons are shared but pulled toward the more electronegative atom, giving partial charges.

Two nonmetals means a covalent bond; ΔEN ≈ 0.9 is too small for ionic but too large for equal sharing.

That puts it in the polar-covalent range: the electrons are pulled toward the more electronegative atom, creating partial charges (δ+ and δ−).

So it is a polar covalent bond (choice B).

Problem 3

In carbon dioxide, CO_2, the central carbon is double-bonded to two oxygen atoms and has no lone pairs. Predict the molecule's shape and whether it is polar overall.
  1. Bent and polar — the two bond dipoles point the same way and add up.
  2. Tetrahedral and nonpolar — four things surround the carbon.
  3. Linear and polar — oxygen is more electronegative, so the whole molecule must be polar.
  4. Linear and nonpolar — the two equal C=O bond dipoles point in opposite directions and cancel.
Hints

Two electron groups around a central atom with no lone pairs push as far apart as possible — what angle is that?

Each C=O bond is polar, but think about whether the two dipoles cancel by symmetry.

Two equal dipoles pointing in exactly opposite directions add to zero.

Answer and explanation

Linear and nonpolar — the two equal C=O bond dipoles point in opposite directions and cancel.

Carbon has two electron groups and no lone pairs, so they point 180° apart: the molecule is linear (O=C=O).

Each C=O bond is polar, but the two equal dipoles point in exactly opposite directions and cancel.

So CO_2 is linear and nonpolar (choice D).

Problem 4

Water (H_2O) and carbon dioxide (CO_2) both contain only polar bonds, yet water is a polar molecule and CO_2 is not. Which explanation is correct?
  1. Water is bent (oxygen has two lone pairs), so its two O–H bond dipoles don't cancel; CO_2 is linear and symmetric, so its bond dipoles cancel.
  2. Water contains ionic bonds while CO_2 contains covalent bonds.
  3. Water molecules are heavier, and heavier molecules are always polar.
  4. Carbon dioxide has lone pairs on the central carbon that cancel its dipole.
Hints

Polar bonds only make a polar molecule if their dipoles don't cancel — geometry decides.

Compare the shapes: one molecule is bent, the other is linear.

Think about whether the bond dipoles are arranged symmetrically.

Answer and explanation

Water is bent (oxygen has two lone pairs), so its two O–H bond dipoles don't cancel; CO_2 is linear and symmetric, so its bond dipoles cancel.

Whether a molecule is polar depends on its shape, not just its bonds: symmetric arrangements let the bond dipoles cancel.

Water is bent (oxygen has two lone pairs), so its O–H dipoles don't cancel and the molecule is polar; CO_2 is linear and symmetric, so its dipoles cancel.

So the correct reason is the shape difference, choice A.

Problem 5

Ammonia (NH_3) and methane (CH_4) have almost the same molar mass, but NH_3 boils at −33 °C while CH_4 boils at −161 °C. What best explains ammonia's much higher boiling point?
  1. NH_3 contains ionic bonds that must be broken for it to boil.
  2. CH_4 is a much larger molecule with far more electrons.
  3. NH_3 molecules hydrogen-bond to one another (N–H), a much stronger intermolecular force than the London dispersion forces between CH_4 molecules.
  4. NH_3 is nonpolar, which lets its molecules pack together very tightly.
Hints

Boiling separates molecules from one another, so it depends on intermolecular forces, not on breaking the bonds inside a molecule.

Which molecule has N–H bonds (N, O, or F bonded to H) that allow hydrogen bonding?

Methane is nonpolar and relies only on weak London dispersion forces.

Answer and explanation

NH_3 molecules hydrogen-bond to one another (N–H), a much stronger intermolecular force than the London dispersion forces between CH_4 molecules.

Boiling overcomes intermolecular forces, and the molar masses are similar, so mass is not the difference.

NH_3 has N–H bonds, so its molecules hydrogen-bond — a relatively strong intermolecular attraction — while nonpolar CH_4 has only weak London dispersion forces.

Stronger attractions need more energy to separate, so NH_3 boils much higher: choice C.

Problem 6

Copper conducts electricity, is malleable (can be hammered into sheets), and can be drawn into wires without shattering. Which bonding model explains ALL of these properties at once?
  1. Ionic bonding — a rigid lattice of alternating positive and negative ions.
  2. Metallic bonding — metal cations in a 'sea' of delocalized electrons that carry charge and let the cations slide past one another without breaking the metal.
  3. Nonpolar covalent bonding — separate small molecules held together only weakly.
  4. Hydrogen bonding between neighboring copper atoms.
Hints

Which model has electrons that are free to move throughout the whole solid?

Conductivity needs mobile charges; malleability needs particles that can shift position without the structure falling apart.

Ionic solids are brittle and don't conduct as solids — does that fit copper?

Answer and explanation

Metallic bonding — metal cations in a 'sea' of delocalized electrons that carry charge and let the cations slide past one another without breaking the metal.

Delocalized (free-moving) electrons carry electric current, which explains the conductivity.

Because the bonding is that shared electron 'sea', layers of cations can slide past one another when struck, so the metal bends instead of shattering (malleable and ductile).

All three properties fit metallic bonding (choice B); ionic solids, by contrast, are brittle and non-conducting when solid.

Problem 7

Aluminum forms Al³⁺ ions and oxygen forms O²⁻ ions. In the neutral ionic compound they make together, how many oxygen atoms are there per formula unit?
Hints

The formula unit must be electrically neutral: total positive charge = total negative charge.

Find the least common multiple of 3 (from Al³⁺) and 2 (from O²⁻).

The total charge to balance is 6: how many Al³⁺ and how many O²⁻ reach +6 and −6?

Answer and explanation

3 atoms

Balance the charges: the least common multiple of 3 and 2 is 6, so you need +6 total and −6 total.

That takes 2 Al³⁺ (2 × 3+ = 6+) and 3 O²⁻ (3 × 2− = 6−), giving the formula Al_2O_3.

So there are 3 oxygen atoms per formula unit.

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