Worked example 1 — solution

Step 1 · Find the electronegativity difference

Worked example 1 — solution

A difference of zero sits at the far left: nonpolar covalent.

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — solution

Step 1 · Find the electronegativity difference

Worked example 2 — solution

A difference of 0.9 is moderate: polar covalent. The pair is shared, but not equally.

Worked example 2 — solution

Worked example 2 — solution

Practice 1 — answer: A

B put δ− on the wrong atom: O (3.5) pulls harder than S (2.5), so the pair shifts toward O; subtracting backwards, 2.5 − 3.5 = −1.0, only flips the sign, not the direction. C ignored the difference: 1.0 is not zero, so the sharing is unequal, not nonpolar. D read sharing as transfer: two nonmetals 1.0 apart still share the pair, polar covalent, not ionic.

Worked example 3 — solution

Step 1 · Find the electronegativity difference

Worked example 3 — solution

1.9 sits just past the 1.7 guideline. The boundary is not a wall: H and F are both nonmetals, and two nonmetals share.

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: A

B read the boundary as a wall: 1.9 clears 1.7, but P and F are both nonmetals, so they share the pair, polar covalent, not ionic. C put δ− on the wrong atom: F (4.0) pulls harder than P (2.1); subtracting backwards, 2.1 − 4.0 = −1.9, flips only the sign. D called it nonpolar: a difference of 1.9 is far from zero, so the sharing is very unequal.

Worked example 1 — solution

Step 2 · Draw the skeleton

Oxygen in the center, one single bond to each H.

Worked example 1 — solution

The outer atoms are hydrogen. A single bond already fills each H's duet, so no lone pairs are added.

Worked example 1 — solution

The outer atoms are hydrogen. A single bond already fills each H's duet, so no lone pairs are added.

Worked example 1 — the oxygen lone pairs

Four electrons sit in the two O–H bonds; the outer hydrogens are done. Four remain.

Step 4 · Place leftovers on the center

Worked example 1 — the oxygen lone pairs

Four electrons sit in the two O–H bonds; the outer hydrogens are done. Four remain.

Step 4 · Place leftovers on the center

Worked example 2 — solution

Step 1 · Count the valence electrons

Worked example 2 — solution

Step 1 · Count the valence electrons

Nitrogen is least electronegative, so it takes the center, with a single bond to each F.

Worked example 2 — solution

Step 1 · Count the valence electrons

Nitrogen is least electronegative, so it takes the center, with a single bond to each F.

Worked example 2 — the fluorine octets

Six electrons sit in the three N–F bonds. Each fluorine still needs three lone pairs.

Step 3 · Complete the outer octets

Worked example 2 — the fluorine octets

Six electrons sit in the three N–F bonds. Each fluorine still needs three lone pairs.

Step 3 · Complete the outer octets

Worked example 2 — the nitrogen lone pair

Six electrons sit in the three N–F bonds and eighteen fill the fluorine octets. Two remain.

Step 4 · Place leftovers on the center

Worked example 2 — the nitrogen lone pair

Six electrons sit in the three N–F bonds and eighteen fill the fluorine octets. Two remain.

Step 4 · Place leftovers on the center

Practice 1 — answer: A

B counted each element once: 6 + 7 = 13. C counted every electron, core included: 8 + 2(9) = 26. D dropped a pair: 20 − 2 = 18.

Worked example 3 — solution

Step 2 · Draw the skeleton

Carbon centered, a single bond to each oxygen.

Worked example 3 — solution

Step 2 · Draw the skeleton

Carbon centered, a single bond to each oxygen.

Worked example 3 — the outer octets

Step 3 · Complete the outer octets

Worked example 3 — the outer octets

Step 3 · Complete the outer octets

Step 4 · Place leftovers on the center

None remain: 16 − 4 − 12 = 0. The carbon is still two pairs short of an octet.

Worked example 3 — the outer octets

Step 3 · Complete the outer octets

Step 4 · Place leftovers on the center

None remain: 16 − 4 − 12 = 0. The carbon is still two pairs short of an octet.

Worked example 3 — two double bonds

Carbon sits two pairs short of an octet. Pull one lone pair from each oxygen into a bond.

Step 5 · Form multiple bonds

Worked example 3 — two double bonds

Carbon sits two pairs short of an octet. Pull one lone pair from each oxygen into a bond.

Step 5 · Form multiple bonds

Practice 2 — answer: A

B stopped at single bonds: 2 × 2 = 4, leaving carbon short. C formed only one double bond: 4 + 2 = 6, and carbon still lacks two electrons. D reported the whole pool of 16, not just the shared electrons.

Worked example 1 — solution

Step 1 · Count the electron domains

Four bonded hydrogens and no lone pairs on carbon. Together that is 4 + 0 = 4 electron domains.

Worked example 1 — solution

Four domains spread as far apart as possible, to 109.5°. The electron-pair geometry is tetrahedral.

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — solution

A common first attempt

A double bond is one region of electrons, so it counts once. Two double bonds are two domains, not four.

Worked example 2 — solution

Two bonded oxygens and no lone pairs on carbon: 1 + 1 = 2 electron domains.

Worked example 2 — solution

Two domains point straight apart, to 180°. The electron-pair geometry is linear.

Worked example 2 — solution

Worked example 2 — solution

Practice 1 — answer: A

B named the electron-pair geometry, not the shape: the lone pair takes a corner, so the atoms are not tetrahedral. C ignored the lone pair; three bonded atoms alone would be trigonal planar, but the fourth corner is filled. D miscounts: four domains, not three, so the base is a tetrahedron, not a triangle.

Worked example 3 — solution

Step 1 · Count the electron domains

Two bonded oxygens count as two domains: the double bond counts once. Add the lone pair: 2 + 1 = 3 electron domains.

Worked example 3 — solution

Three domains spread to 120°. The electron-pair geometry is trigonal planar.

Worked example 3 — solution

Worked example 3 — solution

Worked example 4 — solution

Step 1 · Count the electron domains

Two bonded hydrogens and two lone pairs on oxygen: 2 + 2 = 4 electron domains.

Worked example 4 — solution

Four domains land on the AX₂E₂ row. The electron-pair geometry is tetrahedral.

Worked example 4 — solution

Worked example 4 — solution

Practice 2 — answer: A

B named the electron-pair geometry: three domains are trigonal planar, but one corner is a lone pair, so the atoms are bent. C ignored the lone pair; two bonded atoms alone would be linear, but the lone pair bends them. D counted the double bond twice, reaching four domains and a trigonal pyramid; each bond is one domain, so there are three.

Worked example 1 — solution

Step 1 · Recall the shape

Carbon has two bonding groups and no lone pairs. VSEPR gives a linear molecule: the two oxygens sit 180° apart.

Worked example 1 — solution

Both arrows point outward, away from carbon, toward the oxygens.

Worked example 1 — solution

The two arrows pull in exactly opposite directions and cancel. No net arrow. Nonpolar.

Worked example 1 — solution

Worked example 2 — solution

Step 1 · Recall the shape

Oxygen has two bonding groups and two lone pairs. VSEPR gives a bent molecule; the two O–H bonds meet at about 104.5°.

Worked example 2 — solution

Both arrows point from the hydrogens up toward the oxygen.

Worked example 2 — solution

The two arrows point the same way, so they reinforce instead of cancel. A net arrow runs through the oxygen. Polar.

Worked example 2 — solution

Practice 1 — answer: A

B took polar bonds as proof of a polar molecule and skipped the shape: the symmetric triangle cancels the three equal arrows. C called the symmetric shape lopsided, but three identical arrows 120° apart balance exactly. D denied the bonds are polar. ΔEN(B–Cl) = 3.16 − 2.04 = 1.12, so each bond is polar, and the molecule is nonpolar because of the shape, not because the bonds are nonpolar.

Worked example 3 — shape and dipoles

Step 1 · Recall the shape

Both are tetrahedral: four bonding groups on carbon, no lone pairs, the outer atoms 109.5° apart.

Worked example 3 — shape and dipoles

Each C–Cl arrow points toward its chlorine; the weaker C–H arrow points toward carbon.

Worked example 3 — shape and dipoles

Worked example 3 — adding the arrows

Step 3 · Add the arrows

Practice 2 — answer: B

A applied the tetrahedral shape as if the four outer atoms were identical. They are not: two chlorines and two hydrogens make the arrangement lopsided, so the arrows do not cancel. C denied the bonds are polar, but ΔEN(C–Cl) = 3.16 − 2.55 = 0.61 and ΔEN(C–H) = 2.55 − 2.20 = 0.35, so every bond is polar. D overgeneralized: CCl₄ is tetrahedral and nonpolar, so the shape alone cannot make a molecule polar. The outer atoms must be unequal.

Practice 2 — answer: B