Water (H₂O) has four electron groups around oxygen (two bonding pairs and two lone pairs). What is the correct sequence of electron geometry → molecular geometry?
ABent electron geometry → tetrahedral molecular geometry
BTetrahedral electron geometry → bent molecular geometry
CTetrahedral electron geometry → tetrahedral molecular geometry
DTrigonal planar electron geometry → bent molecular geometry
Four electron groups produce tetrahedral electron geometry (the arrangement minimizing repulsion among four groups). But molecular geometry describes only where the atoms are — two O–H bonds are visible while two lone pairs are not. Since we see only two atoms attached to oxygen, the molecular geometry is bent. This is the essential VSEPR distinction: electron geometry counts everything; molecular geometry counts only atom positions. Confusing the two is the most common VSEPR error.
Question 2 Multiple Choice
Carbon dioxide (CO₂) has the Lewis structure O=C=O with two double bonds. What is the molecular geometry?
ABent — the two double bonds create more electron density and repel each other more strongly
BTrigonal planar — there are two bonding regions plus carbon's lone pair
CLinear — each double bond counts as one electron group, giving two total groups at 180°
DTetrahedral — each double bond contains four electrons, totaling eight electrons around carbon
The critical VSEPR rule: a double bond counts as ONE electron group, regardless of how many electron pairs it contains. CO₂ has two electron groups (two C=O double bonds, no lone pairs on carbon), giving linear geometry (180°). Choosing 'bent' is the classic error from incorrectly thinking double bonds count as two groups or that more electron density always means bending. Compare with SO₂, where a lone pair on sulfur is present, making it bent.
Question 3 True / False
In VSEPR theory, a triple bond counts as three electron groups because it contains three pairs of electrons.
TTrue
FFalse
Answer: False
A triple bond counts as ONE electron group. An electron group is a region of electron density, regardless of how many pairs occupy it. The σ and π electrons in a triple bond all occupy the same general region between two atoms, so they count as a single group. For example, acetylene (HC≡CH) has two electron groups around each carbon (the triple bond plus one single bond), giving linear geometry. Counting triple bonds as three groups would incorrectly predict bent geometry for linear molecules.
Question 4 True / False
Ammonia (NH₃) and boron trifluoride (BF₃) both have three bonds to the central atom, so they is expected to have the same molecular geometry.
TTrue
FFalse
Answer: False
BF₃ has three bonding pairs and no lone pairs on boron (boron has 3 valence electrons, all used in bonds), giving trigonal planar geometry (120°). NH₃ has three bonding pairs plus one lone pair on nitrogen (5 valence electrons, 3 in bonds, 1 remaining as lone pair), giving trigonal pyramidal geometry (~107°). Lone pairs are invisible in molecular geometry naming but they are real electron groups that push bonding pairs together. Ignoring lone pairs when counting groups is the fundamental VSEPR error.
Question 5 Short Answer
Why do water (H₂O) and methane (CH₄) both have tetrahedral electron geometry but different bond angles (104.5° vs 109.5°)? Explain using VSEPR principles.
Think about your answer, then reveal below.
Model answer: Both have four electron groups, giving tetrahedral electron geometry. In methane, all four groups are identical bonding pairs, which repel equally and settle at the ideal 109.5°. In water, two of the four groups are lone pairs, which spread out more than bonding pairs (no second nucleus confines them) and exert greater repulsion on neighboring groups. The two lone pairs push the two O–H bonds closer together, compressing the H–O–H angle from the ideal 109.5° to 104.5°.
The compression is predictable and progressive: each lone pair substituted for a bonding pair compresses bond angles slightly. NH₃ (one lone pair) has angles of ~107°, intermediate between CH₄ (109.5°) and H₂O (104.5°). This regular trend follows directly from the lone-pair repulsion principle.