Question bank — VSEPR theory — geometry from electron pairs (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, etc

True — a multiple bond is one fat domain pointing in one direction, so it counts as one region of density; only the number of directions shapes the molecule.

False — they are identical whenever the central atom has zero lone pairs (e.g. $C{H}_4$, $B{F}_3$, $S{F}_6$), because there are no unseen domains to hide.

False — that is only true with no lone pairs; $N{H}_3$ ($107^{\circ }$) and $H_{2}O$ ($104.5^{\circ }$) share tetrahedral electron geometry but lone pairs squeeze the angle below $109.5^{\circ }$.

False — $C{H}_4$, $N{H}_3$ and $H_{2}O$ all have $SN=4$ but give tetrahedral, pyramidal and bent shapes respectively, because they carry different numbers of lone pairs.

True — a lone pair is held by only one nucleus, so its cloud sits fatter and closer to the central atom and pushes harder, giving the order lone–lone $>$ lone–bond $>$ bond–bond.

False — VSEPR reads lone pairs on the central atom (carbon, which has none here); the oxygens' lone pairs don't bend the O=C=O backbone, so it is linear.

True — $SN=2+0=2$, the two bonding domains point opposite for maximum separation, giving a straight $180^{\circ }$ H–Be–H line just like $C{O}_2$.

False — a single unpaired electron acts as its own domain (a slightly thinner lone pair), so $N{O}_2$ has $SN=3$ (2 bonds + 1 single-electron domain) and comes out bent, wider than $O_3$ because a lone electron pushes less than a full lone pair.

False — you must maximise the minimum separation of all domains at once; you cannot grow one angle without shrinking others, so the true optimum is a balance, not "biggest possible single angle".

Error: the lone pair on N was ignored. $SN=3+1=4$, so electron geometry is tetrahedral and the molecular shape is trigonal pyramidal near $107^{\circ }$.

Error: all four electrons of the double bond point the same way, so they form one domain — count directions, not electrons.

Error: $SN=4$ gives tetrahedral electron geometry, but molecular geometry names only the atoms; with two lone pairs the visible shape is bent/V-shaped at $104.5^{\circ }$.

Error: an axial site has three neighbours at the crowded $90^{\circ }$, an equatorial site only two, so the fat lone pair goes equatorial to dodge the worst repulsions.

Error: Xe keeps 2 lone pairs, so $SN=4+2=6$ (octahedral electron geometry); the two lone pairs go trans and the atoms form a square planar shape.

Error: boron has no lone pair ($SN=3$), so it stays flat trigonal planar at $120^{\circ }$; nitrogen's extra lone pair is what pyramidalises $N{H}_3$.

Error: $C{O}_2$ has zero central-atom domains besides its bonds, but $N{O}_2$ carries one unpaired electron as a third domain, so $SN=3$ and the molecule is bent, not linear.

Error: you cannot fit four symmetric directions at $90^{\circ }$; the true maximum-separation solution is a tetrahedron at $109.5^{\circ }$, a larger angle and lower repulsion.

Error: period-3 atoms like S can hold expanded shells, giving $S{F}_6$ with $SN=6$ (octahedral); the octet limit is not universal.

Water has two lone pairs versus ammonia's one, so it delivers more lone-pair squeezing on the bonds, pushing its angle further below the $109.5^{\circ }$ ideal.

A bonding pair is pulled taut between two nuclei, but a lone pair is anchored by only the central nucleus, so its cloud spreads wider and sits closer, repelling more.

It is still one region of density occupying its own direction, so it must be counted; but one electron carries less charge than a pair, so it pushes neighbouring bonds slightly less hard.

All domains sit at roughly the same distance from the central atom, so their tips lie on a sphere; spreading them for maximum mutual separation is the geometry problem VSEPR solves.

With $SN=5$ and one lone pair placed equatorially (least crowded), the remaining four atoms are forced into two axial and two equatorial positions, giving the lopsided see-saw.

$C{O}_2$ is linear, so its two equal bond dipoles point exactly opposite and cancel; water is bent, so its dipoles add to a net moment — geometry decides polarity, a link to Bond Polarity and Dipole Moment.

Shape is set by where domains point, and each direction is one repelling region; two clouds along the same axis (a double bond) still push in only one direction.

Trans positions are $180^{\circ }$ apart, the farthest an octahedron allows, which minimises the strongest lone–lone repulsion.

Both are linear at $180^{\circ }$: two bonding domains repel by pointing exactly opposite, and this is the genuine multi-atom case where "linear" is a real, angle-bearing prediction (unlike a diatomic).

The electron geometry is trigonal planar, but with one lone pair the two bonded atoms make a bent shape near $120^{\circ }$ (slightly less, since the lone pair squeezes them).

Yes, bent; because a lone electron repels less than a lone pair, $N{O}_2$ opens wider (~$134^{\circ }$) than $O_3$'s pair-squeezed ~$117^{\circ }$.

Both are "bent", but they come from different electron geometries — trigonal planar gives the wider ~$120^{\circ }$ opening, tetrahedral gives the tighter ~$104.5^{\circ }$ opening.

All three lone pairs take the roomy equatorial belt, forcing the two F atoms to the axial line $180^{\circ }$ apart, so the atoms come out linear.

Trivially yes but uselessly: two points can only lie on a line, so any diatomic is "linear" by default with no angle to predict.

There is only a single bonded atom, so there is no angle to describe. By the strict convention VSEPR reports no meaningful molecular shape; some texts still label it "linear" since two points always lie on a line — but that says nothing about geometry, so treat it as degenerate. This differs from $Be{H}_2$, where two bonded atoms make "linear" a genuine result.

Its two equatorial lone pairs press on the axial bonds, bending them slightly toward each other so the real angle dips just below the ideal $90^{\circ }$.

No — molecular geometry names atom positions, and with no bonded atoms there are no atoms to arrange, so the concept doesn't apply.

No within a fixed steric number — extra lone-pair repulsion only closes bond angles; the ideal angle can only rise by lowering SN (fewer domains, wider spread).