301. VSEPR theory predicts molecular shape mainly by considering
ⓐ. central valence-shell pair repulsions
ⓑ. only the atomic mass of the central atom
ⓒ. the mass of one mole of the compound only
ⓓ. lattice enthalpy of gaseous ions
Correct Answer: central valence-shell pair repulsions
Explanation: VSEPR stands for valence shell electron pair repulsion. The theory assumes that electron pairs around a central atom arrange themselves to minimize repulsion. Both bond pairs and lone pairs must be counted as electron domains. The arrangement of these domains gives the electron-pair geometry, and the positions of atoms give the molecular shape. This makes VSEPR especially useful for predicting shapes from Lewis structures.
302. In VSEPR theory, the repulsion order among electron pairs is generally
ⓐ. bond pair-bond pair \(\gt \) lone pair-bond pair \(\gt \) lone pair-lone pair
ⓑ. lone pair-bond pair \(\gt \) bond pair-bond pair \(\gt \) lone pair-lone pair
ⓒ. lone pair-lone pair \(\gt \) lone pair-bond pair \(\gt \) bond pair-bond pair
ⓓ. all electron-pair repulsions are always equal
Correct Answer: lone pair-lone pair \(\gt \) lone pair-bond pair \(\gt \) bond pair-bond pair
Explanation: Lone pairs are localized on the central atom and occupy more effective space than bond pairs. Therefore, two lone pairs repel each other more strongly than a lone pair and a bond pair. Bond pair-bond pair repulsion is usually the weakest of the three because bond pairs are pulled between two nuclei. This order explains why lone pairs compress bond angles in molecules such as \(\mathrm{NH_3}\) and \(\mathrm{H_2O}\). VSEPR predictions become more accurate when lone pairs are counted separately from bonding pairs.
303. In VSEPR theory, a double bond around the central atom is counted as
ⓐ. two separate electron domains
ⓑ. four electron domains
ⓒ. one electron domain
ⓓ. no electron domain
Correct Answer: one electron domain
Explanation: In VSEPR theory, an electron domain means a region of electron density around the central atom. A single bond, double bond, or triple bond each occupies one region in space around the central atom. Therefore, a double bond is counted as one electron domain, not two separate domains. This is why \(\mathrm{CO_2}\) has two electron domains around carbon even though each \(\mathrm{C=O}\) bond contains two shared pairs. VSEPR shape prediction uses regions of repulsion, not the total number of bonding electron pairs inside a multiple bond.
304. The central carbon atom in \(\mathrm{CO_2}\) has two electron domains and no lone pair. The molecular shape and bond angle are
ⓐ. bent, \(104.5^\circ\)
ⓑ. linear, \(180^\circ\)
ⓒ. trigonal planar, \(120^\circ\)
ⓓ. tetrahedral, \(109.5^\circ\)
Correct Answer: linear, \(180^\circ\)
Explanation: The Lewis structure of \(\mathrm{CO_2}\) is \(\mathrm{O=C=O}\). Around carbon, there are two regions of electron density, one for each \(\mathrm{C=O}\) double bond. Two electron domains arrange as far apart as possible, giving a linear arrangement. Since there is no lone pair on carbon, the electron-domain geometry and molecular shape are both linear. The \(\mathrm{O-C-O}\) bond angle is therefore \(180^\circ\).
305. For VSEPR type \(\mathrm{AX_3}\) with no lone pair on the central atom, the ideal shape is
ⓐ. linear
ⓑ. trigonal pyramidal
ⓒ. trigonal planar
ⓓ. tetrahedral
Correct Answer: trigonal planar
Explanation: In the notation \(\mathrm{AX_3}\), the central atom \(\mathrm{A}\) is bonded to three surrounding atoms \(\mathrm{X}\). With no lone pair on the central atom, there are three electron domains. Three domains minimize repulsion by arranging in one plane at angles of about \(120^\circ\). This gives a trigonal planar shape. A trigonal pyramidal shape would require three bond pairs and one lone pair, as in \(\mathrm{NH_3}\).
306. Use the table below to identify the VSEPR description that needs correction.
| Molecule | Electron domains around central atom | Molecular shape |
|---|
| P. \(\mathrm{CO_2}\) | \(2\) | linear |
| Q. \(\mathrm{BF_3}\) | \(3\) | trigonal planar |
| R. \(\mathrm{CH_4}\) | \(4\) | tetrahedral |
| S. \(\mathrm{NH_3}\) | \(4\) | trigonal planar |
The row that needs correction is
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: \(\mathrm{CO_2}\) has two electron domains around carbon and is linear. \(\mathrm{BF_3}\) has three bond pairs around boron and is trigonal planar. \(\mathrm{CH_4}\) has four bond pairs around carbon and is tetrahedral. \(\mathrm{NH_3}\) has four electron domains around nitrogen, but one of them is a lone pair. Its electron-pair geometry is tetrahedral, while its molecular shape is trigonal pyramidal, not trigonal planar.
307. In \(\mathrm{CH_4}\), the electron-pair geometry and molecular shape are both tetrahedral because the central carbon has
ⓐ. two bond pairs and two lone pairs
ⓑ. four bond pairs and no lone pair
ⓒ. three bond pairs and one lone pair
ⓓ. four lone pairs and no bond pair
Correct Answer: four bond pairs and no lone pair
Explanation: The Lewis structure of \(\mathrm{CH_4}\) has four \(\mathrm{C-H}\) single bonds. Each bond is one electron domain around carbon. Carbon has no lone pair in this molecule, so all four electron domains correspond to bonded atoms. Four electron domains arrange tetrahedrally to minimize repulsion. Since no lone pair is hidden from the molecular shape, the electron-pair geometry and molecular shape are both tetrahedral.
308. The shape of \(\mathrm{NH_3}\) is trigonal pyramidal rather than tetrahedral because
ⓐ. nitrogen has only two electron domains
ⓑ. all four electron domains are bond pairs
ⓒ. hydrogen forms double bonds with nitrogen
ⓓ. one nitrogen electron domain is a lone pair
Correct Answer: one nitrogen electron domain is a lone pair
Explanation: Nitrogen in \(\mathrm{NH_3}\) has three \(\mathrm{N-H}\) bond pairs and one lone pair. The four electron domains have a tetrahedral electron-pair arrangement. Molecular shape, however, is described using the positions of atoms, not lone pairs. Since only three surrounding hydrogen atoms are visible in the shape, \(\mathrm{NH_3}\) is trigonal pyramidal. The lone pair still affects the bond angle by repelling the bond pairs more strongly.
309. Water has a bent shape because the oxygen atom has
ⓐ. four bond pairs and no lone pair
ⓑ. two bond pairs and two lone pairs
ⓒ. three bond pairs and one lone pair
ⓓ. two lone pairs and no bond pair
Correct Answer: two bond pairs and two lone pairs
Explanation: Oxygen in \(\mathrm{H_2O}\) forms two \(\mathrm{O-H}\) bonds and has two lone pairs. This gives four electron domains around oxygen, so the electron-pair geometry is tetrahedral. The molecular shape is based only on the positions of bonded atoms, so the two hydrogens and oxygen form a bent structure. The two lone pairs compress the \(\mathrm{H-O-H}\) angle below the ideal tetrahedral value. This is why water is not linear even though it has only two atoms bonded to the central atom.
310. The correct decreasing order of bond angle is
ⓐ. \(\mathrm{H_2O}\gt \mathrm{NH_3}\gt \mathrm{CH_4}\)
ⓑ. \(\mathrm{NH_3}\gt \mathrm{CH_4}\gt \mathrm{H_2O}\)
ⓒ. \(\mathrm{CH_4}\gt \mathrm{NH_3}\gt \mathrm{H_2O}\)
ⓓ. \(\mathrm{CH_4}=\mathrm{NH_3}=\mathrm{H_2O}\)
Correct Answer: \(\mathrm{CH_4}\gt \mathrm{NH_3}\gt \mathrm{H_2O}\)
Explanation: \(\mathrm{CH_4}\), \(\mathrm{NH_3}\), and \(\mathrm{H_2O}\) all have four electron domains around the central atom. \(\mathrm{CH_4}\) has no lone pair on carbon, so its angle is close to \(109.5^\circ\). \(\mathrm{NH_3}\) has one lone pair, so its bond angle is reduced to about \(107^\circ\). \(\mathrm{H_2O}\) has two lone pairs, so the angle is reduced further to about \(104.5^\circ\). More lone pairs on the central atom create stronger compression of bond angles.
311. With five electron domains around the central atom and no lone pair, the ideal electron-domain geometry is
ⓐ. trigonal bipyramidal geometry
ⓑ. tetrahedral electron geometry
ⓒ. octahedral electron geometry
ⓓ. square pyramidal geometry
Correct Answer: trigonal bipyramidal geometry
Explanation: Five electron domains around a central atom arrange in a trigonal bipyramidal geometry. This arrangement has three equatorial positions in one plane and two axial positions perpendicular to that plane. The ideal angles include \(120^\circ\) between equatorial positions, \(90^\circ\) between axial and equatorial positions, and \(180^\circ\) between the two axial positions. A molecule such as \(\mathrm{PF_5}\) is commonly described using this arrangement. Tetrahedral geometry corresponds to four domains, while octahedral geometry corresponds to six.
312. In a trigonal bipyramidal electron-domain arrangement, a lone pair prefers an equatorial position because it has
ⓐ. fewer \(90^\circ\) interactions than axial
ⓑ. more \(90^\circ\) interactions than an axial lone pair
ⓒ. no repulsion with any bond pair
ⓓ. a fixed requirement to stay on the heaviest atom
Correct Answer: fewer \(90^\circ\) interactions than axial
Explanation: In a trigonal bipyramidal arrangement, axial positions are at \(90^\circ\) to three equatorial positions. Equatorial positions are at \(90^\circ\) to only two axial positions and at \(120^\circ\) to the other equatorial positions. Since \(90^\circ\) repulsions are stronger than \(120^\circ\) repulsions, a lone pair is more stable in an equatorial position. Lone pairs occupy more effective space than bond pairs, so reducing close repulsions is important. This explains the shapes of molecules such as \(\mathrm{SF_4}\), \(\mathrm{ClF_3}\), and \(\mathrm{XeF_2}\).
313. The VSEPR type \(\mathrm{AX_4E}\) has five electron domains around the central atom. Its molecular shape is commonly
ⓐ. tetrahedral
ⓑ. square planar
ⓒ. seesaw
ⓓ. linear
Correct Answer: seesaw
Explanation: The notation \(\mathrm{AX_4E}\) means four bonded atoms and one lone pair around the central atom. Five total electron domains first arrange in a trigonal bipyramidal electron-domain geometry. The lone pair occupies an equatorial position to reduce \(90^\circ\) repulsions. The remaining four bonded atoms form a seesaw molecular shape. \(\mathrm{SF_4}\) is a common example of this VSEPR type.
314. Use the arrangement described below. A central atom has five electron domains: three bond pairs and two lone pairs. The two lone pairs occupy equatorial positions in a trigonal bipyramidal arrangement. The molecular shape is
ⓐ. T-shaped
ⓑ. trigonal planar
ⓒ. square pyramidal
ⓓ. tetrahedral
Correct Answer: T-shaped
Explanation: Five electron domains give a trigonal bipyramidal electron-domain arrangement. With two lone pairs, the lone pairs prefer equatorial positions to reduce \(90^\circ\) interactions. The remaining three bond pairs occupy two axial positions and one equatorial position. These three bonded atoms form a T-shaped molecular structure. This is the usual VSEPR description for species such as \(\mathrm{ClF_3}\).
315. A central atom has the VSEPR type \(\mathrm{AX_2E_3}\). The molecular shape is
ⓐ. bent
ⓑ. trigonal planar
ⓒ. trigonal pyramidal
ⓓ. linear
Correct Answer: linear
Explanation: \(\mathrm{AX_2E_3}\) has five electron domains around the central atom. The electron-domain geometry is trigonal bipyramidal. The three lone pairs occupy equatorial positions because this minimizes close repulsions. The two bonded atoms then occupy the two axial positions opposite each other. This gives a linear molecular shape, as commonly described for \(\mathrm{XeF_2}\).
316. Six electron domains around a central atom with no lone pair give the ideal molecular shape
ⓐ. trigonal bipyramidal
ⓑ. square planar
ⓒ. bent
ⓓ. octahedral
Correct Answer: octahedral
Explanation: Six electron domains arrange themselves octahedrally to minimize repulsion. In an octahedral arrangement, the six positions are directed toward the corners of an octahedron. Adjacent positions are separated by \(90^\circ\), and opposite positions are separated by \(180^\circ\). If all six domains are bond pairs, the molecular shape is also octahedral. \(\mathrm{SF_6}\) is a standard example with six \(\mathrm{S-F}\) bonds and no lone pair on sulfur.
317. The VSEPR type \(\mathrm{AX_5E}\) is expected to have the molecular shape
ⓐ. trigonal bipyramidal
ⓑ. square pyramidal
ⓒ. seesaw
ⓓ. linear
Correct Answer: square pyramidal
Explanation: \(\mathrm{AX_5E}\) has six electron domains around the central atom: five bond pairs and one lone pair. The electron-domain arrangement is octahedral. Removing one position from an octahedral set as a lone pair leaves five bonded atoms arranged as a square pyramid. The lone pair is not shown as an atom in the molecular shape, but it still affects repulsions. \(\mathrm{IF_5}\) is commonly described as square pyramidal using this model.
318. The shape of \(\mathrm{XeF_4}\) is square planar because xenon has
ⓐ. four bond pairs and no lone pair in a tetrahedral set
ⓑ. three bond pairs and two lone pairs in a trigonal bipyramidal set
ⓒ. two bond pairs and three lone pairs in a linear set
ⓓ. four bond pairs and two lone pairs in an octahedral set
Correct Answer: four bond pairs and two lone pairs in an octahedral set
Explanation: In \(\mathrm{XeF_4}\), xenon has four \(\mathrm{Xe-F}\) bond pairs and two lone pairs. This gives six electron domains, so the electron-domain arrangement is octahedral. The two lone pairs occupy opposite positions to minimize repulsion. The four fluorine atoms then lie in one square plane around xenon. Thus, the molecular shape is square planar, not tetrahedral.
319. Match the VSEPR type with the molecular shape.
| VSEPR type | Molecular shape |
|---|
| P. \(\mathrm{AX_2}\) | 1. linear |
| Q. \(\mathrm{AX_3E}\) | 2. trigonal pyramidal |
| R. \(\mathrm{AX_4E}\) | 3. seesaw |
| S. \(\mathrm{AX_4E_2}\) | 4. square planar |
The proper matching is
ⓐ. P-2, Q-1, R-3, S-4
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-4, Q-2, R-1, S-3
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: \(\mathrm{AX_2}\) with no lone pair has two electron domains and is linear. \(\mathrm{AX_3E}\) has four electron domains with one lone pair, giving a trigonal pyramidal shape. \(\mathrm{AX_4E}\) has five electron domains with one equatorial lone pair, giving a seesaw shape. \(\mathrm{AX_4E_2}\) has six electron domains with two lone pairs opposite each other, giving a square planar shape. The matching depends on total electron domains first, then on which domains are lone pairs.
320. Four surrounding atoms around a central atom lead a student to conclude that the shape must be tetrahedral. The best evaluation is that the conclusion is
ⓐ. always correct because four bonded atoms always mean tetrahedral geometry
ⓑ. incomplete because four bonds can give seesaw or square planar shapes
ⓒ. wrong because four surrounding atoms always give a linear shape
ⓓ. correct only when the central atom has three lone pairs
Correct Answer: incomplete because four bonds can give seesaw or square planar shapes
Explanation: The number of surrounding atoms alone does not fully determine molecular shape. \(\mathrm{AX_4}\) with no lone pair is tetrahedral, as in \(\mathrm{CH_4}\). But \(\mathrm{AX_4E}\) has five electron domains and gives a seesaw shape. \(\mathrm{AX_4E_2}\) has six electron domains and gives a square planar shape. VSEPR predictions require counting both bond pairs and lone pairs around the central atom.
