College Chemistry Quiz: Vsepr And Hybridization

20 questions · exam conditions

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Vsepr And HybridizationQuestion 1 of 20

Which molecular shape results from tetrahedral electron geometry with two lone pairs present?

Linear

Bent

Trigonal planar

Square planar

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College Chemistry Quiz

College Chemistry Quiz: Vsepr And Hybridization

Practice Vsepr And Hybridization in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vsepr And Hybridization, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which molecular shape results from tetrahedral electron geometry with two lone pairs present?

Linear
Bent (correct answer)
Trigonal planar
Square planar

Explanation: This question tests introductory college chemistry skills in understanding VSEPR theory and hybridization. VSEPR theory predicts the shape of molecules based on electron pair repulsion around a central atom, while hybridization explains the mixing of atomic orbitals to form new hybrid orbitals. In this question, tetrahedral electron geometry with two lone pairs results in bent molecular shape. The correct answer is B because it accurately describes the bent shape, reflecting the positions of atoms. Choice A is incorrect because it misapplies linear shape, a common mistake when ignoring lone pair repulsion. To help students, reinforce the connection between electron pair geometry and molecular shape, and practice identifying hybridization states through examples of common molecules. Encourage visualization using molecular models.

Question 2

How many lone pairs are on the central oxygen atom in water (H2O) for VSEPR?

0
1
2 (correct answer)
3

Explanation: This question tests introductory college chemistry skills in understanding VSEPR theory and hybridization. VSEPR theory predicts the shape of molecules based on electron pair repulsion around a central atom, while hybridization explains the mixing of atomic orbitals to form new hybrid orbitals. In this question, water (H2O) demonstrates two lone pairs on oxygen. The correct answer is C because it accurately describes the two lone pairs, contributing to the bent shape. Choice A is incorrect because it misapplies zero lone pairs, a common mistake when confusing with molecules like CO2. To help students, reinforce the connection between electron pair geometry and molecular shape, and practice identifying hybridization states through examples of common molecules. Encourage visualization using molecular models.

Question 3

How many lone pairs are on the central carbon atom in methane (CH4) for VSEPR?

0 (correct answer)
1
2
4

Explanation: This question tests introductory college chemistry skills in understanding VSEPR theory and hybridization. VSEPR theory predicts the shape of molecules based on electron pair repulsion around a central atom, while hybridization explains the mixing of atomic orbitals to form new hybrid orbitals. In this question, methane (CH4) demonstrates zero lone pairs on carbon. The correct answer is A because it accurately describes no lone pairs, leading to tetrahedral shape. Choice C is incorrect because it misapplies two lone pairs, a common mistake when confusing with water. To help students, reinforce the connection between electron pair geometry and molecular shape, and practice identifying hybridization states through examples of common molecules. Encourage visualization using molecular models.

Question 4

A compound $AX_3E_2$ (where E represents a lone pair) has what molecular geometry and bond angle?

Trigonal planar geometry with 120° bond angles
T-shaped geometry with bond angles less than 90° (correct answer)
Trigonal pyramidal geometry with bond angles less than 109.5°
Linear geometry with 180° bond angles
Tetrahedral geometry with 109.5° bond angles

Explanation: When you encounter molecular geometry problems with notation like

AX_3E_2

, you're dealing with VSEPR (Valence Shell Electron Pair Repulsion) theory. The key is understanding that both bonding pairs (X) and lone pairs (E) occupy space around the central atom, but only the bonding pairs determine the final molecular geometry. With

AX_3E_2

, you have 5 total electron pairs around the central atom (3 bonding + 2 lone pairs). These 5 pairs arrange themselves in a trigonal bipyramidal electron geometry to minimize repulsion. However, the lone pairs preferentially occupy the equatorial positions because this minimizes repulsion with other electron pairs. This leaves the three bonding pairs in a T-shaped arrangement. The bond angles are compressed below 90° because lone pairs take up more space than bonding pairs, pushing the bonded atoms closer together. Option A describes trigonal planar geometry, which would be

AX_3

with no lone pairs. Option C suggests trigonal pyramidal (

AX_3E

), which has only one lone pair and bond angles less than 109.5°. Option D describes linear geometry (

AX_2

AX_2E_3

), which is completely different. Option B correctly identifies the T-shaped geometry with compressed bond angles. Study tip: Always count total electron pairs first, then remove lone pairs to find molecular geometry. Remember that lone pairs are "invisible" in the final shape but still influence bond angles by taking up more space than bonding pairs.

Question 5

A molecule has the formula $XF_4$ and exhibits a square planar geometry. What is the hybridization of the central atom X and how many lone pairs are present on X?

$sp^3d^2$ hybridization with 2 lone pairs (correct answer)
$sp^3d$ hybridization with 1 lone pair
$sp^3d^2$ hybridization with 1 lone pair
$sp^3$ hybridization with 2 lone pairs
$sp^3d$ hybridization with 2 lone pairs

Explanation: When you encounter molecular geometry problems, you need to determine both the steric number (bonding + lone pairs) and the actual molecular shape to find the correct hybridization. For

XF_4

with square planar geometry, start by analyzing the structure. Square planar molecules have four bonding pairs arranged in a flat square around the central atom. However, this geometry only occurs when there are also lone pairs present that force this specific arrangement. The key insight is that square planar geometry actually derives from an octahedral electron geometry with two lone pairs. The steric number is 6 (4 bonding pairs + 2 lone pairs), which corresponds to

sp^3d^2

hybridization. The two lone pairs occupy positions opposite each other in the octahedral arrangement, leaving the four fluorine atoms in a square plane. Looking at the incorrect options: Option B suggests

sp^3d

hybridization with 1 lone pair, but this would give trigonal bipyramidal electron geometry and see-saw molecular geometry, not square planar. Option C proposes

sp^3d^2

with only 1 lone pair, which would result in square pyramidal geometry instead. Option D suggests

sp^3

hybridization with 2 lone pairs, but

sp^3

only accommodates 4 electron pairs total, giving tetrahedral electron geometry and bent molecular shape. Therefore, A is correct:

sp^3d^2

hybridization with 2 lone pairs. Study tip: Remember that square planar geometry is a "reduced" octahedral arrangement. When you see square planar, immediately think octahedral electron geometry minus two opposite lone pairs, which means

sp^3d^2

hybridization.

Question 6

Which molecule has a linear molecular geometry but does not have $sp$ hybridization at the central atom?

$BeF_2$
$CO_2$
$XeF_2$ (correct answer)
$C_2H_2$
$HCN$

Explanation: This question tests your understanding of the relationship between molecular geometry and hybridization, specifically looking for cases where they don't follow the typical pattern. To solve this, you need to analyze both the molecular geometry and the hybridization state of each central atom. Linear geometry can arise from different hybridization states depending on the number of electron domains around the central atom.

XeF_2

(choice C) has a linear molecular geometry because the central xenon atom has 5 electron domains: 2 bonding pairs (to the fluorine atoms) and 3 lone pairs. These arrange in a trigonal bipyramidal electron geometry with

sp^3d

hybridization, but the molecular geometry is linear because the lone pairs occupy the equatorial positions, leaving the two fluorine atoms in a straight line. Choice A (

BeF_2

) is linear with

sp

hybridization at beryllium, which has only 2 electron domains. Choice B (

CO_2

) is also linear with

sp

hybridization at carbon, involving two double bonds that count as 2 electron domains total. Choice D (

C_2H_2

) has

sp

hybridization at each carbon atom, creating the linear geometry around the triple bond. The key insight is recognizing that linear geometry doesn't always mean

sp

hybridization. When lone pairs are present, they can force a linear arrangement even with higher hybridization states. Always count all electron domains (bonding and non-bonding) to determine hybridization, then consider how lone pairs affect the final molecular shape.

Question 7

Consider the molecules $PCl_5$ and $SF_4$ . How do their molecular geometries compare?

Both have trigonal bipyramidal molecular geometry
$PCl_5$ is trigonal bipyramidal and $SF_4$ is seesaw (correct answer)
Both have seesaw molecular geometry
$PCl_5$ is seesaw and $SF_4$ is trigonal bipyramidal
$PCl_5$ is octahedral and $SF_4$ is tetrahedral

Explanation: When you encounter molecular geometry questions, you need to determine both the electron geometry (based on all electron pairs) and the molecular geometry (based on only atoms). The key is using VSEPR theory to count bonding pairs and lone pairs around the central atom. For

PCl_5

: Phosphorus has 5 valence electrons and forms 5 bonds with chlorine atoms, using all its electrons with no lone pairs remaining. This gives 5 electron pairs total, creating a trigonal bipyramidal electron geometry. Since all electron pairs are bonding pairs, the molecular geometry is also trigonal bipyramidal. For

SF_4

: Sulfur has 6 valence electrons and forms 4 bonds with fluorine atoms, leaving 2 electrons as 1 lone pair. This gives 5 electron pairs total (4 bonding + 1 lone), so the electron geometry is trigonal bipyramidal. However, the molecular geometry only considers atom positions, not lone pairs. With one lone pair occupying an equatorial position, the molecular shape becomes seesaw. Looking at the wrong answers: Choice A incorrectly assumes

SF_4

ignores its lone pair. Choice C wrongly applies the seesaw geometry to both molecules, missing that

PCl_5

has no lone pairs. Choice D completely reverses the geometries of both molecules. The correct answer is B:

PCl_5

is trigonal bipyramidal and

SF_4

is seesaw. Study tip: Always count lone pairs carefully—they determine electron geometry but are "invisible" in molecular geometry, often creating shapes with specific names like seesaw, T-shaped, or square pyramidal.

Question 8

In the molecule $ClF_5$ , the axial Cl-F bonds are longer than the equatorial Cl-F bonds because:

Axial fluorines experience more electron-electron repulsion from equatorial fluorines (correct answer)
Equatorial fluorines are more electronegative than axial fluorines
The lone pair preferentially interacts with axial fluorines
Axial positions require different hybridization than equatorial positions
The octahedral geometry naturally creates unequal bond lengths

Explanation: When analyzing molecular geometry and bond lengths, you need to consider how electron-electron repulsions affect the spatial arrangement and bonding in molecules.

ClF_5

has a square pyramidal geometry with one lone pair and five bonding pairs around the central chlorine atom. The axial Cl-F bonds are indeed longer because axial fluorines experience more electron-electron repulsion from equatorial fluorines. In this geometry, each axial fluorine is positioned at 90° angles to three equatorial fluorines, creating significant repulsive interactions. The equatorial fluorines, however, are at 90° angles to only two other fluorines (the axial ones), experiencing less overall repulsion. This increased repulsion around the axial positions weakens and lengthens those bonds. Looking at the incorrect options: Option B is wrong because electronegativity is an intrinsic atomic property that doesn't change based on molecular position. Option C incorrectly focuses on the lone pair's interaction with axial fluorines, but the primary effect is the fluorine-fluorine repulsions, not lone pair interactions. Option D misunderstands hybridization—all positions in

ClF_5

use the same

sp^3d^2

hybrid orbitals from the central atom. Remember this pattern: in molecules with lone pairs or when comparing axial versus equatorial positions, always consider electron-electron repulsion as the driving force behind bond length differences. The more crowded the electron environment around a bond, the longer and weaker that bond becomes.

Question 9

Which of the following molecules exhibits both $sp^3$ hybridization and a bent molecular geometry?

$SO_2$
$H_2O$ (correct answer)
$BF_2H$
$NO_2^-$
$CO_2$

Explanation: When analyzing molecular structure, you need to determine both the hybridization of the central atom and the molecular geometry, which depends on both bonding and lone pairs around that atom. For

H_2O

, the oxygen atom has 6 valence electrons. It forms two bonds with hydrogen atoms and has two lone pairs, giving it four electron domains total. Four electron domains require

sp^3

hybridization. However, the molecular geometry only considers the positions of atoms, not lone pairs. With two bonding pairs and two lone pairs, water adopts a bent geometry as the lone pairs push the hydrogen atoms closer together (bond angle ~104.5°). Let's examine why the other options don't work. Option A,

SO_2

, does have bent geometry due to a lone pair on sulfur, but sulfur uses

sp^2

hybridization (three electron domains: two bonding, one lone pair). Option C,

BF_2H

, has

sp^3

hybridization around boron (four bonding pairs), but this creates a tetrahedral geometry, not bent. Option D,

NO_2^-

, is bent due to a lone pair on nitrogen, but nitrogen uses

sp^2

hybridization (three electron domains total). Remember that hybridization depends on the total number of electron domains (bonding + lone pairs), while molecular geometry depends on atom positions only. Look for molecules where the central atom has exactly four electron domains but fewer than four bonding pairs to find

sp^3

hybridization with non-tetrahedral geometry.

Question 10

A student claims that $PF_5$ and $ClF_5$ have the same molecular geometry because they both have 5 fluorine atoms bonded to the central atom. Evaluate this claim.

Correct, both have trigonal bipyramidal molecular geometry
Incorrect, $PF_5$ is trigonal bipyramidal while $ClF_5$ is square pyramidal (correct answer)
Incorrect, $PF_5$ is square pyramidal while $ClF_5$ is trigonal bipyramidal
Correct, both have square pyramidal molecular geometry
Incorrect, both have different hybridizations leading to different geometries

Explanation: When determining molecular geometry, you need to consider both bonding pairs AND lone pairs of electrons around the central atom, not just the number of bonded atoms. The key is applying VSEPR (Valence Shell Electron Pair Repulsion) theory correctly. For

PF_5

: Phosphorus has 5 valence electrons, all of which form bonds with the 5 fluorine atoms. This gives 5 bonding pairs and 0 lone pairs around phosphorus, creating a trigonal bipyramidal molecular geometry. For

ClF_5

: Chlorine has 7 valence electrons. Five form bonds with fluorine atoms, leaving 2 electrons as one lone pair. This gives 5 bonding pairs and 1 lone pair around chlorine. The electron geometry is octahedral, but the molecular geometry (considering only atom positions) is square pyramidal because the lone pair occupies one position. Answer choice A is wrong because while

PF_5

is trigonal bipyramidal,

ClF_5

is not. Answer choice C reverses the correct geometries. Answer choice D is incorrect because

PF_5

is not square pyramidal. Answer choice B correctly identifies that

PF_5

is trigonal bipyramidal while

ClF_5

is square pyramidal, making the student's claim incorrect. Remember: Always count both bonding AND lone pairs when determining molecular geometry. The number of atoms bonded to the central atom alone doesn't determine molecular shape—lone pairs significantly affect geometry by taking up space around the central atom.

Question 11

In a molecule with $sp^3d$ hybridization, lone pairs preferentially occupy which positions?

Axial positions only
Equatorial positions only (correct answer)
Any position with equal probability
Axial positions first, then equatorial positions
The position depends on the identity of the central atom

Explanation: When you encounter questions about molecular geometry and lone pair placement, you need to understand that electron pairs arrange themselves to minimize repulsion according to VSEPR theory, with lone pairs requiring more space than bonding pairs. In

sp^3d

hybridization, you have a trigonal bipyramidal electron geometry with five positions: three equatorial (in a triangular plane) and two axial (above and below the plane). The key insight is that equatorial positions are separated by 120° angles, while axial positions are only 90° apart from equatorial positions. Since lone pairs occupy more space and create stronger repulsions than bonding pairs, they preferentially occupy equatorial positions where they have maximum separation from other electron pairs. Looking at the wrong answers: Choice A suggests lone pairs prefer axial positions, which is incorrect because the 90° angles create more crowding and repulsion. Choice C implies random placement, ignoring the fundamental principle that electron arrangements minimize energy through optimal spacing. Choice D reverses the actual preference order – lone pairs avoid axial positions rather than filling them first. This preference is so strong that even in molecules like

SF_4

(which has

sp^3d

hybridization with one lone pair), the lone pair always occupies an equatorial position, creating a see-saw molecular shape rather than placing the lone pair axially. Study tip: Remember "LEP" – Lone pairs prefer Equatorial Positions in trigonal bipyramidal geometries. This pattern appears frequently in molecular geometry problems and directly affects predicted molecular shapes.

Question 12

The Lewis structure of $SO_2$ shows sulfur with one lone pair. Based on VSEPR theory, what is the molecular geometry and approximate bond angle?

Linear geometry with 180° bond angle
Bent geometry with bond angle less than 120° (correct answer)
Trigonal planar geometry with 120° bond angles
Bent geometry with bond angle greater than 120°
Tetrahedral geometry with 109.5° bond angles

Explanation: When you encounter molecular geometry questions, you need to combine Lewis structures with VSEPR (Valence Shell Electron Pair Repulsion) theory to predict three-dimensional shapes. For

SO_2

, start by drawing the Lewis structure. Sulfur has 6 valence electrons, and each oxygen contributes 6, giving 18 total electrons. After forming two S=O double bonds, sulfur retains one lone pair. This gives you three electron domains around sulfur: two bonding pairs and one lone pair. VSEPR theory tells us that electron pairs repel each other and arrange to minimize repulsion. With three electron domains, the electron geometry is trigonal planar, but the molecular geometry only considers atoms, not lone pairs. Since one domain is a lone pair, the molecular shape is bent. Crucially, lone pairs occupy more space than bonding pairs, compressing the O-S-O bond angle below the ideal 120° trigonal planar angle to approximately 119°. Answer A is wrong because linear geometry requires only two electron domains around the central atom. Answer C incorrectly describes the electron geometry rather than molecular geometry—trigonal planar ignores the effect of the lone pair on the actual shape. Answer D fails because lone pair repulsion compresses bond angles below, not above, the ideal trigonal planar angle. Remember this pattern: when lone pairs are present, they always reduce bond angles from ideal values due to their greater spatial requirements. Practice identifying electron domains first, then determine how lone pairs affect the final molecular shape.

Question 13

The hybridization of the central atom in $ICl_4^-$ is:

$sp^3$
$sp^3d$
$sp^3d^2$ (correct answer)
$sp^2$
$sp$

Explanation: When you encounter hybridization questions, you need to determine the electron geometry around the central atom by counting both bonding pairs and lone pairs of electrons. For

ICl_4^-

, start by drawing the Lewis structure. Iodine has 7 valence electrons, each chlorine contributes 1 electron (4 total), and the negative charge adds 1 more electron, giving you 12 total electrons or 6 electron pairs around iodine. Four pairs form bonds with the chlorine atoms, leaving 2 lone pairs on the central iodine atom. With 6 electron pairs total (4 bonding + 2 lone pairs), the electron geometry is octahedral, which requires

sp^3d^2

hybridization. This uses one s orbital, three p orbitals, and two d orbitals from iodine to accommodate all six electron pairs. Choice A (

sp^3

) only accounts for 4 electron pairs in a tetrahedral arrangement - this would work if iodine had no lone pairs, but it has two. Choice B (

sp^3d

) accommodates 5 electron pairs in a trigonal bipyramidal geometry, which falls short of the 6 pairs present. Choice D (

sp^2

) only handles 3 electron pairs in a trigonal planar arrangement, far too few for this ion. Remember this pattern: count all electron pairs around the central atom (bonding and lone pairs), then match to the appropriate hybridization. Four pairs =

sp^3

, five pairs =

sp^3d

, six pairs =

sp^3d^2

. Don't let the molecular shape distract you - hybridization depends on electron geometry, not molecular geometry.

Question 14

The hybridization of carbon in $CO_2$ is $sp$ , while the hybridization of sulfur in $SO_2$ is $sp^2$ . This difference is primarily due to:

Carbon being smaller than sulfur
Different numbers of lone pairs on the central atoms (correct answer)
Oxygen being more electronegative when bonded to carbon
Sulfur having access to d-orbitals while carbon does not
Different formal charges on the central atoms

Explanation: When determining molecular geometry and hybridization, you need to count both bonding pairs and lone pairs around the central atom, as both occupy space and influence the electron geometry. In

CO_2

, carbon forms two double bonds with oxygen atoms and has no lone pairs. This gives carbon a total of 2 electron domains (each double bond counts as one domain). With 2 electron domains arranged linearly, carbon uses

sp

hybridization. In

SO_2

, sulfur forms two double bonds with oxygen atoms but also has one lone pair. This gives sulfur 3 electron domains total. With 3 electron domains arranged in a trigonal planar electron geometry, sulfur uses

sp^2

hybridization. The molecular shape is bent due to the lone pair, but the hybridization is determined by the total electron domains. Choice A is incorrect because atomic size doesn't directly determine hybridization—electron domain geometry does. Choice C is wrong because oxygen's electronegativity affects bond polarity, not the hybridization scheme of the central atom. Choice D represents a common misconception; while sulfur can access d-orbitals in some compounds,

SO_2

doesn't require d-orbital participation. The

sp^2

hybridization explains its geometry perfectly using only s and p orbitals. Remember: always count total electron domains (bonding + lone pairs) around the central atom first. This determines electron geometry and hybridization, regardless of what atoms are bonded or their relative sizes.

Question 15

Which of the following molecules has a trigonal pyramidal molecular geometry and $sp^3$ hybridization at the central atom?

$BCl_3$
$SO_3$
$NH_3$ (correct answer)
$BF_3$
$CO_3^{2-}$

Explanation: When determining molecular geometry, you need to consider both the electron geometry around the central atom and the actual arrangement of atoms in space. The key is using VSEPR theory to count electron domains (bonding pairs plus lone pairs) and then identifying the molecular shape. For

NH_3

(ammonia), nitrogen has 5 valence electrons. It forms three single bonds with hydrogen atoms and has one lone pair of electrons. This gives us 4 electron domains around the central nitrogen atom. Four electron domains correspond to

sp^3

hybridization and a tetrahedral electron geometry. However, since one of these domains is a lone pair (not a bonding pair), the molecular geometry is trigonal pyramidal - the three hydrogen atoms form a pyramid with nitrogen at the apex. Option A (

BCl_3

) has boron with 3 electron domains (no lone pairs), creating trigonal planar geometry with

sp^2

hybridization. Option B (

SO_3

) also has trigonal planar geometry in its most common form, though sulfur can exhibit

sp^2

hybridization. Option D (

BF_3

) is identical to

BCl_3

in geometry - trigonal planar with

sp^2

hybridization at boron. The crucial distinction is that trigonal pyramidal geometry only occurs when you have 4 electron domains with one lone pair, while trigonal planar occurs with 3 electron domains and no lone pairs. Study tip: Always count lone pairs when determining molecular geometry - they occupy space and change the shape even though they don't involve additional atoms.

Question 16

A molecule $AX_2E_3$ has what molecular geometry and what hybridization at atom A?

Bent geometry with $sp^3d$ hybridization
Linear geometry with $sp^3d$ hybridization (correct answer)
T-shaped geometry with $sp^3d^2$ hybridization
Trigonal planar geometry with $sp^2$ hybridization
Linear geometry with $sp$ hybridization

Explanation: When you encounter molecular geometry problems with notation like

AX_2E_3

, you're working with VSEPR theory. The subscripts tell you the electron domain geometry: A is the central atom, X represents bonding pairs, and E represents lone pairs of electrons. For

AX_2E_3

, atom A has 2 bonding pairs and 3 lone pairs, giving 5 total electron domains around the central atom. With 5 electron domains, the electron domain geometry is trigonal bipyramidal, requiring

sp^3d

hybridization. However, molecular geometry describes only the arrangement of atoms, not lone pairs. When you have 2 bonding pairs and 3 lone pairs in a trigonal bipyramidal arrangement, the lone pairs occupy the three equatorial positions (they're bulkier and prefer the roomier equatorial spots). This leaves the two bonding pairs in axial positions, creating a linear molecular geometry with a 180° bond angle. Choice A incorrectly suggests bent geometry, which occurs with different lone pair arrangements. Choice C proposes T-shaped geometry (which happens with

AX_3E_2

) and wrong hybridization -

sp^3d^2

corresponds to 6 electron domains, not 5. Choice D gives trigonal planar geometry with

sp^2

hybridization, which would be correct for

AX_3E_0

(3 electron domains), not our 5-domain system. Remember: always count total electron domains first to determine hybridization, then consider how lone pairs affect the molecular geometry. Lone pairs "hide" in the final shape but still influence hybridization.

Question 17

A central atom with $sp^3d^2$ hybridization could exhibit which of the following molecular geometries?

Only octahedral geometry
Only square planar geometry
Octahedral, square pyramidal, or square planar geometries (correct answer)
Trigonal bipyramidal or seesaw geometries
Tetrahedral or trigonal pyramidal geometries

Explanation: When you encounter questions about hybridization and molecular geometry, remember that the same hybridization can produce multiple geometries depending on how many lone pairs are present on the central atom.

sp^3d^2

hybridization involves six hybrid orbitals arranged octahedrally around the central atom. However, the actual molecular geometry depends on how these orbitals are occupied. If all six positions contain bonding pairs, you get octahedral geometry. If one position has a lone pair, the five bonding pairs arrange in square pyramidal geometry. If two positions have lone pairs (positioned opposite each other to minimize repulsion), the four bonding pairs form square planar geometry. Let's examine each option: Choice A incorrectly limits

sp^3d^2

to only octahedral geometry, ignoring the effect of lone pairs. Choice B makes the opposite error, restricting it to only square planar geometry. Choice D confuses

sp^3d^2

with

sp^3d

hybridization—trigonal bipyramidal and seesaw geometries arise from

sp^3d

hybridization with five hybrid orbitals, not six. Choice C correctly identifies that

sp^3d^2

hybridization can produce octahedral (6 bonding pairs), square pyramidal (5 bonding pairs, 1 lone pair), or square planar (4 bonding pairs, 2 lone pairs) geometries. Study tip: Always consider lone pairs when predicting molecular geometry from hybridization. The hybridization tells you the arrangement of electron domains, but lone pairs "hide" in the geometry, creating different molecular shapes than you might initially expect.

Question 18

The bond angles in $NH_3$ (107°) are smaller than those in $CH_4$ (109.5°) because:

Nitrogen is more electronegative than carbon, causing electron pairs to be pulled closer
The lone pair on nitrogen occupies more space than a bonding pair, compressing the bonding pairs (correct answer)
Nitrogen forms weaker bonds than carbon, allowing more flexibility in bond angles
The hydrogen atoms in $NH_3$ are larger than those in $CH_4$
Nitrogen uses different hybridization than carbon in these compounds

Explanation: When you encounter questions about molecular geometry and bond angles, you need to apply VSEPR (Valence Shell Electron Pair Repulsion) theory, which explains how electron pairs around a central atom arrange themselves to minimize repulsion. Both

NH_3

and

CH_4

have four electron pairs around their central atoms, but they differ in composition.

CH_4

has four bonding pairs arranged in a perfect tetrahedral geometry (109.5°).

NH_3

has three bonding pairs and one lone pair, creating a trigonal pyramidal shape with compressed bond angles (107°). The key insight is that lone pairs occupy more space than bonding pairs because they're attracted to only one nucleus instead of being shared between two atoms. This extra spatial requirement forces the lone pair in

NH_3

to "squeeze" the three bonding pairs closer together, reducing the bond angles from the ideal tetrahedral angle. Answer B correctly identifies this lone pair repulsion effect. Looking at the incorrect options: A misunderstands electronegativity's role—while nitrogen is more electronegative, this doesn't directly cause the angle compression. C incorrectly suggests bond strength affects geometry in this way; both molecules have stable covalent bonds. D makes no sense because the hydrogen atoms are identical in both compounds. Study tip: Remember the VSEPR hierarchy of repulsion: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. Whenever you see molecules with the same number of electron pairs but different numbers of lone pairs, expect the lone pairs to compress bond angles.

Question 19

Which statement about hybridization and molecular geometry is correct for $ClF_3$ ?

$sp^3$ hybridization with trigonal pyramidal geometry
$sp^3d$ hybridization with T-shaped geometry (correct answer)
$sp^3d^2$ hybridization with octahedral geometry
$sp^2$ hybridization with trigonal planar geometry
$sp^3d$ hybridization with trigonal bipyramidal geometry

Explanation: When you encounter molecular geometry questions, you need to determine both the hybridization and molecular shape by analyzing the central atom's electron domain geometry and accounting for lone pairs. For

ClF_3

, start with chlorine as the central atom. Chlorine has 7 valence electrons, and each fluorine contributes 1 electron for bonding, giving you 3 bonding pairs. This leaves 4 electrons (2 lone pairs) on chlorine. The total is 5 electron domains around the central atom: 3 bonding pairs + 2 lone pairs. Five electron domains require

sp^3d

hybridization, which creates a trigonal bipyramidal electron geometry. However, molecular geometry describes only the arrangement of atoms, not lone pairs. The two lone pairs occupy equatorial positions (to minimize repulsion), leaving the three fluorine atoms in a T-shaped arrangement. Choice A incorrectly suggests

sp^3

hybridization, which only accommodates 4 electron domains, not the 5 needed here. Choice C proposes

sp^3d^2

hybridization, which would require 6 electron domains and create an octahedral electron geometry—this is overkill for

ClF_3

. Choice D suggests

sp^2

hybridization with trigonal planar geometry, which completely ignores the lone pairs and undercounts the electron domains. The correct answer is B:

sp^3d

hybridization with T-shaped geometry. Study tip: Always count total electron domains first (bonding + lone pairs) to determine hybridization, then consider only the atomic positions for molecular geometry. Lone pairs significantly affect molecular shape by occupying space while remaining invisible in the final geometry description.

Question 20

Consider the series: $CF_4$ , $NF_3$ , $OF_2$ . How does the F-X-F bond angle change across this series?

Bond angles increase from $CF_4$ to $OF_2$
Bond angles decrease from $CF_4$ to $OF_2$ (correct answer)
Bond angles remain approximately constant across the series
$NF_3$ has the largest bond angle in the series
$OF_2$ has the smallest bond angle in the series

Explanation: When you encounter questions about bond angles in molecular series, you need to apply VSEPR (Valence Shell Electron Pair Repulsion) theory, which predicts molecular geometry based on electron pair arrangements around the central atom. Let's analyze each molecule's geometry.

CF_4

has 4 bonding pairs and 0 lone pairs around carbon, giving it a tetrahedral geometry with F-C-F bond angles of 109.5°.

NF_3

has 3 bonding pairs and 1 lone pair around nitrogen, creating a trigonal pyramidal shape.

OF_2

has 2 bonding pairs and 2 lone pairs around oxygen, forming a bent geometry. The key insight is that lone pairs occupy more space than bonding pairs because they're attracted only to one nucleus instead of being shared between two. This causes lone pairs to compress the bond angles between bonding pairs. In

NF_3

, the lone pair reduces the F-N-F angle to about 102°. In

OF_2

, the two lone pairs compress the F-O-F angle even further to about 103°, but the overall trend shows decreasing angles as lone pairs increase. Answer B is correct because bond angles decrease across the series: 109.5° → ~102° → ~103°. Answer A incorrectly suggests angles increase. Answer C is wrong because the angles change significantly due to varying numbers of lone pairs. Answer D is incorrect because

CF_4

has the largest bond angle at 109.5°, not

NF_3

. Study tip: Remember that lone pairs are "bulkier" than bonding pairs and compress bond angles. Count lone pairs first to predict how much compression will occur.