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WorksheetsChapter 11 Introduction & VSEPR Theory Basics
Total questions: 115
Worksheet time: 58mins
Which statement best describes the basis of the VSEPR approach?
Electron groups around a central atom arrange to maximize attraction.
Electron groups around a central atom arrange to minimize repulsions.
Atoms arrange to maximize atomic number.
Bonds form to equalize electronegativity across the molecule.
In the context of artificial sweeteners, which statement distinguishes taste from caloric value?
Taste depends on electron geometry; caloric value depends on molecular geometry.
Taste is a sensory perception; caloric value measures energy provided by metabolism.
Taste equals the number of calories; caloric value equals the flavor intensity.
Taste is determined by bond angles; caloric value is determined by lone pairs.
How do humans primarily "taste" food according to the section focus?
By measuring caloric value.
By receptors that interact with molecular structures producing sensory signals.
By counting electron groups.
By VSEPR predictions of geometry.
What does the acronym VSEPR stand for?
Valence Shell Electron Pair Repulsion
Variable Shape Electron Pair Rotation
Valence State Energy Pyramid Rule
Vibrational Symmetry Electron Pair Repulsion
Which list correctly names the electron groups upon which VSEPR is based?
Single atoms only
Bonding pairs and lone pairs
Protons and neutrons
Ions and radicals
Which statement explains how VSEPR works?
It predicts molecular polarity from electronegativity tables.
It arranges electron groups to minimize repulsion, determining geometry.
It calculates reaction rates from collision theory.
It uses caloric values to determine flavor.
For molecules with one interior (central) atom, what determines the molecular geometry?
The number of atoms in the periodic table row
The arrangement of electron groups around the central atom
The caloric value of the molecule
The mass of the central atom
What geometry is predicted by VSEPR for a central atom with two electron groups?
Bent
Trigonal planar
Linear
Tetrahedral
What geometry is predicted by VSEPR for a central atom with three electron groups (all bonding pairs)?
Trigonal planar
Trigonal pyramidal
T-shaped
Octahedral
How does bonding to a highly electronegative atom affect a three-electron-group arrangement around a central atom?
It increases repulsions equally among all groups.
It can alter bond angles due to stronger bond pair localization, subtly affecting geometry.
It eliminates lone pairs.
It forces the geometry to become linear.
As the number of electron groups around a central atom reaches four or more, how does VSEPR prediction change?
Geometry becomes independent of repulsions.
Three-dimensional arrangements become necessary to minimize repulsions.
All molecules become planar.
Electronegativity no longer matters.
What is the three-dimensional electron-group geometry predicted for four electron groups around a central atom?
Trigonal planar
Tetrahedral
Trigonal bipyramidal
Octahedral
What is the three-dimensional electron-group geometry predicted for five electron groups around a central atom?
Square planar
Trigonal bipyramidal
Tetrahedral
Pentagonal pyramidal
What is the three-dimensional electron-group geometry predicted for six electron groups around a central atom?
Octahedral
Trigonal planar
Linear
Seesaw
Which statement best defines electron geometry?
The shape formed by only the atoms.
The arrangement of all electron groups (bonding pairs and lone pairs) around a central atom.
The orientation of orbitals used in bonding only.
The overall polarity of the molecule.
Which statement best defines molecular geometry?
Arrangement of electrons only.
Three-dimensional arrangement of atoms (ignoring lone pairs) determined by electron-group geometry.
Energy distribution across bonds.
Planar projection of the electron cloud.
How do lone pairs influence bonding pairs in VSEPR?
They attract bonding pairs, increasing bond angles.
They repel more strongly than bonding pairs, compressing adjacent bond angles.
They have no effect on geometry.
They convert single bonds to double bonds.
What is the typical repulsion order among electron groups in VSEPR?
Bonding pair–bonding pair > lone pair–bonding pair > lone pair–lone pair
Lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair
Bonding pair–lone pair > lone pair–lone pair > bonding pair–bonding pair
All interactions have equal repulsion
Which description defines a seesaw molecular geometry arising from five electron groups with one lone pair?
Derived from trigonal planar with one lone pair; angles near 120°.
Derived from trigonal bipyramidal with one lone pair; axial–equatorial angle adjustments compress some bond angles.
Derived from octahedral with two lone pairs; 90° angles only.
Derived from tetrahedral with one lone pair; ~109.5° angles.
Which molecular geometry results from six electron groups with one lone pair?
Square pyramidal
Trigonal pyramidal
Seesaw
T-shaped
According to VSEPR theory, what primarily determines a molecule’s geometry around a central atom?
Number of electron groups (bonding pairs and lone pairs) around the central atom
Atomic masses of bonded atoms
Total number of electrons in the molecule
Electronegativity of the central atom
Which molecular geometry is associated with four electron groups arranged with all positions equivalent but with two lone pairs creating a planar shape?
Tetrahedral
Trigonal planar
Square planar
Trigonal bipyramidal
In the context of predicting molecular shapes, what is the first simple procedure you should perform?
Assign formal charges to all atoms
Determine the number of electron groups around the central atom
Calculate bond enthalpies
Identify the oxidation state of the central atom
Which statement best summarizes VSEPR theory?
Electron groups repel and arrange to minimize repulsions, determining molecular geometry
Bond lengths determine geometry, independent of electron repulsion
Geometry depends only on the central atom’s hybridization
Molecular geometry is set by crystal packing forces
For a molecule with five electron groups and no lone pairs on the central atom, VSEPR predicts which electron-group geometry?
Octahedral
Trigonal bipyramidal
Tetrahedral
Square planar
When predicting molecular geometries for larger molecules compared to smaller ones, which approach is most appropriate?
Treat the entire molecule as one electron group
Analyze each central atom separately using VSEPR
Ignore lone pairs due to averaging effects
Use only bond lengths to infer 3D structure
Which step is essential for determining a molecule’s shape and polarity?
Count electron groups, determine geometry, then assess bond dipole vector addition
Calculate molar mass and compare to a reference
Measure boiling point to infer polarity
Assign oxidation numbers to each atom
A molecule is considered nonpolar when which condition is met?
All bonds are ionic
Bond dipoles cancel through vector addition, yielding no net dipole
It has at least one lone pair on the central atom
Its central atom is carbon
Which best describes vector addition in the context of molecular polarity?
Adding atomic numbers to obtain total polarity
Summing individual bond dipole vectors to determine the net molecular dipole
A method for calculating formal charge
Subtracting electronegativities to get polarity
Which geometry typically leads to complete cancellation of identical bond dipoles, making the molecule nonpolar when all peripheral atoms are the same?
Bent
Trigonal pyramidal
Linear
Seesaw
If a central atom has four electron groups with one lone pair, which molecular shape is most likely?
Trigonal pyramidal
Trigonal planar
Tetrahedral (molecular)
Square planar
Which factor directly influences whether a molecule with polar bonds is overall polar?
Distribution of bond dipoles relative to geometry
Molar mass of the molecule
Number of resonance structures
Presence of double bonds only
When applying VSEPR, electron groups include which items?
Only single bonds
Bonding regions (single, double, triple) and lone pairs
Only lone pairs
Ionic interactions
Which statement correctly contrasts predicting shapes for larger versus smaller molecules?
Larger molecules cannot be analyzed by VSEPR
For larger molecules, determine the local geometry around each central atom rather than one global geometry
Smaller molecules require hybridization while larger ones do not
Larger molecules always have square planar geometry
In determining polarity, which sequence is most accurate?
Determine geometry, identify bond polarities, perform vector addition, conclude net dipole
Find molar mass, compute electronegativity averages, decide polarity
Measure density, compute partial charges, decide polarity
Assign formal charges, check resonance, decide polarity
Which molecule type is most likely polar based on shape considerations alone?
Trigonal planar with identical substituents
Linear with identical substituents
Bent with two different polar bonds
Square planar with identical substituents
What is the role of lone pairs in VSEPR predictions of molecular shape?
They are ignored because they do not repel
They count as electron groups and can distort ideal geometries, affecting polarity
They only affect bond lengths, not angles
They convert any geometry to square planar
According to Valence Bond Theory, what is the fundamental criterion for a covalent bond to form between two atoms?
Electrostatic attraction between nuclei only
Constructive overlap of half‑filled atomic orbitals with opposite spins
Destructive overlap of filled orbitals
Transfer of electrons to create ions
In Valence Bond Theory, what happens when completely filled orbitals attempt to overlap significantly?
A strong sigma bond forms
A pi bond forms
Overlap is unfavorable due to electron‑electron repulsion
Hybridization increases the bond strength
Which statement best summarizes Valence Bond Theory?
Bonds arise from the mixing of entire molecules
Bonds form when atomic orbitals on different atoms overlap and the paired electrons occupy the region of overlap
Bonds exist only when electrons are delocalized over the whole molecule
Bonds are explained solely by electrostatic forces without considering orbitals
What is meant by orbital overlap in the context of Valence Bond Theory?
The physical collision of nuclei
The spatial intersection of electron density from atomic orbitals on adjacent atoms
The transfer of electrons from one atom to another
The expansion of an atom’s electron cloud due to heating
Define hybridization as introduced in Valence Bond Theory.
The process of forming ions by electron transfer
The combination of atomic orbitals on the same atom to form new, equivalent hybrid orbitals suited for bonding
The rearrangement of nuclei to minimize repulsion
The delocalization of electrons over many atoms
What is a hybrid orbital?
An orbital created by overlap between two different atoms
A new orbital formed by the linear combination of s, p, or d orbitals on the same atom, oriented for bonding
A high‑energy excited state of an electron
An orbital that contains two nuclei
Which scenario most likely leads to bond formation under Valence Bond Theory?
Overlap of two filled orbitals with the same spin
Overlap of a half‑filled orbital with a vacant orbital on the same atom
Constructive overlap between half‑filled orbitals on different atoms with opposite spins
No overlap but strong electrostatic attraction
Which statement about hybridization is consistent with Valence Bond Theory?
Hybridization changes atomic identity
Hybridization explains observed molecular geometries by reorienting electron density into equivalent bonding directions
Hybridization occurs only in ionic compounds
Hybridization eliminates the need for orbital overlap
Which statement best describes sp³ hybridization in a central atom?
Combination of one s and three p orbitals forming four equivalent hybrid orbitals arranged tetrahedrally
Combination of one s and two p orbitals forming three planar hybrid orbitals
Combination of one s and one p orbital forming two linear hybrid orbitals
Overlap of two p orbitals forming a pi bond
In sp² hybridization, what is the geometry and number of hybrid orbitals formed?
Tetrahedral geometry with four hybrids
Trigonal planar geometry with three hybrids
Linear geometry with two hybrids
Bent geometry with two hybrids
What remains unhybridized in an sp²-hybridized atom and is typically used to form a pi bond?
An s orbital
One p orbital perpendicular to the plane
Two p orbitals in the plane
One d orbital
Which hybridization corresponds to linear geometry around the central atom?
sp³
sp²
sp
None; linear geometry does not arise from hybridization
Sigma bonds are characterized by which type of orbital overlap?
Side-by-side overlap of p orbitals
End-to-end overlap along the internuclear axis
Overlap of d orbitals only
No orbital overlap
Pi bonds result from which interaction?
Overlap along the internuclear axis
Side-by-side overlap of unhybridized p orbitals
Overlap of s orbitals only
Overlap of hybrid orbitals only
Relative to a corresponding hybrid orbital, an unhybridized p orbital typically has which property?
Lower energy and more spherical shape
Higher energy and lobes oriented perpendicular to hybrid orbitals
Same energy and identical orientation
No role in bonding
Which statement correctly compares rotation about single and double bonds in Valence Bond Theory?
Rotation is free about both single and double bonds
Rotation is restricted about single bonds but free about double bonds
Rotation is free about single bonds but restricted about double bonds due to pi bond overlap
Rotation is impossible about any covalent bond
What unique feature of sigma bonds contributes to bond strength?
They arise from side-by-side overlap which is weaker
They have electron density concentrated along the internuclear axis allowing maximal overlap
They involve only d orbitals
They cannot form between hybrid orbitals
Which hybridization state is most consistent with a trigonal planar central atom and one pi bond present?
sp³ with four sigma bonds
sp² with three sigma bonds and one unhybridized p orbital
sp with two sigma bonds and two pi bonds
No hybridization is needed to describe trigonal planar geometry
In a molecule of formaldehyde (H₂CO), which bonding description aligns with Valence Bond Theory?
Carbon is sp³-hybridized with all sigma bonds
Carbon is sp²-hybridized forming three sigma bonds and one pi bond with oxygen
Carbon is sp-hybridized forming two sigma bonds and two pi bonds
Oxygen contributes only s orbitals to bonding
Which set correctly pairs hybridization with ideal bond angle?
sp³ — 109.5°, sp² — 120°, sp — 180°
sp³ — 120°, sp² — 109.5°, sp — 180°
sp³ — 90°, sp² — 120°, sp — 109.5°
sp³ — 104.5°, sp² — 120°, sp — 180°
Which statement best differentiates sigma and pi bonds in terms of formation?
Sigma bonds form only from unhybridized orbitals; pi bonds form only from hybrid orbitals
Sigma bonds can form from s, p, or hybrid orbitals via axial overlap; pi bonds form from unhybridized p orbitals via lateral overlap
Both sigma and pi bonds form from hybrid orbitals only
Pi bonds are stronger due to axial overlap while sigma bonds are weaker due to lateral overlap
Which combination of atomic orbitals forms sp hybridization on a central atom?
one s and one p orbital
one s and two p orbitals
one s and three p orbitals
one s, three p, and one d orbital
In sp hybridization, what is the ideal electron-group geometry around the central atom?
trigonal planar
linear
tetrahedral
trigonal bipyramidal
Acetylene (C2H2) features which hybridization on each carbon in its Lewis/valence bond description?
sp
sp2
sp3
sp3d
When describing acetylene with valence bond theory, which statement is correct?
Each C uses sp2 orbitals to form three sigma bonds.
Each C uses sp3 orbitals to form four sigma bonds.
Each C uses sp orbitals to form two sigma bonds and the remaining unhybridized p orbitals form two pi bonds.
Each C uses sp3d orbitals to form five sigma bonds.
Identify the bond types present in acetylene (C2H2).
one C–C sigma, two C–C pi, and two C–H sigma bonds
one C–C sigma and one C–C pi only
two C–C sigma and two C–H pi bonds
one C–C pi and two C–H sigma bonds
Which set of atomic orbitals mixes to give sp3d hybridization?
one s and one p
one s and two p
one s and three p
one s, three p, and one d
The ideal electron-group geometry associated with sp3d hybridization is:
tetrahedral
square planar
trigonal bipyramidal
octahedral
Arsenic pentafluoride (AsF5) has a central arsenic atom with which hybridization in a valence bond model?
sp2
sp3
sp3d
sp3d2
In AsF5, which arrangement of bonding pairs best describes the geometry around arsenic?
four equivalent positions forming a tetrahedron
three equatorial and two axial positions of a trigonal bipyramid
six equivalent positions forming an octahedron
square planar
Which combination of orbitals forms sp3d2 hybridization on a central atom?
one s and three p
one s, two p, and one d
one s, three p, and one d
one s, three p, and two d
The ideal electron-group geometry associated with sp3d2 hybridization is:
trigonal bipyramidal
octahedral
tetrahedral
linear
Sulfur hexafluoride (SF6) is best described by which hybridization at sulfur in the valence bond model?
sp3
sp3d
sp3d2
sp2
In SF6, the arrangement of six S–F sigma bonds corresponds to which geometry?
trigonal planar
octahedral
trigonal bipyramidal
square pyramidal
According to VSEPR, which hybridization is typically assigned to a central atom with four electron groups?
sp
sp2
sp3
sp3d
According to VSEPR, a central atom with five electron groups most commonly corresponds to which hybridization?
sp2
sp3
sp3d
sp3d2
According to VSEPR, a central atom with six electron groups most commonly corresponds to which hybridization?
sp3d
sp3d2
sp2
sp
Which statement best contrasts sp3d and sp3d2 hybridizations?
Both use one s, three p, and one d orbital; sp3d2 uses one fewer d orbital.
sp3d uses one s, three p, and one d orbital to give trigonal bipyramidal geometry, whereas sp3d2 uses one s, three p, and two d orbitals to give octahedral geometry.
sp3d and sp3d2 are identical and both give tetrahedral geometry.
sp3d uses only p orbitals while sp3d2 uses s and d orbitals.
Which statement about sigma and pi bonding is consistent with the valence bond descriptions in this section?
Sigma bonds form from side-by-side overlap of p orbitals; pi bonds form from end-to-end overlap.
Sigma bonds form from end-to-end (head-on) overlap of orbitals; pi bonds form from side-by-side overlap of unhybridized p orbitals.
Pi bonds form from hybrid orbitals only; sigma bonds form from unhybridized orbitals.
Sigma bonds and pi bonds are identical in orientation.
Which statement best describes a limitation of Valence Bond Theory addressed by Molecular Orbital Theory?
It cannot describe ionic bonding at all
It treats electrons as localized between specific atoms and struggles to explain delocalization
It predicts that all bonds are the same strength
It assumes atoms have no orbitals
Molecular Orbital Theory primarily explains bonding by:
Hybridizing atomic orbitals on a single atom
Overlapping atomic orbitals to form delocalized molecular orbitals over the entire molecule
Assigning resonance structures to localized bonds
Using electron pairs confined between two nuclei
In the Linear Combination of Atomic Orbitals (LCAO) approach, molecular orbitals are formed by:
Adding and subtracting atomic orbital wavefunctions
Multiplying atomic orbital energies
Averaging atomic electronegativities
Pairing electrons by spin only
A bonding molecular orbital is formed when the LCAO combination leads to:
Increased electron density between nuclei and lower energy
A node between nuclei and higher energy
No change in electron density and same energy
Electrons localized on one atom
An antibonding molecular orbital (often labeled with an asterisk, e.g., σ*) is characterized by:
Lower energy than the corresponding bonding orbital
Electron density concentrated between the nuclei
A nodal plane between nuclei and higher energy
Complete absence of electron density around the molecule
What is the standard formula for calculating bond order in Molecular Orbital Theory?
(number of bonding electrons − number of antibonding electrons) ÷ 2
(number of bonding orbitals − number of antibonding orbitals) ÷ 2
Total valence electrons ÷ 2
Electrons in σ orbitals − electrons in π orbitals
According to the bond order criterion for stability, a molecule is generally considered stable when:
Bond order equals zero
Bond order is negative
Bond order is greater than zero
Bond order is greater than two only
Which statement best summarizes LCAO-MO Theory?
Electrons are strictly localized in hybrid orbitals on each atom
Atomic orbitals combine linearly to produce molecular orbitals that can be bonding or antibonding, populated according to energy
Electrons pair only in σ bonds while π bonds are unpaired
Molecules form by transferring electrons to the lowest-energy atomic orbital
When constructing a generic MO energy diagram for a diatomic molecule, which step is correct?
Place molecular orbitals at the same energy as the higher atomic orbital
Combine atomic orbitals of similar symmetry and comparable energy to yield bonding (lower) and antibonding (higher) MOs
Fill antibonding orbitals before bonding orbitals
Ignore electron spin when filling MOs
If a diatomic molecule has 8 electrons in bonding MOs and 6 electrons in antibonding MOs, what is its bond order?
0
1
2
3
Which statement best explains why Li2 is predicted to be stable while Be2 is not, according to simple MO theory for homonuclear diatomics?
Li2 has electrons occupying a bonding σ2s orbital without filling the corresponding antibonding σ*2s orbital, giving bond order 1.
Li2 has electrons only in antibonding orbitals, giving bond order 0.
Be2 has a half-filled π2p bonding orbital, giving bond order 1.
Be2 has unpaired electrons in σ*2s that increase stability.
In MO diagrams for second-period homonuclear diatomics, why do some molecules use the ordering where σ2p is lower than π2p while others use the reverse?
Because experimental bond lengths differ randomly.
Due to s–p orbital mixing being significant for lighter elements (B, C, N), which raises σ2p relative to π2p; for O and F, s–p mixing diminishes and σ2p lies below π2p.
Because π orbitals cannot be antibonding.
Because electronegativity prevents any mixing of orbitals.
Paramagnetism in a molecule is most directly related to which feature of its molecular orbital electron configuration?
Presence of equal numbers of bonding and antibonding electrons.
Presence of unpaired electrons in any MO.
High bond order only.
Electrons localized on the more electronegative atom.
Which observation would indicate a molecule is paramagnetic when placed in a magnetic field?
It is repelled strongly by the field.
It shows no interaction with the field.
It is weakly attracted to the field due to unpaired electrons.
It becomes diamagnetic because of paired electrons.
How does electronegativity influence the location (distribution) of electrons in heteronuclear diatomic MOs?
Electrons preferentially localize on the less electronegative atom.
Electrons prefer the more electronegative atom, lowering the energy of MOs with greater contribution from that atom.
Electronegativity has no effect on MO energies or electron distribution.
It forces all electrons into antibonding orbitals.
For a heteronuclear diatomic molecule AB, which statement about MO energy levels is generally correct?
Atomic orbitals from the more electronegative atom contribute more to lower-energy MOs.
Atomic orbitals always mix equally regardless of electronegativity.
Only p orbitals form MOs while s orbitals remain atomic.
Lower-energy MOs are purely from atom A.
Which factor most directly determines whether a molecular orbital diagram predicts a nonzero bond order?
Difference in atomic masses of the bonded atoms.
Net excess of electrons in bonding orbitals compared to antibonding orbitals.
Presence of d orbitals.
Magnitude of electronegativity difference alone.
Which pair correctly matches a property with the MO feature that causes it?
High bond order — significant occupation of antibonding orbitals.
Long bond length — higher occupation of bonding orbitals.
Paramagnetism — presence of unpaired electrons.
Diamagnetism — presence of unpaired electrons.
Why do MO diagrams vary across polyatomic molecules compared with diatomics?
Polyatomics require consideration of symmetry-adapted linear combinations of atomic orbitals and more interactions, leading to different MO patterns.
Polyatomics do not have antibonding orbitals.
Polyatomics always follow the same ordering as O2.
MO diagrams are arbitrary and do not reflect physical reality.
Consider O2 and F2. Which statement best captures the difference in MO ordering compared to B2, C2, and N2?
In O2 and F2, σ2p lies below π2p due to reduced s–p mixing; in B2–N2, π2p lies below σ2p due to stronger s–p mixing.
All second-period diatomics share identical MO ordering.
O2 and F2 have no π orbitals.
B2–N2 have σ2p below π2p because of no mixing.
Which theory predicts molecular shapes based on electron-pair repulsions around a central atom?
Valence Bond Theory
Molecular Orbital Theory
VSEPR theory
Hybridization
According to VSEPR, what is the electron-pair geometry around a central atom with four electron domains?
Trigonal planar
Tetrahedral
Trigonal bipyramidal
Octahedral
Which molecular geometry results when one of four electron domains is a lone pair?
Bent
Trigonal pyramidal
Linear
Square planar
Valence Bond Theory explains covalent bonding primarily through which concept?
Electrons delocalized over the entire molecule
Overlap of atomic orbitals to form localized bonds
Electrostatic attraction between ions
Resonance structures only
Hybridization that corresponds to a trigonal planar electron geometry is:
sp
sp2
sp3
sp3d
Which hybridization is associated with linear molecular geometry?
sp
sp2
sp3
sp3d2
In Molecular Orbital (MO) Theory, a bond order of 0 implies:
A single bond
A double bond
No net bonding; the molecule is unstable
Presence of resonance
Which statement best distinguishes MO Theory from Valence Bond Theory?
MO Theory localizes electrons between two atoms; VBT delocalizes them
MO Theory delocalizes electrons over the whole molecule; VBT localizes bonding via orbital overlap
Both theories describe electrons as localized between nuclei
Neither uses wave functions
How is MO Theory applied to polyatomic molecules?
By assuming only sigma bonds exist
By constructing molecular orbitals from symmetry-adapted linear combinations of atomic orbitals across the molecule
By counting lone pairs only
By using only hybrid orbitals
For a central atom with five electron domains and one lone pair, the predicted molecular geometry is:
Trigonal bipyramidal
Seesaw
T-shaped
Square pyramidal
Which hybridization corresponds to octahedral electron geometry?
sp3
sp3d
sp3d2
sp2
In VSEPR, which arrangement minimizes repulsions for three electron domains?
Linear
Trigonal planar
Tetrahedral
Square planar
Which molecular geometry results from two bonding pairs and two lone pairs around a central atom?
Bent (≈104.5°)
Trigonal planar
T-shaped
Square planar
Which of the following best describes a sigma bond in Valence Bond Theory?
Side-by-side overlap of p orbitals
End-to-end overlap along the internuclear axis
Overlap that creates antibonding orbitals
Delocalized electron cloud over multiple atoms
A pi bond arises from:
End-to-end overlap of s orbitals
Side-by-side overlap of parallel p orbitals
Mixing of s and p to form hybrid orbitals
Electron transfer between ions
Which statement about hybridization is correct?
Hybridization is a physical mixing of electrons
Hybridization forms orbitals that better align for bonding and explain observed geometries
Hybridization always increases bond order
Hybridization is only used in ionic compounds
In MO Theory for diatomics, which occupancy yields a bond order of 1?
Two electrons in a bonding MO and two in an antibonding MO
Two electrons in a bonding MO and zero in an antibonding MO
Zero electrons in bonding and two in antibonding
Equal electrons in bonding and antibonding
Which geometry is expected for sp3d hybridization with no lone pairs?
Trigonal planar
Tetrahedral
Trigonal bipyramidal
Octahedral
Which concept connects symmetry and orbital mixing when extending MO Theory to molecules beyond diatomics?
Crystal field splitting
Symmetry-adapted linear combinations (SALCs)
Electron affinity trends
Lattice energy models
