W1 summary + mc qs

Question 1

Atomic Orbitals and Electron Distribution

Answer 1

Atomic Orbitals: Define the probability of locating electrons around a nucleus.
Types: 1s, 2s, 2p, etc., differing in size and shape.
1s vs. 2s Orbital:
Both spherical; 2s is larger, with electrons more likely found farther from the nucleus.
2p Orbitals: Two lobes separated by a nodal plane (zero probability of electron presence).
Orbitals have fixed energy levels (quantized).

Question 2

Valence Orbitals and Electrons:

Answer 2

Outer orbitals involved in bonding are less stable.
Electrons in these orbitals are termed valence electrons.

Question 3

bonding in organic molecules

Answer 3

Electrostatic Attraction: Holds atoms together in molecules.
Two types of bonds
Ionic Bonds: Transfer of electrons (e.g., NaCl).
Covalent Bonds: Sharing of electrons.

Question 4

lewis structures

Answer 4

Lewis Structures:
Represent atom connectivity and formal charges:
Formal Charge = (# Valence Electrons) - (# Bonds) - (# Non-bonded Electrons).
Octet rule exceptions often indicate reactivity.

Question 5

covalent bond formation

Answer 5

Sigma (σ) Bonds:
Formed by head-on overlap of orbitals (e.g., s-orbitals).
Strong due to direct electron sharing between nuclei.

Pi (π) Bonds:
Formed by side-by-side overlap of p-orbitals.
Weaker than sigma bonds due to less orbital overlap.

Bond Strength: σ > π.

Question 6

electronegativity

Answer 6

Electronegativity: Atom’s ability to pull electrons in a bond.

Creates bond dipoles in polar covalent bonds.

Dipole vectors point from less electronegative (partial +) to more electronegative (partial -) atom.

Question 7

geometries based on bonding

Answer 7

Geometries Based on Bonding

VSEPR Theory: Valence shell electron pairs arrange to minimize repulsion, dictating geometry.

Tetrahedral: 109° bond angle (e.g., CH4).
Trigonal Planar: 120° bond angle (e.g., BF3).
Linear: 180° bond angle (e.g., CO2).

Question 8

hybrid orbitals

Answer 8

Hybrid Orbitals: Mixed orbitals to explain molecular geometries.

Question 9

sp³ hybridization

Answer 9

Formed by mixing one s and three p orbitals.
Creates four equivalent orbitals for tetrahedral geometry

Question 10

sp² Hybridization:

Answer 10

One s and two p orbitals mixed.
Results in trigonal planar geometry and a π bond.

Question 11

sp Hybridization:

Answer 11

One s and one p orbital mixed.
Produces linear geometry.

Question 12

Pi Bonds and Hybridization:

Answer 12

Pi bonds arise from unhybridized p-orbitals.
Hybrid orbitals (σ bonds) direct toward surrounding atoms.

Question 13

Why are hybrid orbitals necessary for molecular bonding geometries?

Answer 13

Hybrid orbitals are necessary to align atomic orbitals with the observed molecular geometries (e.g., tetrahedral, trigonal planar, linear), maximize orbital overlap for stronger sigma bonds, and minimize electron repulsion, ensuring stable molecular shapes and bond angle

Question 14

What defines a nodal plane in an atomic orbital?
A. A region of maximum probability of finding an electron
B. A region where the probability of finding an electron is zero
C. The boundary of an orbital’s shape
D. A region between atomic nuclei

Answer 14

b

Question 15

What distinguishes valence electrons from core electrons?
A. They have higher energy and participate in bonding.
B. They are located closer to the nucleus.
C. They do not influence molecular geometry.
D. They are found in the 1s orbital.

Answer 15

a

Question 16

Why are sigma bonds stronger than pi bonds?
A. Sigma bonds are formed by s orbitals only.
B. Sigma bonds involve direct head-on overlap, while pi bonds have less overlap.
C. Sigma bonds have fewer electrons.
D. Sigma bonds are non-polar.

Answer 16

b

Question 17

Which bond angle is associated with a trigonal planar geometry?
A. 90°
B. 109°
C. 120°
D. 180°

Answer 17

c

Question 18

What does hybridization accomplish in atomic orbitals?
A. It changes their energy levels.
B. It combines orbitals to match molecular geometries.
C. It removes electrons from outer shells.
D. It forms ionic bonds

Answer 18

b

Question 19

what is the significance of “degenerate orbitals”?
A. They have the same energy level.
B. They contain only one electron.
C. They are unoccupied.
D. They are hybrid orbitals.

Answer 19

a

Question 20

How does electronegativity influence bond polarity?
A. It decreases bond strength.
B. It creates partial charges due to unequal electron sharing.
C. It causes electrons to be equally distributed.
D. It increases bond length.

Answer 20

b

Question 21

Why is hybridization necessary for tetrahedral geometries?
A. To align orbitals along bonding directions.
B. To increase bond strength in pi bonds.
C. To stabilize ionic bonds.
D. To create non-polar molecules.

Answer 21

a

Question 22

What happens if a molecule violates the octet rule?
A. It becomes more stable.
B. It becomes reactive and unstable.
C. It forms ionic bonds exclusively.
D. It cannot exist in nature.

Answer 22

b

Question 23

How does VSEPR theory explain molecular shapes?
A. By predicting orbital overlap.
B. By arranging electron groups to minimize repulsion.
C. By calculating dipole moments.
D. By analyzing hybrid orbitals.

Answer 23

b

Question 24

Why is a 2s orbital larger than a 1s orbital?
A. It has more electrons.
B. It is farther from the nucleus, increasing its energy level.
C. It forms stronger bonds.
D. It overlaps with p orbitals.

Answer 24

b

Question 25

if an atom has sp² hybridization, what is its molecular geometry? A. Linear B. Trigonal planar C. Tetrahedral D. Bent

Answer 25

b

Question 26

Why is the formal charge of an atom significant? A. It determines bond length. B. It affects molecular polarity and reactivity. C. It influences atomic mass. D. It creates resonance structures

Answer 26

b

Question 27

what role do pi bonds play in molecular structure? A. They provide strength to single bonds. B. They allow flexibility in rotation. C. They form side-by-side orbital overlaps, contributing to double or triple bonds. D. They are weaker because they involve hybrid orbitals.

Answer 27

c

Question 28

How does the number of groups of electrons determine hybridization? A. Each group corresponds to an atomic number. B. The total groups match the superscripts in hybridization (e.g., sp³ = 4 groups). C. It limits the formation of pi bonds. D. It determines the energy level of the atom.

Answer 28

b

Question 29

What would happen if pi bonds replaced sigma bonds in molecular structures? A. Molecules would become more flexible. B. Molecules would weaken due to less orbital overlap. C. Molecules would exhibit stronger bonds. D. Hybridization would increase.

Answer 29

b

Question 30

If a molecule has a formal charge of -1, what can be inferred? A. It has gained an electron relative to its neutral state. B. It has an incomplete octet. C. It forms ionic bonds exclusively. D. It is non-polar.

Answer 30

a

Question 31

Why are polar covalent bonds significant in molecular interactions? A. They reduce the molecule's reactivity. B. They create partial charges that enable hydrogen bonding and dipole-dipole interactions. C. They prevent hybridization. D. They make molecules non-polar.

Answer 31

b

Question 32

How does VSEPR theory predict the shape of NH₃? A. Linear B. Tetrahedral C. Trigonal pyramidal due to lone-pair repulsion D. Trigonal planar

Answer 32

c

Question 33

What would happen to a molecule's geometry if lone pairs were ignored in predictions? A. The predicted geometry would match the observed geometry. B. The geometry would deviate due to unaccounted electron repulsion. C. Lone pairs do not affect molecular geometry. D. The molecule would become unstable.

Answer 33

b

Question 34

if sp hybridization occurs, what type of geometry does the molecule exhibit? A. Linear B. Tetrahedral C. Trigonal planar D. Bent

Answer 34

a

Question 35

Why does electronegativity decrease down a group in the periodic table? A. Atomic radii decrease. B. Shielding increases, reducing the effective nuclear charge felt by bonding electrons. C. Atoms lose their valence electrons. D. Nuclear charge decreases.

Answer 35

b

Question 36

How does hybridization influence bond strength? A. It weakens bonds by reducing electron density. B. It strengthens bonds by optimizing orbital overlap. C. It has no effect on bond strength. D. It forms pi bonds exclusively.

Answer 36

b

Question 37

Why are lone pairs placed farther apart than bonding pairs in VSEPR theory? A. Lone pairs have higher energy. B. Lone pairs exert greater repulsive forces due to higher electron density. C. Lone pairs attract bonding pairs. D. Bonding pairs repel lone pairs equally.

Answer 37

b

Question 38

What would happen if atomic orbitals did not hybridize? A. Molecular geometries would remain unchanged. B. Orbital overlap would decrease, weakening bonds and disrupting geometries. C. Bonds would form stronger sigma bonds. D. Molecules would exhibit ionic character.

Answer 38

b

Question 39

What feature of sigma bonds prevents rotation around double bonds? A. Sigma bonds alone do not prevent rotation. B. Pi bonds restrict rotation by forming regions of electron density above and below the bond axis. C. Sigma bonds overlap side-by-side, locking the bond. D. Sigma bonds form polar covalent bonds.

Answer 39

b

Question 40

How does the geometry of CO₂ differ from H₂O? A. CO₂ is linear due to no lone pairs on the central atom, while H₂O is bent due to two lone pairs. B. Both have linear geometries. C. CO₂ is bent due to lone pairs, while H₂O is tetrahedral. D. Both have trigonal planar geometries.

Answer 40

a

Question 41

Why are hybrid orbitals used in sigma bonding? A. They increase orbital overlap along the bond axis for stronger bonds. B. They are weaker and more flexible than atomic orbitals. C. They prevent lone-pair repulsions. D. They restrict bonding to pi bonds only.

Answer 41

a

Question 42

What would occur if a pi bond formed from hybrid orbitals? A. The bond would be stronger than a typical pi bond. B. Pi bonds cannot form from hybrid orbitals; only unhybridized p orbitals participate. C. The molecule would exhibit increased flexibility. D. The molecule would not form.

Answer 42

b

Question 43

if an atom is sp² hybridized, what type of bonds can it form? A. Three sigma bonds and one pi bond B. Four sigma bonds C. Two sigma bonds and two pi bonds D. Three pi bonds only

Answer 43

a

Question 44

why do degenerate orbitals have the same energy? A. They arise from unhybridized p orbitals with equal spatial distribution. B. They are hybridized to form specific geometries. C. They are influenced by lone pairs. D. They always align with the bond axis.

Answer 44

a

Question 45

. How does a nodal plane affect molecular bonding? A. It enhances orbital overlap. B. It indicates regions of zero electron density, reducing bonding likelihood. C. It stabilizes ionic bonds. D. It increases electronegativity.

Answer 45

b

Question 46

How does the formal charge help predict reactivity? A. Molecules with high formal charges are more reactive. B. Formal charge prevents electron repulsion. C. Formal charge stabilizes nonpolar molecules. D. Molecules with zero formal charge are always reactive.

Answer 46

a

Question 47

Why is the bond dipole in CH₃OH directed toward oxygen? A. Oxygen is less electronegative. B. Oxygen is more electronegative, creating a partial negative charge. C. The hydrogen atoms repel electrons. D. The dipole points away from the lone pairs on oxygen.

Answer 47

b

Question 48

what role does the hashed bond in VSEPR diagrams serve? A. It represents lone pairs. B. It indicates bonds pointing away from the viewer, into the page. C. It marks the most polar bond. D. It shows pi bonds exclusively.

Answer 48

b

Question 49

true or false: propanol is more soluble than hexane

Answer 49

true; propanol is a polar organic molecule that will dissolve in a polar solvent like water

Question 50

true or false: the CF3H molecule has a net dipole moment

Answer 50

true the CF3H molecule is not symmetric, and each C-F bond has a bond dipole that combine to generate a net dipole moment

Question 51

true or false: 2-bromopentane is a hydrogen bond donor

Answer 51

false 2-bromopentane does not contain a hydrogen atom connected to an electronegative atom like oxygen or nitrogen

Question 52

true or false: boiling points decrease with increasing dispersion forces

Answer 52

false boiling points increase with increasing dispersion forces