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UNE - CHEM 1011 - MIDTERM STUDY GUIDE ACTUAL 2026 QUESTIONS AND VERIFIED ANSWERS ACTUAL!!!

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UNE - CHEM 1011 - MIDTERM STUDY GUIDE ACTUAL 2026 QUESTIONS AND VERIFIED ANSWERS ACTUAL!!!

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UNE - CHEM 1011
Course
UNE - CHEM 1011

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UNE - CHEM 1011 - MIDTERM
STUDY GUIDE ACTUAL 2026
QUESTIONS AND VERIFIED ANSWERS
ACTUAL!!!




Question 1
A hydrogen bond is characterized by:
A. the attraction of temporary dipoles produced by the random asymmetries in electron motion
B. the electrostatic attraction between the permanent dipoles in any polar molecule
C. the highly concentrated partial charge between an H atom and F, O, or N atoms
D. the covalent sharing of electrons between two atoms in a molecule


ANSWER: C. the highly concentrated partial charge between an H atom and F, O, or N atoms


Rationale: Option A describes London dispersion forces (LDFs), which arise from temporary
fluctuations in electron distribution. Option B describes dipole-dipole interactions between
polar molecules. Option C correctly defines hydrogen bonding, a special type of dipole-dipole
interaction that occurs when hydrogen is bonded to highly electronegative atoms (F, O, or N),
creating a highly concentrated partial positive charge. Option D describes covalent bonds,
which involve the sharing of electrons. Hydrogen bonds are intermolecular forces, not covalent
bonds.

,Question 2
Which of the following will have the strongest dipole-dipole attractions?
A. NCl3
B. CCl4
C. BCl3
D. Cl2


ANSWER: A. NCl3


Rationale: NCl3 is polar due to the lone pair on nitrogen creating an asymmetrical molecular
geometry (trigonal pyramidal). This results in a net dipole moment. CCl4 is nonpolar due to its
symmetrical tetrahedral geometry with identical C-Cl bonds that cancel each other out. BCl3 is
nonpolar with a symmetrical trigonal planar geometry. Cl2 is nonpolar as a homonuclear
diatomic molecule. Only NCl3 possesses a permanent dipole, making it the only choice with
dipole-dipole attractions.


Question 3
Which intermolecular force is present in all molecules regardless of polarity?
A. Hydrogen bonding
B. Dipole-dipole forces
C. London dispersion forces
D. Ionic bonding


ANSWER: C. London dispersion forces


Rationale: London dispersion forces (LDFs) exist in all molecules because they arise from
temporary fluctuations in electron distribution. Option A is incorrect because hydrogen bonding
requires specific conditions (H bonded to F, O, or N). Option B is incorrect because dipole-dipole
forces only occur in polar molecules. Option D is incorrect because ionic bonding is an
intramolecular force, not an intermolecular force.

,Question 4
What is the electron geometry of a molecule with 4 bonding domains and 0 lone pairs?
A. Tetrahedral
B. Trigonal pyramidal
C. Bent
D. Trigonal planar


ANSWER: A. Tetrahedral


Rationale: With 4 bonding domains and no lone pairs, the electron geometry is tetrahedral. This
is based on VSEPR theory, which states that electron domains arrange themselves to minimize
repulsion. Four domains adopt a tetrahedral arrangement with bond angles of approximately
109.5°. Option B (trigonal pyramidal) would have 3 bonding domains and 1 lone pair. Option C
(bent) would have 2 bonding domains and 1 or 2 lone pairs. Option D (trigonal planar) would
have 3 bonding domains and 0 lone pairs.


Question 5
Which of the following molecules is polar?
A. CO2
B. BeCl2
C. SO2
D. CCl4


ANSWER: C. SO2


Rationale: SO2 has a bent molecular geometry due to the lone pair on sulfur, resulting in an
asymmetrical charge distribution and a net dipole moment. CO2 is linear and nonpolar because
the bond dipoles cancel. BeCl2 is linear and nonpolar with bond dipoles that cancel. CCl4 is
tetrahedral and nonpolar because the identical C-Cl bond dipoles cancel symmetrically.

, Question 6
What is the molecular shape of H2O according to VSEPR theory?
A. Linear
B. Bent
C. Trigonal planar
D. Tetrahedral


ANSWER: B. Bent


Rationale: H2O has 4 electron domains (2 bonding pairs with hydrogen and 2 lone pairs on
oxygen). The electron geometry is tetrahedral, but the molecular shape is bent because only
the positions of atoms are considered, not lone pairs. Option A is incorrect because linear
geometry would require no lone pairs. Option C would require 3 bonding domains. Option D is
the electron geometry, not the molecular shape.


Question 7
Which type of bonding involves the sharing of electron pairs between atoms?
A. Ionic bonding
B. Covalent bonding
C. Metallic bonding
D. Hydrogen bonding


ANSWER: B. Covalent bonding


Rationale: Covalent bonding involves the sharing of electron pairs between atoms to achieve
stable electron configurations. Ionic bonding (Option A) involves the transfer of electrons.
Metallic bonding (Option C) involves delocalized electrons in a "sea" of electrons. Hydrogen
bonding (Option D) is an intermolecular force, not a type of chemical bonding.

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