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CHEM 233 Lesson 1 Exam (PDF) | 2026 Exam Questions and Answers + Rationales | Study Guide | 100% Correct

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INSTANT PDF DOWNLOAD – Comprehensive CHEM 233 Lesson 1 Exam study guide featuring practice questions, verified answers, and detailed answer rationales. Covers atomic structure, periodic trends, chemical bonding, molecular geometry, chemical nomenclature, stoichiometry, mole conversions, balancing chemical equations, significant figures, dimensional analysis, laboratory safety, problem-solving techniques, and foundational general chemistry concepts designed to help students prepare confidently for the CHEM 233 Lesson 1 examination.

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CHEM 233 LESSON 1 EXAM (PDF) | 2026 EXAM QUESTIONS
AND ANSWERS + RATIONALES | STUDY GUIDE | 100%
CORRECT
1. Which of the following electron configurations represents the electron configuration for a
magnesium cation?
A) 1s²2s²2p⁶3s²3p²
B) 1s²2s²2p⁶
C) 1s²2s²2p⁴
D) 1s²2s²2p⁶3s²
E) None of the choices are correct.
Correct Answer: B) 1s²2s²2p⁶
Rationale: A magnesium cation (Mg²⁺) forms when a neutral magnesium atom loses two valence
electrons. The neutral magnesium atom has the electron configuration 1s²2s²2p⁶3s². When the
two 3s electrons are removed, the remaining configuration is 1s²2s²2p⁶, which is the same as
neon .
2. What are the hybridizations of carbons 1 and 2 respectively in cyclopentene?
A) sp² and sp²
B) sp² and sp³
C) sp³ and sp
D) sp³ and sp²
E) None of the choices are correct.
Correct Answer: B) sp² and sp³
Rationale: In cyclopentene, carbon 1 is the double-bonded carbon, which is sp² hybridized due to
the presence of a pi bond. Carbon 2 is a single-bonded carbon in the ring, which is sp³
hybridized. The remaining ring carbons are also sp³ hybridized .
3. How many lone pairs of electrons will be present in the following molecule? CH₃–
N=N=N
A) 4
B) 0
C) 3
D) 1
E) 2
Correct Answer: E) 2
Rationale: The molecule CH₃–N=N=N (methyl azide) has a total of two lone pairs of electrons.
The central nitrogen atom in the azide group has one lone pair, and the terminal nitrogen atom
has one lone pair. The methyl group and the other nitrogen atoms do not have lone pairs .

, 4. What is the total number of σ bonds found in the following compound? (A compound
with 15 carbons, 22 hydrogens, and various functional groups)
A) 15
B) 22
C) 11
D) 8
E) 10
Correct Answer: B) 22
Rationale: In organic compounds, sigma (σ) bonds are single bonds. The total number of σ bonds
is calculated by counting all single bonds, including C-C, C-H, and C-O bonds, plus one σ bond
for each multiple bond. In this specific compound, the total number of σ bonds is 22 .
5. What is the best estimate of the magnitude of the bond angle around atom D in
Cinnamoside?
A) 109.5°
B) 180°
C) 104.5°
D) 120°
E) 107.3°
Correct Answer: D) 120°
Rationale: Atom D in Cinnamoside is a carbon atom that is sp² hybridized, which corresponds to
a trigonal planar geometry and a bond angle of approximately 120°. The other bond angles listed
correspond to other hybridization states: sp³ (109.5°), linear (180°), and bent (104.5°) .
6. Identify the term for any species that can accept electrons.
A) Electronegativity
B) Lewis Acid
C) Brønsted-Lowry Acid
D) Brønsted-Lowry Base
E) Lewis Base
Correct Answer: B) Lewis Acid
Rationale: A Lewis acid is defined as any species that can accept a pair of electrons to form a
covalent bond. This is a more general definition than the Brønsted-Lowry definition, which
specifically involves proton transfer .
7. What are the orbitals of the reactants which interact in the Lewis acid-base reaction
between BH₃ and H₂O?
A) an empty p orbital on boron and a non-bonding lone pair on oxygen
B) a filled σ orbital on boron and an empty σ* orbital on oxygen
C) an empty sp³ orbital on boron and a non-bonding lone pair on oxygen
D) an empty s orbital on the hydrogen and a filled σ orbital on boron

, Correct Answer: C) an empty sp³ orbital on boron and a non-bonding lone pair on oxygen
Rationale: In the Lewis acid-base reaction between BH₃ and H₂O, boron is the Lewis acid and
has an empty p orbital that can accept electrons. However, in its trigonal planar geometry, the
empty p orbital is perpendicular to the plane. When considering the orbital interaction, the non-
bonding lone pair on oxygen (H₂O) interacts with the empty sp³ orbital of boron .
8. Identify the term for the ability of an atom to attract the shared electrons in a covalent
bond.
A) Lewis Base
B) Electronegativity
C) Brønsted-Lowry Base
D) Brønsted-Lowry Acid
E) Lewis Acid
Correct Answer: B) Electronegativity
Rationale: Electronegativity is the measure of the ability of an atom to attract shared electrons in
a covalent bond. This property is fundamental to understanding bond polarity and the behavior of
atoms in chemical reactions .
9. Which of the following statements is not true for ethylene?
A) C=C bond length is shorter than the C-C length in ethane.
B) The entire molecule has a planar geometry.
C) Both carbons are sp² hybridized.
D) The two bonds in C=C are equally strong.
E) The molecule has a dipole moment of zero.
Correct Answer: D) The two bonds in C=C are equally strong.
Rationale: In ethylene (C₂H₄), the two bonds in the C=C double bond are not equally strong. The
pi bond is weaker than the sigma bond, which is why alkenes undergo addition reactions more
readily than alkanes. The other statements are all true for ethylene .
10. Which of the following would represent the strongest acid?
A) Ka = 2.5 x 10⁻¹
B) pKa = 4.60
C) pKa = 14.5
D) Ka = 2.5 x 10⁻⁵
Correct Answer: A) Ka = 2.5 x 10⁻¹
Rationale: The strength of an acid is determined by its acid dissociation constant (Ka). A larger
Ka value indicates a stronger acid because it dissociates more completely in solution. The
relationship between Ka and pKa is pKa = -log(Ka). Therefore, the acid with the largest Ka (2.5
x 10⁻¹) is the strongest acid .

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