CHM 101/CHM101
NUR2356 MDC 1 Final Exam Review
2025/2026 - Multidimensional Care I Quiz
Bank
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EXAM 3 – Mid-Semester Proficiency (50 questions)
SECTION A – GENERAL CHEMISTRY (20 questions)
A1. Chemical Bonding & Molecular Structure (7)
Question 1
Which set of molecules is correctly ranked in order of increasing bond polarity?
A. HF < HCl < HBr
,B. HBr < HCl < HF
C. HCl < HBr < HF
D. HF < HBr < HCl
Answer: B – Electronegativity difference ΔEN: 0.7 (H–Br) < 0.9 (H–Cl) < 1.9 (H–F).
Rationale: Polarity scales with ΔEN. Fluorine is the most electronegative element (3.98
Pauling). Choices A & C invert the order; D mis-orders HBr/HCl.
Question 2
The Lewis structure of the sulfite ion, SO₃²⁻, contains:
A. 1 double bond, 2 single bonds, zero formal charges on S
B. 1 double bond, 2 single bonds, formal charge –1 on S
C. 3 double bonds, formal charge –2 on S
D. 3 equivalent S–O bonds, resonance, average –⅓ formal charge on each O
Answer: D – Resonance gives 3 identical bonds (1⅓ bond order) and –⅔ per O,
summing to –2.
Rationale: Expanded octet on S allows delocalization; A & B ignore resonance; C violates
octet on O atoms.
,Question 3
Which hybridization best describes the central atom in XeF₄?
A. sp
B. sp²
C. sp³d
D. sp³d²
Answer: D – Octahedral electron geometry (4 bonds + 2 lone pairs) requires six hybrid
orbitals.
Rationale: VSEPR predicts square-planar molecular geometry; sp³d² is the 2025 IUPAC
preferred symbol for d²sp³/octahedral.
Question 4
The molecule CO has a bond order of:
A. 1
B. 1.5
C. 2
, D. 3
Answer: C – MO diagram gives (σ2s)²(σ2s*)²(π2p)⁴(σ2p)²; BO = ½(10 – 4) = 3, but
heteronuclear correction lowers effective bond order to 3 (still integer).
Rationale: Experimental bond length 112 pm matches triple bond; students often
miscount antibonding electrons.
Question 5
Which statement about intermolecular forces is true at 298 K?
A. CH₃OH exhibits only London dispersion.
B. BF₃ exhibits hydrogen bonding.
C. CH₃CH₂OH has a higher surface tension than CH₃OCH₃ due to H-bonding.
D. All molecules with dipole moments exhibit hydrogen bonding.
Answer: C – Hydrogen bonding among alcohol molecules increases cohesive energy.
Rationale: A misses H-bonding in methanol; B lacks H bound to N,O,F; D conflates dipole
with H-bond donor.
Question 6
NUR2356 MDC 1 Final Exam Review
2025/2026 - Multidimensional Care I Quiz
Bank
======================================================
============
EXAM 3 – Mid-Semester Proficiency (50 questions)
SECTION A – GENERAL CHEMISTRY (20 questions)
A1. Chemical Bonding & Molecular Structure (7)
Question 1
Which set of molecules is correctly ranked in order of increasing bond polarity?
A. HF < HCl < HBr
,B. HBr < HCl < HF
C. HCl < HBr < HF
D. HF < HBr < HCl
Answer: B – Electronegativity difference ΔEN: 0.7 (H–Br) < 0.9 (H–Cl) < 1.9 (H–F).
Rationale: Polarity scales with ΔEN. Fluorine is the most electronegative element (3.98
Pauling). Choices A & C invert the order; D mis-orders HBr/HCl.
Question 2
The Lewis structure of the sulfite ion, SO₃²⁻, contains:
A. 1 double bond, 2 single bonds, zero formal charges on S
B. 1 double bond, 2 single bonds, formal charge –1 on S
C. 3 double bonds, formal charge –2 on S
D. 3 equivalent S–O bonds, resonance, average –⅓ formal charge on each O
Answer: D – Resonance gives 3 identical bonds (1⅓ bond order) and –⅔ per O,
summing to –2.
Rationale: Expanded octet on S allows delocalization; A & B ignore resonance; C violates
octet on O atoms.
,Question 3
Which hybridization best describes the central atom in XeF₄?
A. sp
B. sp²
C. sp³d
D. sp³d²
Answer: D – Octahedral electron geometry (4 bonds + 2 lone pairs) requires six hybrid
orbitals.
Rationale: VSEPR predicts square-planar molecular geometry; sp³d² is the 2025 IUPAC
preferred symbol for d²sp³/octahedral.
Question 4
The molecule CO has a bond order of:
A. 1
B. 1.5
C. 2
, D. 3
Answer: C – MO diagram gives (σ2s)²(σ2s*)²(π2p)⁴(σ2p)²; BO = ½(10 – 4) = 3, but
heteronuclear correction lowers effective bond order to 3 (still integer).
Rationale: Experimental bond length 112 pm matches triple bond; students often
miscount antibonding electrons.
Question 5
Which statement about intermolecular forces is true at 298 K?
A. CH₃OH exhibits only London dispersion.
B. BF₃ exhibits hydrogen bonding.
C. CH₃CH₂OH has a higher surface tension than CH₃OCH₃ due to H-bonding.
D. All molecules with dipole moments exhibit hydrogen bonding.
Answer: C – Hydrogen bonding among alcohol molecules increases cohesive energy.
Rationale: A misses H-bonding in methanol; B lacks H bound to N,O,F; D conflates dipole
with H-bond donor.
Question 6
