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Quizzes > Science > Chemistry

Molecular Structure Quiz: Test Your Chemistry Mastery

Ready for a molecular geometry quiz? Challenge your electron configuration skills!

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Layered paper art molecules atoms electrons orbitals resonance arrows on teal background for molecular structure quiz

Use this molecular structure quiz to practice molecular geometry: apply VSEPR to predict shapes and bond angles, check polarity and resonance, and review hybridization. You get instant feedback to spot gaps before a test; start with the free practice set or continue with the shape quiz .

What is the molecular geometry of CH4?
Tetrahedral
Bent
Linear
Trigonal planar
CH4 has four bonding pairs around the central carbon and no lone pairs, which arranges the atoms as far apart as possible in three dimensions, resulting in a tetrahedral geometry. This shape minimizes electron pair repulsion according to VSEPR theory. Carbon in CH4 is sp3 hybridized, leading to bond angles of approximately 109.5°. .
Which electron pair geometry describes NH3?
Trigonal pyramidal
Tetrahedral
Bent
Trigonal planar
NH3 has three bonding pairs and one lone pair around nitrogen, giving four regions of electron density. VSEPR theory classifies four electron domains as tetrahedral electron pair geometry. The lone pair occupies one vertex of the tetrahedron, while the three hydrogens occupy the others. .
What hybridization does the central carbon in CO2 exhibit?
sp
sp3
sp2
dsp2
CO2 is a linear molecule with two regions of electron density around the central carbon atom. VSEPR theory indicates that two domains correspond to sp hybridization. The 180° bond angle and linear arrangement arise from mixing one s and one p orbital. .
What is the bond angle in H2O?
104.5°
109.5°
90°
120°
Water has two bonding pairs and two lone pairs on the central oxygen atom. The lone pairs repel more strongly than bonding pairs, compressing the H - O - H angle to about 104.5°. This gives water a bent molecular shape. .
Which of the following molecules exhibits resonance?
Water (H2O)
Benzene (C6H6)
Methane (CH4)
Ammonia (NH3)
Benzene has a cyclic conjugated ? system in which electrons are delocalized over all six carbon atoms. This delocalization is represented by resonance structures. Neither methane, ammonia, nor water have such conjugated ? networks. .
Which molecule contains a central atom that expands its octet?
CH4
BF3
SF6
NH3
In SF6, sulfur forms six S - F bonds, using 12 valence electrons and exceeding the octet rule. This is possible because sulfur has available d orbitals to accommodate extra electron density. BF3 actually has an incomplete octet, while CH4 and NH3 follow the octet rule. .
Which of the following molecules has a net dipole moment?
BF3
SO2
CH4
CO2
SO2 has a bent geometry with unequal distribution of electron density, giving it a net dipole moment. CO2 and BF3 are linear and trigonal planar respectively with symmetric charge distribution, resulting in no net dipole. CH4 is also nonpolar due to its tetrahedral symmetry. .
What is the molecular geometry of XeF4?
See-saw
Square planar
Square pyramidal
Tetrahedral
XeF4 has six electron domains: four bonding pairs and two lone pairs on xenon. VSEPR theory places the lone pairs opposite each other in an octahedral arrangement, leaving the fluorine atoms in a square plane. This gives a square planar molecular geometry. .
What is the formal charge on the central nitrogen atom in the nitrate ion, NO3-?
-1
+2
+1
0
In one resonance form of NO3-, nitrogen is bonded to three oxygens (one double bond and two single bonds) with no lone pairs. The formal charge is calculated as valence electrons (5) minus nonbonding electrons (0) minus half the bonding electrons (6 pairs = 12 electrons/2 = 6), giving +1. The overall negative charge resides on the oxygens. .
Predict the molecular shape and electron pair geometry of ClF3.
T-shaped (trigonal bipyramidal)
See-saw (trigonal bipyramidal)
T-shaped (octahedral)
Trigonal planar (trigonal planar)
ClF3 has five regions of electron density (three bonding pairs and two lone pairs) around the central chlorine, leading to a trigonal bipyramidal electron geometry. According to VSEPR theory, the lone pairs occupy equatorial positions to minimize repulsion, resulting in a T-shaped molecular geometry. Bond angles are compressed slightly from idealized positions due to lone pair repulsions. .
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Study Outcomes

  1. Analyze molecular geometries -

    Identify and distinguish shapes such as trigonal bipyramidal, tetrahedral, and linear in the molecular geometry quiz to better grasp three-dimensional structures.

  2. Predict bond angles -

    Estimate bond angles using VSEPR theory and understand how electron pair repulsion influences molecular shape and geometry.

  3. Determine electron configurations -

    Assign electron configurations for atoms within molecules and relate these configurations to overall structure and reactivity.

  4. Interpret resonance structures -

    Compare and evaluate resonance forms to assess their contribution to molecular stability and delocalized electron distribution.

  5. Apply VSEPR principles -

    Use valence shell electron pair repulsion theory to deduce three-dimensional arrangements and predict the most stable molecular geometry.

  6. Predict molecular polarity -

    Determine molecule polarity by combining knowledge of geometry, bond polarity, and electron distribution to anticipate dipole moments.

Cheat Sheet

  1. VSEPR Shapes and Bond Angles -

    Dive into VSEPR theory (IUPAC guidelines) to predict molecular geometries by minimizing electron-pair repulsions, like the trigonal bipyramidal structure with 120° equatorial and 90° axial bonds. A catchy mnemonic - "three in the equator, two at the poles" - helps cement those angles. This trick transforms even the trickiest molecular structure quiz questions into quick recall.

  2. Hybridization and Steric Number -

    Determine an atom's hybridization by counting its steric number: the sum of bonded atoms and lone pairs dictates sp, sp², sp³, sp³d or sp³d² hybridization (as detailed in MIT OpenCourseWare). For example, PCl₅ exhibits sp³d hybridization forming a trigonal bipyramid, while CH₄ is classic sp³. Mastering this link makes any molecular geometry quiz smoother.

  3. Electron Configuration & Molecular Orbitals -

    Apply the Aufbau principle and Hund's rule from reputable sources like the University of California to fill atomic and molecular orbitals - 1s→2s→2p→σ2p<π2p→π*2p<σ*2p. Remember O₂'s paramagnetism arises from two unpaired electrons in π*2p, a common electron configuration quiz staple. Sketching MO diagrams boosts confidence for molecular structure quiz sections.

  4. Resonance Structures and Delocalization -

    Use resonance to depict electron delocalization in molecules such as benzene (C₆H₆) and carbonate (CO₃²❻), following arrow-pushing rules from ACS resources. Favor structures with full octets and minimal charge separation to identify the major contributor. This strategy reinforces your answers in resonance quiz problems.

  5. Bond Order, Length & Strength -

    Understand that bond order = (bonding electrons − antibonding electrons)/2 (as explained by Chemistry LibreTexts); increasing bond order shortens and strengthens bonds (e.g., C≡C vs C=C vs C - C). Recognizing this trend helps predict molecular stability and reactivity in both molecular geometry and resonance quizzes. Drawing correlations between bond order and bond length makes your quiz answers stand out.

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