Molecular Shapes (VSEPR): Quiz

Molecular Shapes (VSEPR): Quiz

7 questions

1. Question 1EasyWhat is the shape of methane, $\ce{CH4}$?

   ASquare planarBTetrahedralCTrigonal pyramidalDTrigonal planar
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   Answer: Tetrahedral

   4 bonding domains and no lone pairs on carbon spread out to the corners of a tetrahedron, 109.5° apart.

2. Question 2EasyHow many electron domains are on the carbon atom in $\ce{CO2}$?

   A2B4C6D8
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   Answer: 2

   Each C=O double bond counts as one domain, and carbon has no lone pairs: 2 domains, linear.

3. Question 3EasyWhat is the shape of water, $\ce{H2O}$?

   ALinearBBentCTetrahedralDTrigonal planar
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   Answer: Bent

   4 domains (2 bonds, 2 lone pairs) give tetrahedral electron geometry, but the atoms form a bent shape.

4. Question 4MediumWhich molecule is trigonal pyramidal?

   A$\ce{BF3}$B$\ce{NH3}$C$\ce{CH4}$D$\ce{SO3}$
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   Answer: $\ce{NH3}$

   N has 3 bonds and 1 lone pair. $\ce{BF3}$ and $\ce{SO3}$ have 3 domains and no lone pairs on the central atom (trigonal planar).

5. Question 5MediumPut the bond angles in order: $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$.

   A$\ce{CH4}$ > $\ce{NH3}$ > $\ce{H2O}$B$\ce{H2O}$ > $\ce{NH3}$ > $\ce{CH4}$CAll are 109.5°D$\ce{NH3}$ > $\ce{CH4}$ > $\ce{H2O}$
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   Answer: $\ce{CH4}$ > $\ce{NH3}$ > $\ce{H2O}$

   109.5° > about 107° > about 104.5°: each extra lone pair squeezes the angle further.

6. Question 6HardWhich molecule is polar?

   A$\ce{CO2}$B$\ce{CCl4}$C$\ce{BF3}$D$\ce{CHCl3}$
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   Answer: $\ce{CHCl3}$

   The others are symmetrical, so their bond polarities cancel. In $\ce{CHCl3}$ one bond (C–H) differs, so they do not cancel.

7. Question 7MediumWhat is the shape of $\ce{SF6}$?

   ATrigonal bipyramidalBOctahedralCHexagonal planarDSquare planar
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   Answer: Octahedral

   6 bonding domains and no lone pairs on sulfur point to the corners of an octahedron, 90° apart.