AP Chemistry : VSEPR and Bond Hybridization

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

Example Question #854 : Ap Chemistry

What is the molecular shape of ?

Possible Answers:

Trigonal pyramidal

Tetrahedral

Octahedral

Trigonal bipyramidal

Trigonal planar

Correct answer:

Trigonal pyramidal

Explanation:

NH3 is trigonal pyramidal because it has 4 electron domains, one of them being a lone pair of electrons and the other three being H atoms. When this is arranged in a three-dimensional space, it is trigonal pyramidal in shape.

Example Question #12 : Vsepr And Bond Hybridization

What is the bond angle in the following structure?

 

COH2

Possible Answers:
145
120
120 and 109.5
180
109.5
Correct answer: 120
Explanation:

CO2H is trigonal planar, so the bond angles will be 120 only. 

Example Question #11 : Vsepr And Bond Hybridization

What is/are the approximate bond angle(s) in the following molecule?

COH2

Possible Answers:

90, 120

90, 109.5

180

120

109.5

Correct answer:

120

Explanation:

COH2:        

               :O:

                ||

           H—C—H

 

This is a trigonal planar molecule, which only has bond angles of 120o

Example Question #12 : Vsepr And Bond Hybridization

Consider the following reactions that lead to the production of dichloromethane, CH_2Cl_2:

CH_4+Cl_2\rightarrow CH_3Cl +HCl

CH_3Cl+Cl_2\rightarrow CH_2Cl_2+HCl

 

What is the Cl-Cl bond angle of the product, dichloromethane?

Possible Answers:

180^{\circ}

109.4^{\circ}

120^{\circ}

90^{\circ}

None of the available answers.

Correct answer:

109.4^{\circ}

Explanation:

CH_2Cl_2 has four bonding regions and no lone pairs, so it has a tetrahedral geometry. The bond angles between all of the bonded atoms is 109.4^{\circ}.

Example Question #12 : Vsepr And Bond Hybridization

What are the flourine-flourine bond angles in ?

Possible Answers:

Correct answer:

Explanation:

has three bonds and one lone pair, so it has a tetrahedral geometry with a trigonal pyramidal shape. Due to its shape, the bond angles are .

Example Question #14 : Vsepr And Bond Hybridization

Which of the following molecules has the largest bond angle between atoms?

Possible Answers:

Correct answer:

Explanation:

In a compound, the atoms (and electron lone pairs) will be as far away from each other as possible due to the repulsive forces between electrons.

Methane  will have a tetrahedral geometry with bond angles of 109.5o. Both ammonia and water have lone electron pairs, which will decrease the bond angle between atoms. Water will be bent, while ammonia will be trigonal pyramidal. Bent and trigonal pyramidal are both permutations of tetrahedral, with slightly decreased bond angles.

 does not have lone electron pairs and will have a trigonal planar geometry. This results in bond angles of 120o. As a result,  has the largest bond angles out of the options.

Example Question #13 : Vsepr And Bond Hybridization

What are the bond angles in a molecule of ?

Possible Answers:

The molecule has multiple bond angle measures

Correct answer:

Explanation:

The carbon in the molecule is  hybridized and contains no lone pairs attached to it. This will give the molecule a tetrahedral geometry, standard of a central atom with four single bonds. Bond angles in a tetrahedral molecule are all .

Example Question #14 : Vsepr And Bond Hybridization

Rank the following compounds in order of increasing bond angle:

CO_2, CH_4, SF_6, NH_3, BF_3

If there are multiple bond angles in a single molecule, only consider the smaller bond angle

Possible Answers:

SF_6, \hspace{1 mm}BF_3, \hspace{1 mm}CH_4, \hspace{1 mm}NH_3, \hspace{1 mm}CO_2

CO_2, \hspace{1 mm}NH_3,\hspace{1 mm} CH_4,\hspace{1 mm} BF_3,\hspace{1 mm} SF_6

CH_4,\hspace{1 mm} CO_2,\hspace{1 mm} SF_6,\hspace{1 mm} NH_3,\hspace{1 mm} BF_3

SF_6,\hspace{1 mm}NH_3,\hspace{1 mm}CH_4\hspace{1 mm},BF_3,\hspace{1 mm}CO_2

CO_2,\hspace{1 mm}BF_3,\hspace{1 mm}NH_3, \hspace{1 mm}CH_4,\hspace{1 mm}SF_6

Correct answer:

SF_6,\hspace{1 mm}NH_3,\hspace{1 mm}CH_4\hspace{1 mm},BF_3,\hspace{1 mm}CO_2

Explanation:

CO_2 has two bonding regions and no lone pairs around the central atom, so it has a linear geometry, so its bond angle is 180^{\circ}.

CH_4 has four bonding regions and no lone pairs around the central atom, so it has a tetrahedral geometry with bond angles of 109.5^{\circ}.

SF_6 has six bonding regions and no lone pairs around the central atom, so it has an octahedral geometry with bond angles of 90^{\circ} and 180^{\circ}. For the sake of ranking, we will only consider the smaller angle.

NH_3 has three bonding regions and one lone pair around the central atom, so it has a trigonal pyramidal geometry with bond angles of 107^{\circ}.

BF_3 has three bonding regions and no lone pairs around the central atom, so it has a trigonal planar geometry with bond angles of 120^{\circ}.

 

Example Question #15 : Vsepr And Bond Hybridization

Consider the following reaction, called the Sabatier reaction:

CO_2+4H_2\rightarrow CH_4\hspace{1 mm}+2H_2O

What are the molecular geometries of the products?

Possible Answers:

Tetrahedral and bent

Square planar and bent

Square planar and linear

Tetrahedral and linear

Tetrahedral and tetrahedral

Correct answer:

Tetrahedral and bent

Explanation:

CH_4 has four bonding regions and no lone pairs, so it has a tetrahedral geometry. H_2O has two bonding regions, but it also has two lone pairs, so it has a bent geometry.

Example Question #18 : Vsepr And Bond Hybridization

What is the hybridization state of the central atom in ?

Possible Answers:

Correct answer:

Explanation:

According to hybrid orbital bonding theories, the hybridization state will use a number of orbitals equal to the steric number. The steric number is the number of atoms bonding to the central atom, plus any additional lone pairs.

Nitrogen (N) is the central atom because hydrogen atoms are never central. 

Nitrogen has 5 valence electrons, and in ,  it is bonded with 3 hydrogen atoms, which each borrow one of nitrogen's electrons to create a covalent bond. Subtracting those 3 borrowed electrons from its original 5 leaves 2 left, or 1 lone pair.

In total, the nitrogen atom has 3 bonds and 1 lone pair for a steric number of 4. As a result, its hybridization state must use 4 orbitals. The lowest energy way to do that is to use the lowest energy orbitals that are available. In this case, those are the s and p orbitals. So it uses one s and three p orbitals, giving the sp3 hybridization state. 

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