# Molecular Geometry And Hybridization Quiz #2

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| By Ashley Vickers
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Ashley Vickers
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Quizzes Created: 7 | Total Attempts: 5,327
Questions: 10 | Attempts: 290

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• 1.

### In order to produce sp hybrid orbitals, __________ s atomic orbitals(s) and ___________ p atomic orbital(s) must be mixed.

• A.

One, two

• B.

One, three

• C.

One, one

• D.

Two, two

• E.

Two, three

B. One, three
Explanation
To produce sp hybrid orbitals, one s atomic orbital and three p atomic orbitals must be mixed. This is because the sp hybrid orbitals are formed by combining one s orbital with three p orbitals, resulting in four sp hybrid orbitals. This hybridization allows for the formation of stronger and more stable bonds in certain molecules.

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• 2.

### The angles between sp orbitals are __________.

• A.

45

• B.

180

• C.

90

• D.

109.5

• E.

120

E. 120
Explanation
The angles between sp orbitals are 120. In sp hybridization, one s orbital and one p orbital combine to form two sp hybrid orbitals. These orbitals are arranged in a trigonal planar geometry, with an angle of 120 degrees between them. This arrangement allows for maximum overlap and bonding with other atoms or orbitals.

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• 3.

### There are ____________ sigma bonds and ___________ pi bonds in the H--CC--H molecule.

• A.

Three, two

• B.

Three, four

• C.

Four, three

• D.

Two, three

• E.

Five, zero

A. Three, two
Explanation
In the H--CC--H molecule, there are three sigma bonds and two pi bonds. Sigma bonds are formed by the overlap of atomic orbitals along the bonding axis, while pi bonds are formed by the sideways overlap of p orbitals. Since there are two carbon atoms connected by a double bond (CC), there is one sigma bond and one pi bond between them. Additionally, each carbon atom is bonded to a hydrogen atom (H--C and C--H), resulting in two sigma bonds. Therefore, the molecule has three sigma bonds and two pi bonds.

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• 4.

### The total number of pi bonds in the H--CC--CC--CN molecule is _______.

• A.

3

• B.

4

• C.

6

• D.

9

• E.

12

C. 6
Explanation
In the given molecule, H--CC--CC--CN, there are three carbon-carbon double bonds (CC) and one carbon-nitrogen triple bond (CN). Each double bond consists of one pi bond, and each triple bond consists of two pi bonds. Therefore, the total number of pi bonds in the molecule is 3 (from the double bonds) + 2 (from the triple bond) = 5.

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• 5.

### The O--C--O bond angle in the CO ion is approximately ___________.

• A.

90

• B.

109.5

• C.

120

• D.

180

• E.

60

C. 120
Explanation
The O--C--O bond angle in the CO ion is approximately 120 degrees. This is because the CO ion has a linear molecular geometry, meaning that the two oxygen atoms are directly opposite each other with the carbon atom in the middle. In a linear molecular geometry, the bond angle is always 180 degrees, but since the CO ion has a charge, it experiences electron-electron repulsion which pushes the oxygen atoms slightly closer together, resulting in a bond angle of approximately 120 degrees.

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• 6.

### The central iodine atom in the ICl ion has _________ nonbonded electron pairs and ___________ bonded electron pairs in its valence shell.

• A.

Two, two

• B.

Three, four

• C.

One, three

• D.

Three, two

• E.

Two, four

E. Two, four
Explanation
The central iodine atom in the ICl ion has two nonbonded electron pairs and four bonded electron pairs in its valence shell.

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• 7.

### The central Xe atom in the XeF molecule has ________ unbonded electron pairs and __________ bonded electron pairs in its valence shell.

• A.

One, four

• B.

Two, four

• C.

Four, zero

• D.

Four, one

• E.

Four, two

B. Two, four
Explanation
The central Xe atom in the XeF molecule has two unbonded electron pairs and four bonded electron pairs in its valence shell. This means that there are two pairs of electrons that are not involved in any chemical bonds and four pairs of electrons that are shared with other atoms in the molecule.

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• 8.

### An electron domain consists of ____________.                             a) a nonbonding pair of electrons                             b) a single bond                             c) a multiple bond

• A.

A only

• B.

B only

• C.

C only

• D.

A, b, and c

• E.

B and c

D. A, b, and c
Explanation
An electron domain consists of a nonbonding pair of electrons, a single bond, and a multiple bond. This means that it can include any of these three types of electron configurations.

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• 9.

### According to VSEPR theory, if there are three electron domains on a central atom, they will be arranged such that the angles between the domains are _____________.

• A.

90

• B.

180

• C.

109.5

• D.

360

• E.

120

E. 120
Explanation
According to VSEPR theory, if there are three electron domains on a central atom, they will be arranged such that the angles between the domains are 120 degrees. This is because when there are three electron domains, they will arrange themselves in a trigonal planar geometry, with each domain pointing towards the corners of an equilateral triangle. The angles between the domains in this arrangement are 120 degrees.

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• 10.

### The electron-domain geometry and the molecular geometry of a molecule of the general formula AB are _________.

• A.

Never the same

• B.

Always the same

• C.

Sometimes the same

• D.

Not related

• E.

Mirror images of one another

C. Sometimes the same
Explanation
The electron-domain geometry and the molecular geometry of a molecule of the general formula AB can be sometimes the same. This means that in certain cases, the arrangement of the electron domains around the central atom can result in the same overall molecular shape. However, it is important to note that this is not always the case, as there are instances where the electron-domain geometry and molecular geometry can differ.

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• Current Version
• Mar 20, 2023
Quiz Edited by
ProProfs Editorial Team
• Nov 12, 2014
Quiz Created by
Ashley Vickers

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