Molecular Geometry And Hybridization Quiz #1

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| By Ashley Vickers
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Ashley Vickers
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Molecular Geometry And Hybridization Quiz #1 - Quiz

Hybridization is the mixing of atomic orbitals into new hybrid orbitals, suitable for the pairing of electrons. Molecular Geometry highly uses this concept. The quiz below is on the subject. May seem hard, but try it out.


Questions and Answers
  • 1. 

    For a molecule with the formula AB the molecular shape is _________.

    • A.

      Linear or bent

    • B.

      Linear or trigonal planar

    • C.

      Linear or T-shaped

    • D.

      T-shaped

    • E.

      Trigonal planar

    Correct Answer
    A. Linear or bent
    Explanation
    The molecular shape of a molecule with the formula AB can be either linear or bent. This is because the central atom (A) can have two or three bonding pairs of electrons, resulting in either a linear shape if there are two bonding pairs, or a bent shape if there are three bonding pairs. The actual shape depends on the arrangement of the bonding and non-bonding electron pairs around the central atom.

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  • 2. 

    According to VSEPR theory, if there are four electron domains in the valence shell of an atom, they will be arranged in a(n) _____________ geometry.

    • A.

      Octahedral

    • B.

      Linear

    • C.

      Tetrahedral

    • D.

      Trigonal planar

    • E.

      Trigonal bipyramidal

    Correct Answer
    C. Tetrahedral
    Explanation
    According to VSEPR theory, if there are four electron domains in the valence shell of an atom, they will be arranged in a tetrahedral geometry. In this arrangement, the four electron domains are positioned as far apart as possible, creating a three-dimensional shape with a central atom and four surrounding atoms or lone pairs. This geometry is commonly observed in molecules like methane (CH4), where the central carbon atom is surrounded by four hydrogen atoms.

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  • 3. 

    The electron-domain geometry and molecular geometry of iodine trichloride are _____________ and ______________, respectively.

    • A.

      Trigonal bipyramidal, trigonal planar

    • B.

      Tetrahedral, trigonal pyramidal

    • C.

      Trigonal bipyramidal, T-shaped

    • D.

      Octahedral, trigonal planar

    • E.

      T-shaped, trigonal planar

    Correct Answer
    C. Trigonal bipyramidal, T-shaped
    Explanation
    The electron-domain geometry of iodine trichloride is trigonal bipyramidal because it has five electron domains around the central iodine atom. The molecular geometry is T-shaped because it has three bonded atoms and two lone pairs of electrons, resulting in a T-shaped arrangement.

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  • 4. 

    The molecular geometry of _________ is square planar.

    • A.

      CCl4

    • B.

      XeF4

    • C.

      PH3

    • D.

      XeF2

    • E.

      ICl3

    Correct Answer
    B. XeF4
    Explanation
    XeF4 is the correct answer because it has a square planar molecular geometry. In XeF4, xenon (Xe) is surrounded by four fluorine (F) atoms in a square planar arrangement. This arrangement allows for the bond angles to be 90 degrees, resulting in a square shape. The other options, CCl4, PH3, XeF2, and ICl3, have different molecular geometries and do not have a square planar shape.

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  • 5. 

    The molecular geometry of the HO ion is ________.

    • A.

      Linear

    • B.

      Tetrahedral

    • C.

      Bent

    • D.

      Trigonal pyramidal

    • E.

      Octahedral

    Correct Answer
    D. Trigonal pyramidal
    Explanation
    The HO ion consists of one oxygen atom bonded to one hydrogen atom. The oxygen atom has two lone pairs of electrons, causing the molecule to have a trigonal pyramidal molecular geometry. In this geometry, the oxygen atom is at the center, and the three atoms (two lone pairs and one hydrogen) are arranged in a pyramid shape around it.

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  • 6. 

    The F-B-F bond angle in the BF molecule is __________.

    • A.

      90

    • B.

      109.5

    • C.

      120

    • D.

      180

    • E.

      60

    Correct Answer
    C. 120
    Explanation
    The F-B-F bond angle in the BF molecule is 120 degrees. This is because the BF molecule has a trigonal planar molecular geometry, with the central boron atom bonded to three fluorine atoms. In a trigonal planar arrangement, the bond angles between the atoms are all equal and measure 120 degrees.

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  • 7. 

    The O-S-O bond angle in SO is slightly less than ________.

    • A.

      90

    • B.

      180

    • C.

      109.5

    • D.

      120

    • E.

      60

    Correct Answer
    D. 120
    Explanation
    The O-S-O bond angle in SO is slightly less than 120. This is because the molecule has a bent or V-shaped geometry due to the presence of two lone pairs of electrons on the central sulfur atom. These lone pairs repel the bonding pairs, causing the bond angle to be slightly less than the ideal tetrahedral angle of 109.5 degrees.

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  • 8. 

    According to valence bond theory, which orbitals on bromine atoms overlap in the formation of the bond in Br?

    • A.

      3s

    • B.

      4p

    • C.

      3p

    • D.

      3d

    • E.

      4s

    Correct Answer
    B. 4p
    Explanation
    In valence bond theory, the formation of a bond involves the overlapping of atomic orbitals. In the case of bromine (Br), the correct answer is 4p. This is because bromine has an electron configuration of [Ar] 3d10 4s2 4p5. The valence electrons in bromine are in the 4p orbital, and during bond formation, these orbitals overlap with orbitals from other atoms to create the bond. Therefore, the 4p orbitals on bromine atoms overlap in the formation of the bond in Br.

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  • 9. 

    The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal. The hybridization of the central sulfur atom is ________.

    • A.

      Sp

    • B.

      Sp^2

    • C.

      Sp^3

    • D.

      Sp^3d

    • E.

      Sp^3d^2

    Correct Answer
    D. Sp^3d
    Explanation
    The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal, which means that there are five electron domains around the central sulfur atom. In order to accommodate these five electron domains, the central sulfur atom must undergo hybridization. The sp^3d hybridization allows the central sulfur atom to form five hybrid orbitals, which are arranged in a trigonal bipyramidal shape. Therefore, the correct hybridization of the central sulfur atom in this compound is sp^3d.

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  • 10. 

    The hybridization of orbitals on the central atom in a molecule is sp. The electron-domain geometry around this central atom is _________.

    • A.

      Octahedral

    • B.

      Linear

    • C.

      Trigonal planar

    • D.

      Trigonal bipyramidal

    • E.

      Tetrahedral

    Correct Answer
    B. Linear
    Explanation
    The hybridization of orbitals on the central atom in a molecule is sp. This means that the central atom has two hybridized orbitals and two unhybridized p orbitals. In a linear electron-domain geometry, the central atom is surrounded by two electron domains, which corresponds to the sp hybridization. Therefore, the correct answer is linear.

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  • Current Version
  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 12, 2014
    Quiz Created by
    Ashley Vickers
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