Download Midterm II Example (Key)

Midterm II Example (Key)
useful information
1
1
Electronegativity and polarity
a Rank the following elements from largest to smallest electronegativity: Cl, Al, P
Answer: Cl > P > Al
b Rank the following bonds from smallest to largest polarity: O-O, O-H, O-N
Answer: O-O < O-N < O-H
c Do the following molecules have a net dipole moment (yes/no)?
BF3 : No
ICl3 : Yes
XeF2 : No
TeF5 : Yes
2
2
Lewis Structures
a Draw the Lewis structure of SO3 , where all the atoms obey the octet rule. If the structure is
a resonance structure, then give all the resonance structures. If there are any formal charges,
indicate them as well.
b Draw the Lewis structure of IF5
c For the resonance structures below, encircle only those that are stable. Note that the formal
charges on the atoms are not indicated in these structures.
• A: Violation octet rule on lower C, unstable.
• B: No violation, stable.
• C: Violation octet rule on central C, unstable.
• D: No violation, stable.
3
3
VSEPR theory
a For the following molecule, describe the electron geometry, the molecular geometry, and
whether there are any deviating bond angles: SO2
Electron geometry: trigonal planar
Molecular geometry: bent
Deviating bond angles (yes/no): Yes
b For the following molecule, describe the electron geometry, the molecular geometry, and
whether there are any deviating bond angles: BrF3
Electron geometry: trigonal bipyramidal
Molecular geometry: T-shaped
Deviating bond angles (yes/no): Yes
c Circle all that are true according to VSEPR theory: (Answer: (a) and (c))
(a) Molecular geometry is determined by the repulsion between electrons that make up
bonds and lone pairs.
(b) Molecular geometry is only determined by the repulsion between electrons that make up
bonds.
(c) The repulsion due to a double bond is approximately equal to the repulsion due to a
single bond.
(d) The repulsion due to a double bond is always higher than the repulsion due to a single
bond.
4
4
Valence bond theory
a Circle all that are true: (Answer: (a) and (b))
(a) The hybridized orbitals on a given atom all have the same energy.
(b) The angles between sp3 hybridized orbitals are 109.5o .
(c) If carbon is sp hybridized, it always has a double bond.
(d) In a linear molecule of three atoms, the central atom must be sp hydridized.
b Give the hybridization state of the central atom in the following compounds:
CO2 : sp
PCl3 : sp3
SF6 : sp3 d2
c Give the hybridization for the circled atoms. Write ”unhybridized” if an atom is not hybridized.
i: sp2
ii: sp2
iii: sp3
5
5
Organic molecules
a Draw the line structure for this molecule.
FCH2 CH2 CH2 CHCH2
b Determine the mass percentage of fluorine in the molecule above.
molecular molar mass: C5 H9 F = 88.12 g/mol
mass percentage:
19.00 g/mol
88.12 g/mol
× 100% = 21.56%
Answer: 21.56%
c Write the molecular formula for this molecule in the form Cx Hy Oz , where x, y, z are integers.
Answer: C9 H6 O3
6
7