Chem201X In-Class Worksheet #12: Hybrid Orbitals and Gas Laws Answer Key
1. Dopamine is a neurotransmitter that helps control the brain's reward and pleasure centers. Its
Lewis structure is shown below with some of the atoms labeled with numbers. Referring to
this structure complete the table below.
Geometry
Angle
Hybridization
O1
bent
<109.5º
sp3
C2
trignal planar
120º
sp2
C3
tetrahedral
109.5º
sp3
N
trigonal pyramidal
<109.5º
sp3
2. What is the hybridization on the central atom in TeF4? Draw orbital diagrams for Te alone
and Te in TeF4.
Te in TeF4
Te alone
5d
5d
sp3d hybridized
5p
sp3d
5s
3. Draw the Lewis structure for H2CNH.
a. What is the hybridization on the C and on the N? C is sp2 and N is sp2
b. In what orbital are the lone pair of electrons on the N located? sp2 hybrid orbital
c. Which orbitals from each atom overlap to form the σ bond between C and N?
C (sp2) – N(sp2)
d. Which orbitals from each atom overlap to form the π bond between C and N?
C(p)–N(p)
Chem201X In-Class Worksheet #12: Hybrid Orbitals and Gas Laws Answer Key
4. Aluminum reacts with oxygen gas according to the reaction, 4 Al(s) + 3 O2(g) → 2 Al2O3(s).
PV = nRT
R = 0.08206 L·atm/mol·K
a. How many moles of oxygen are required to react with aluminum to produce 15.2 g of
aluminum oxide? (keep all digits of your answer in your calculator for the next part)
[0.223617 mol O2]
15.2 g Al2O3 | mol Al2O3
| 3 mol O2
= 0.223617 mol O2
| 101.96 g Al2O3 | 2 mol Al2O3
b. What volume of oxygen gas at 741 mmHg and 19.5 °C will produce 15.2 g of aluminum
oxide? [5.51 L]
741 mmHg | 1 atm
= 0.975 atm
| 760 mmHg
PV = nRT
19.5 ºC + 273.15 = 292.65 K
V = nRT/V = (0.223617 mol O2)(0.08206 Latm/molK)(292.65 K)/0.975 atm
= 5.5078 L = 5.51 L
c. If 3.70 g of Al is placed in a 2.15 L container of oxygen at STP (0ºC, 1 atm), assuming a
complete reaction, which compound will be the limiting reactant? How many grams of
aluminum oxide will be produced? [O2 is LR; 6.52 g Al2O3]
3.70 g Al | mol Al | 2 mol Al2O3 | 101.96 g Al2O3 = 6.99 g Al2O3
| 26.98 g Al | 4 mol Al
| 1 mol Al2O3
n = PV/RT = (1 atm)(2.15 L)/[(0.08206 L·atm/mol·K)(273.15 K)] = 0.095919 mol O2
0.095919 mol O2 | 2 mol Al2O3 | 101.96 g Al2O3
| 3 mol O2 | 1 mol Al2O3
= 6.52 g Al2O3 and O2 is LR