Unit 7-9 Test Review
APC MM
12-13
1. 14 lone pairs missing2. See picture3. See picture4. 7 pi bonds (7 double bonds)
3. Shapes aren’t the same. Carbon lacks a lone pair so its shape is trigonal planar. Nitrogen has a lone pair so it is trigonal pyramidal.
1. 8 lone pairs (green circled)
2. 4 pi bonds (4 double bonds)
Tell the hybridization for a carbon in the following…
1. CH4 – sp3
2. C2H2 – sp
3. C2H4 – sp2
4. C2H6 – sp3
• Predict whether the following molecules will be polar (have a dipole moment) or nonpolar (have a zero dipole moment.)
1. PCl3 – polar
2. SO3 – nonpolar
3. SF6 – nonpolar
• Tell the hybridization for the following…
1. XeF4 – sp3d2
2. XeF2 – sp3d
3. SF6 – sp3d2
4. SF4 – sp3d
Chlorines are 180° to each other in the para isomer which will cancel out the dipoles.
Chlorines are in the opposite directions which will cancel out each other
1. IF – polar
2. CS2 – nonpolar
3. IF5 – polar
1. Linear (3), trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar
2. Linear (2)/dumbbell, bent, trigonal pyramidal, seesaw, t-shaped, square pyramidal
3. Polar
4. Nonpolar
(614 kJ/mol + 146 kJ/mol) – (2•358 kJ/mol +348 kJ/mol) = -304 kJ/mol
• Although I3- is known, F3
- does not form.
Explain your answer.
Fluorine doesn’t have an available d-orbital to put extra electrons in like Iodine does because Fluorine is in the 2nd period and Iodine is in the 5th period of the PT.
• Pick the element with the larger value depending on the trend presented.1. atomic size: Rb or Sr
2. ionization energy: Cl or Br
3. electronegativiy: As or S
4. cation size: Ca2+ or K+
5. anion size: O2- or F-