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-