Chem 101 Book Answers (Ch.6)

Chem 101 Book Answers (Ch.6) - Chapter 6 Molecular...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6 Molecular Structure 6-1 FC l F F F Xe F F F F Cl F F F F 1. Draw the Lewis structure of each of the following ions, showing all nonzero formal charges. Indicate whether each ion is linear or bent. If the ion is bent, what is the bond angle? a) NO 2 1- b) N 3 1- c) ClO 2 1- A lone pair on the N results in three electron groups and a bent structure with ~120 o bond angle. ~120 o ONO NNN No lone pair on the central N result in two electron groups and a linear structure with a 180 o bond angle. OC lO O Cl O ~109 o Two bonding and two lone pairs results in four electron groups and a ~109 o bond angle O N O 3. Draw Lewis structures for the following molecules. Indicate nonzero formal charges and whether each is trigonal planar or trigonal pyramidal. a) PF 3 b) COCl 2 c) BF 3 FP F F Cl C O Cl F B F F P F F F O C Cl Cl F B FF One lone pair on P results in four electron groups and trigonal pyramidal structure with ~109 o bond angles. No lone pairs on central atom, so it has only three electron groups. The molecules are trigonal planar with 120 o bond angles. ~109 o ~120 o 120 o Note that BF 3 does not have a double bond because a double bond to a halogen puts positive formal charge on the halogen. Instead, the boron is electron deficient having only six electrons. 5. Describe the shapes of the following: a) ClF 3 SP = ½(32-28) = 2 shared pair, which is not sufficient for three Cl-F bonds. Cl uses an expanded valence shell, so use Eq. 6.1 to get lone pairs. Cl is in Group 7 and is in the +3 oxidation state, so LP = ½ (7-3) = 2. There are three bonds and two lone pairs, so the five electron regions are situated in a trigonal bipyramid arrangement about Cl. The lone pairs go into the equatorial positions, so ClF 3 is T-shaped. b) XeF 4 SP= ½ (40-36) = 2 shared pair, which is insufficient for four Xe-F bonds. Xe is in Group 8 and has a +4 oxidation state, so Eq 6.1 yields LP = ½ (8 – 4) = 2 lone pairs. Six electron regions assume octahedral geometry with the lone pairs opposite one another. XeF 4 is square planar. c) ClF 5 SP= ½ (48-42) = 3 shared pair, which is insufficient for five Cl-F bonds. Cl is in Group 7 and has a +5 oxidation state, so Eq 6.1 yields LP = ½ (7 – 5) = 1 lone pair. Six electron regions assume octahedral geometry with the lone pair in any position. ClF 5 is a square pyramid. 7. What is the hybridization on the central atom in each of the ions in Exercise 1? Central atom electron regions hybridization a) NO 2 1- 2 bonding + 1 lone pairs = 3 sp 2 b) N 3 1- 2 bonding + 0 lone pairs = 2 sp c) ClO 2 1- 2 bonding + 2 lone pairs = 4 sp 3
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Molecular Structure 6-2 9. What is the hybridization on the central atom in each of the molecules in Exercise 3?
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/30/2009 for the course CH Ch101 taught by Professor Bigham during the Fall '08 term at N.C. State.

Page1 / 5

Chem 101 Book Answers (Ch.6) - Chapter 6 Molecular...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online