Molecular Structure Information Sheet (Page 2 of 3) in pdf

Chemistry 111, section 2, Spring 2006

Answers to even # questions

Chapter 10

formal charge on each F is zero. On O it is -1 and on S it is +1. We cannot make a double bond

between S and O because there are no available 2p orbitals on S (they have been used in the

hybridization). Since O is more electronegative than S, we want the negative charge on O and

the positive charge on S. This is the correct Lewis structure. There are five electron “lumps”

around S, making the electron geometry trigonal bipyramidal. All

positions are occupied by F and O atoms, so the molecular geometry is

:F:

also trigonal bipyramidal.

.. F:

The central S has five sigma bonds and no

lone pairs, or a total of 5 hybrid orbitals.

:F: ..

:O:

That would require 1 s, 3 ps and 1 d orbital

from S, making the hybridization sp

d) Br

One of the Br is the central atom; there are 3(7) + 1 = 22 valence electrons. Make

the 2 sigma bonds and fill the outer atoms. That leaves 6 electrons, so put them on the central

Br. The two outer Br have filled octets and zero formal charge. The central Br has one extra

electron pair and a formal charge of -1. There is no way to improve the charge; this is as close to

zero as we can get with this structure, so this is the correct Lewis structure.

There are five electron “lumps”, making the electron geometry trigonal

bipyramidal. Two positions are occupied by Br atoms with the three lone

:Br:

pairs in the trignonal plane, maximizing their separation at 120°, so the

. .

molecular geometry is linear.

:Br

. .

The central Br has two sigma bonds and 3 lone

:Br:

pairs, or a total of 5 hybrid orbitals. That would

require 1 s, 3 ps, and 1 d atomic orbitals, making

the hybridization sp

10.30. a) Carbon 1 has three sigma bonds and no lone pairs, so it is sp

hybridized. Carbon 2,

likewise, is sp

hybridized.

b) Around carbon 1 the three sp

orbitals are in a planar trigonal arrangement, thus the

bond angle A is that for a planar trigonal geometery, or 120°. The same is true for the other two

angels for the same reason.

c) No, there is only possible configuration of acrolein. You might think that rotation

around the C=C double bond would put the –CHO group “up” instead of “down” as it’s drawn in

the book, but notice that the far left-hand carbon has two identical H atoms, one “up” and one

“down”. Thus, twisting the C=C double bond by 180° by “holding on” to the –CHCHO part of

the molecule will produce the same –CH

on the left-hand side of acrolein. Both “ends” that are

connected by a double bond must be asymmetric in order for cis and trans isomers to be possible,

and that’s not the case here.

10.32. a) A: 120° because the central C has trigonal planar electronic geometry.

B: 109.5° because the central O has tetrahedral electronic geometry.

C: 109.5° because the central C has tetrahedral electronic geometry.

Page 2

Molecular Structure Information Sheet Page 2

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