Humboldt State University ® Department of Chemistry

Richard A. Paselk

General Chemistry
Spring 2008

Exercise: VSEPR Theory and Molecular Geometry
© R. Paselk 2008

Discussion Modules

Use Back Button to Return to Notes

*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
	General Chemistry	Spring 2008
Exercise: VSEPR Theory and Molecular Geometry	© R. Paselk 2008
	Discussion Modules
	Use Back Button to Return to Notes

VSEPR Theory and Molecular Geometry

Periodic Table of the Elements

IA IIA IIIA IVA VA VIA VIIA VIIIA

H He

2 Li Be B C N O F Ne

3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

1. Methane, CH₄

Looking at the Periodic Table, carbon and hydrogen are both Intermediate, therefore covalent;

valence electrons = 4 + 4x1= 8

four bonds possible, since only 4 pairs, single bonds because only have H's bound to C.

LS: from symmetry C will be central atom, therefore=

Considering C as the central atom, have 4 bonded atoms and no lone-pairs, therefore

steric number = 4, so tetrahedral electronic geometry,

4 atoms so tetrahedral molecular geometry

rotated to a different angle =

Periodic Table of the Elements

IA IIA IIIA IVA VA VIA VIIA VIIIA

H He

2 Li Be B C N O F Ne

3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

2. Hydrogen sulfide

Looking at the Periodic Table, H is Intermediate and S is Hi, therefore covalent;

valence electrons = 6 + 2x1= 8

only 4 pairs, single bonds because only have H's bound to O, 2 bonds, since only 2 H's

LS: from symmetry O will be central atom, therefore =

Considering S as the central atom, have 2 bonded atoms and 2 lone-pairs, therefore

steric number = 4, so tetrahedral electronic geometry,

two lone pairs, so bent molecular geometry

Periodic Table of the Elements

IA IIA IIIA IVA VA VIA VIIA VIIIA

H He

2 Li Be B C N O F Ne

3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

3. Carbon monoxide, CO

Looking at the Periodic Table, carbon is Intermediate and oxygen is Hi, therefore covalent;

valence electrons = 4 + 6 = 10

6y + 2 = 14, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)

LS = :C:::O:

Considering C as the central atom, have one bonded atom and one lone-pair, therefore

steric number = 2, so linear electronic geometry, and two atoms so

linear molecular geometry

Polarity:

C and O have significantly different Electronegativity values (2.5& 3.5), so the C-O bond will be polar, with O partially negative.

Carbon monoxide is polar as shown with the dipole arrow in the image:

Periodic Table of the Elements

IA IIA IIIA IVA VA VIA VIIA VIIIA

H He

2 Li Be B C N O F Ne

3 Na Mg IIIB IVB VB VI VIIB VIIIB IB IIB Al Si P S Cl Ar

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe

6 Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

4. Ammonia, NH₃

Looking at the Periodic Table, hydrogen is Intermediate and nitrogen is Hi, therefore covalent;

valence electrons = 5 + 3x1= 8

only 4 pairs, single bonds because only have H's bound to N, 3 bonds, since only 3 H's

LS: from symmetry N will be central atom, therefore=

Considering N as the central atom, have 3 bonded atoms and one lone-pair, therefore

steric number = 4, so tetrahedral electronic geometry,

only 3 atoms so trigonal pyramidal molecular geometry

rotated to a different angle =

Polarity:

H and N have significantly different Electronegativity values (2.1& 3.0), so the H-N bond will be polar, with N partially negative.

Ammonia is polar as shown with the dipole arrow in the image below:

The polar contributions of the angled polar H-O bonds along the central axis add up so overall the molecule is polar.

In addition to these exercises you should familiarize yourself with the text materials.

VSEPR Theory & Molecular Geometry

Engravings of geometric solids from Encyclopedia Britannica 11^th edition (1910) vol 7.

© R A Paselk

Last modified 27 October 2008

	IA	IIA											IIIA	IVA	VA	VIA	VIIA	VIIIA
H																		He
2	Li	Be											B	C	N	O	F	Ne
3	Na	Mg	IIIB	IVB	VB	VI	VIIB	VIIIB			IB	IIB	Al	Si	P	S	Cl	Ar
4	K	Ca	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn	Ga	Ge	As	Se	Br	Kr
5	Rb	Sr	Y	Zr	Nb	Mo	Tc	Ru	Rh	Pd	Ag	Cd	In	Sn	Sb	Te	I	Xe
6	Cs	Ba	Lu	Hf	Ta	W	Re	Os	Ir	Pt	Au	Hg	Tl	Pb	Bi	Po	At	Rn

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Discussion Modules

VSEPR Theory and Molecular Geometry

1. Methane, CH4

2. Hydrogen sulfide

3. Carbon monoxide, CO

Polarity:

4. Ammonia, NH3

Polarity:

VSEPR Theory & Molecular Geometry

1. Methane, CH₄

4. Ammonia, NH₃