Humboldt State University ® Department of Chemistry

Richard A. Paselk

General Chemistry
Spring 2011

Exercise: Lewis Structures & VSEPR Theory
© R. Paselk 2008

Discussion Modules

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*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
	General Chemistry	Spring 2011
Exercise: Lewis Structures & VSEPR Theory	© 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. Determine the Geometry of Phosphorus pentachloride, PCl₅.

First determine whether this is an ionic or covalent compound.

Looking at the Periodic Table we see that P is Intermediate and Cl is Hi , so we expect this to be a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, P is in group V so it has 5 valence electrons while Cl is in group VII so each has 7 valence electrons,

valence electrons = 5 + 5(7) = 40

For Clark's rules we see that there are 1(P) and 5(Cl) atoms or a total of 6 atoms other than hydrogen,

6y + 2 = 6(6) + 2 = 38

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 40 - 38 = 2, we have two extra electrons around the central atom!

This compound has an expanded valence shell with 10 electrons in 5 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 5 bonded atoms + 0 lone pairs = 5

For SN = 5 the Electronic Geometry is then Trigonal bipyramidal:

Since there are NO lone pairs the Molecular Geometry is identical to the electronic geometry, and the molecule is Trigonal bipyramidal:

= =

Polarity:

P and Cl have significantly different Electronegativity values (2.1& 3.0), so the P-Cl bonds will be polar, with Cl partially negative.

Of course the equatorial chlorine-iodine bonds will cancel since they are completely symmetrically arranged.

The axial chlorine-iodine bonds will also cancel since they are opposed, thus the molecule is non-polar.

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

**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

**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. Determine the Geometry of Sulfur tetrafluoride, SF₄.

First determine whether this is an ionic or covalent compound.

Looking at the Periodic Table we see that S is Intermediate and F is Hi , so we expect this to be a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, S is in group VI so it has 6 valence electrons while F is in group VII so each has 7 valence electrons,

valence electrons = 6 + 4(7) = 34

For Clark's rules we see that there are 1(S) and 5(F) atoms or a total of 5 atoms other than hydrogen,

6y + 2 = 6(5) + 2 = 32

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 34 - 32 = 2, we have two extra electrons around the central atom!

This compound has an expanded valence shell with 10 electrons in 5 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 4 bonded atoms + 1 lone pairs = 5

For SN = 5 the Electronic Geometry is then Trigonal bipyramidal:

Since there is one lone pair the Molecular Geometry is different from the electronic geometry, and the molecule is Seesaw shaped:

=

Polarity:

S and F have significantly different Electronegativity values (2.5 & 4.0), so the S-F bond will be polar, with F partially negative.

Of course the axial sulfur-fluorine bonds will cancel since they completely oppose each other.

However, the two equatorial bonds are unbalanced, making Sulfur tetafluoride a polar molecule, as seen in the diagram below:

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

**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. Determine the Geometry of Chlorine trifluoride, ClF₃.

First determine whether this is an ionic or covalent compound.

Looking at the Periodic Table we see that both Cl and F are Hi , so we expect this to be a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, Cl and F are in group VII so each has 7 valence electrons,

valence electrons = 4(7) = 28

For Clark's rules we see that there are 1(Cl) and 3(F) atoms or a total of 4 atoms other than hydrogen,

6y + 2 = 6(4) + 2 = 26

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 28 -26 = 2, we have two extra electrons around the central atom!

This compound has an expanded valence shell with 10 electrons in 5 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 3 bonded atoms + 2 lone pairs = 5

For SN = 5 the Electronic Geometry is then Trigonal bipyramidal:

Since there are two lone pairs the Molecular Geometry is different from the electronic geometry, and the molecule is T-shaped:

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

**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. Determine the Geometry of Triiodide ion, I₃^-.

First determine whether this is an ionic or covalent compound.

All of the atoms are identical, so this is a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, I is in group VII so each has 7 valence electrons and we have one extra electron because of the -1 charge,

valence electrons = 3(7) + 1= 22

For Clark's rules we see that there are 1(Cl) and 3(F) atoms or a total of 4 atoms other than hydrogen,

6y + 2 = 6(3) + 2 = 20

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 22 - 20 = 2, we have two extra electrons around the central atom!

This compound has an expanded valence shell with 10 electrons in 5 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 2 bonded atoms + 3 lone pairs = 5

For SN = 5 the Electronic Geometry is then Trigonal bipyramidal:

Since there are three lone pairs the Molecular Geometry is different from the electronic geometry, and the molecule is linear:

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

**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

5. Determine the Geometry of Arsenic hexafluoride ion, AsF₆^-.

First determine whether this is an ionic or covalent compound.

Arsenic is intermediate while fluorine is Hi, so this is a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, As is in group V so it has 5 valence electrons while F is in group VII with 7 electrons each, and we have one extra electron because of the -1 charge,

valence electrons = 5 + 6(7) + 1= 48

For Clark's rules we see that there are 1(Cl) and 3(F) atoms or a total of 4 atoms other than hydrogen,

6y + 2 = 6(7) + 2 = 44

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 48 - 44 = 4, we have four extra electrons around the central atom!

This compound has an expanded valence shell with 12 electrons in 6 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 6 bonded atoms + 0 lone pairs = 6

For SN = 6 the Electronic Geometry is then Octahedral:

Since there are no lone pairs the Molecular Geometry is identical to the electronic geometry, and the molecule is Octahedral:

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

**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

6. Determine the Geometry of Iodine pentachloride, ICl₅.

First determine whether this is an ionic or covalent compound.

Both Cl and I are Hi, so this is a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, both Cl and I are in group VII with 7 electrons each,

valence electrons = 6(7) = 42

For Clark's rules we see that there are 1(I) and 5(Cl) atoms or a total of 6 atoms other than hydrogen,

6y + 2 = 6(6) + 2 = 38

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 42 - 38 = 4, we have four extra electrons around the central atom!

This compound has an expanded valence shell with 12 electrons in 6 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 5 bonded atoms + 1 lone pairs = 6

For SN = 6 the Electronic Geometry is then Octahedral:

Since there is one lone pair the Molecular Geometry is different from the electronic geometry, and the molecule is tetragonal pyramidal:

=

Polarity:

I and Cl have somewhat different Electronegativity values (2.5 & 3.0), so the I-Cl bond will be polar, with Cl partially negative.

Of course the equatorial chlorine-iodine bonds will cancel since they are completely symmetrically arranged.

However, the remaining bond is unbalanced, making Iodine pentachloride a polar molecule, as seen in the diagram below:

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

**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

7. Determine the Geometry of Xenon tetrafluoride, XeF₄.

First determine whether this is an ionic or covalent compound.

Xenon is intermediate while fluorine is Hi, so this is a covalent compound.

Next determine the number of valence electrons.

Looking at the Periodic Table, Xe is in group VIII so it has 8 valence electrons while F is in group VII with 7 electrons each,

valence electrons = 8 + 4(7) = 36

For Clark's rules we see that there are 1(Xe) and 4(F) atoms or a total of 5 atoms other than hydrogen,

6y + 2 = 6(5) + 2 = 32

Comparing the number of valence electrons needed for eight valance electrons around each atom (6y +2) to the actual we find = 36 - 32 = 4, we have four extra electrons around the central atom!

This compound has an expanded valence shell with 12 electrons in 6 pairs.

We can now draw a correct Lewis Structure

The Steric Number (SN) = 4 bonded atoms + 2 lone pairs = 6

For SN = 6 the Electronic Geometry is then Octahedral:

Since there are two lone pairs the Molecular Geometry is not the same as the electronic geometry, and keeping the lone pair as far apart as possible, the molecule is Square planar:

=

Polarity:

Due to the symmetry of this molecule it cannot be polar, regardless of electronegativity differences. Any bond dipoles which may exist will cancel completely.

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

Sections 7.6 (pp 293-4) in your textbook-you should be able to do the examples and exercises in the assigned problems listed on the Schedule.

VSEPR Theory & Molecular Geometry

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

© R A Paselk

Last modified 2 April 2008

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Discussion Modules

VSEPR Theory and Molecular Geometry

1. Determine the Geometry of Phosphorus pentachloride, PCl5.

Polarity:

2. Determine the Geometry of Sulfur tetrafluoride, SF4.

Polarity:

3. Determine the Geometry of Chlorine trifluoride, ClF3.

4. Determine the Geometry of Triiodide ion, I3-.

5. Determine the Geometry of Arsenic hexafluoride ion, AsF6-.

6. Determine the Geometry of Iodine pentachloride, ICl5.

Polarity:

7. Determine the Geometry of Xenon tetrafluoride, XeF4.

Polarity:

VSEPR Theory & Molecular Geometry

1. Determine the Geometry of Phosphorus pentachloride, PCl₅.

2. Determine the Geometry of Sulfur tetrafluoride, SF₄.

3. Determine the Geometry of Chlorine trifluoride, ClF₃.

4. Determine the Geometry of Triiodide ion, I₃^-.

5. Determine the Geometry of Arsenic hexafluoride ion, AsF₆^-.

6. Determine the Geometry of Iodine pentachloride, ICl₅.

7. Determine the Geometry of Xenon tetrafluoride, XeF₄.