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. Formaldehyde, CH₂O

Looking at the Periodic Table, C and H are Intermediate while O is Hi, therefore covalent;valence electrons = 4 + 2x1 + 6 = 12

6y + 2 = 6 x 2 + 2 = 14; so molecule has 2 fewer electrons than required for all single bonds, 1 double bond

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

Considering C as the central atom, have 3 bonded atoms and no lone-pairs, therefore

steric number = 3, so trigonal planar electronic geometry, and 3 atoms so

trigonal planar molecular geometry

or, showing the double bond

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. C and H differ only slightly in EN (2.5 & 2.1), so the C-H bonds will be only slightly polar.

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

The polar contributions of the angled slightly polar C-H bonds will not cancel the C-O bond, so overall the molecule is 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

2. Carbon dioxide, CO₂

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

valence electrons = 4 + 2x6 = 16

6y + 2 = 20, thus 4 fewer electrons than required for all single bonds, 4/2 = 2 multi-bonds (2 double or 1 triple)

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

Considering C as the central atom, have 2 bonded atoms and no lone-pairs, therefore

steric number = 2, so linear electronic geometry, and

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.

However, the two polar bonds exactly cancel each other since they point in opposite directions, so the molecule is not 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

3. Chloromethane, CH₃Cl

Looking at the Periodic Table, C and H are Intermediate and Cl is Hi, therefore covalent;

valence electrons = 4 + 3x1 + 7 = 14

6y + 2 = 6(2) + 2 = 14, thus 8 electrons in single bonds around central atom.

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,

Since no lone pairs, molecular geometry = electronic 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

4. Acetic acid, CH₃CO₂H

Looking at the Periodic Table, C and H are Intermediate and O is Hi, therefore covalent;

valence electrons = 2x4 + 4x1 + 2x6 = 24

6y + 2 = 6(4) + 2 = 26, thus 2 fewer electrons than required for all single bonds, 2/2 = 1 double bond.

LS: from formula can see that three hydrogens and one carbon are bonded to first carbon, with th etwo oxygens bonded to the second giving =

Consider the two carbons individually as central atoms.

C-1 as the central atom, have 4 bonded atoms (3H & C) and no lone-pairs, therefore

steric number = 4, so tetrahedral electronic geometry,

Since no lone pairs, molecular geometry = electronic geometry

C-2 as central atom, have 3 bonded atoms (C & 2O) and no lone pairs therefore trigonal planar electronic and molecular geometries.

Overall the molecule will then look like:

rotated =

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. Ozone, O₃

All three atoms identical, therefore covalent;

valence electrons = 3x6 = 18

6y + 2 = 6(3) + 2 = 20, thus 2 fewer electrons than required for all single bonds, 2/2 = 1 double bond.

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

However, since three atoms identical the bonding must also be identical - need resonance structures:

Considering the middle O as the central atom have 2 bonded atoms and one lone pair,

steric number = 3, so trigonal planar electronic geometry,

But one lone pair, so V-shape or Bent 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

6. Sulfite ion, SO₃^2-

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

valence electrons = 6 + 3x6 + 2= 26

6y + 2 = 6(4) + 2 = 26, thus atoms have enough electrons for octets with no multibonding.

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

Considering S as the central atom, have 4 bonded atoms and 1 lone-pair, therefore

steric number = 4, so tetrahedral electronic geometry,

But one lone pair so trigonal pyramidal Molecular geometry,

rotated =

Polarity:

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

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

The polar contributions of the angled polar S-O bonds along the central axis add up so overall the molecule is polar. (Note that the dipole is in the opposite direction to that seen earlier in ammonia.)

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. Water, H₂O

Looking at the Periodic Table, H is Intermediate and O 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 O 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

rotated =

Polarity:

H and O have significantly different Electronegativity values (2.1& 3.5), so the H-O bond will be polar, with O partially negative.

The two dipoles add together in the water molecule which 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

8. Carbocation, CH₃⁺

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

valence electrons = 4 + 3x1 - 1 = 6

three bonds possible, since only three 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 3 bonded atoms and no lone-pairs, therefore

steric number = 3, so trigonal planar electronic geometry, and 3 atoms, so trigonal planar 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

**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

**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

**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

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 5 November 2009

VSEPR Theory and Molecular Geometry

1. Formaldehyde, CH2O

Polarity:

2. Carbon dioxide, CO2

Polarity:

3. Chloromethane, CH3Cl

4. Acetic acid, CH3CO2H

5. Ozone, O3

6. Sulfite ion, SO3 2-

Polarity:

7. Water, H2O

Polarity:

8. Carbocation, CH3+

VSEPR Theory & Molecular Geometry

1. Formaldehyde, CH₂O

2. Carbon dioxide, CO₂

3. Chloromethane, CH₃Cl

4. Acetic acid, CH₃CO₂H

5. Ozone, O₃

6. Sulfite ion, SO₃^2-

7. Water, H₂O

8. Carbocation, CH₃⁺