Humboldt State University ® Department of Chemistry

Richard A. Paselk

General Chemistry
Fall 2008

Exercise: Lewis Structures
© R. Paselk 2008

Supplemental Study Modules

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*Humboldt State University* ® *Department of Chemistry*

Richard A. Paselk
	General Chemistry	Fall 2008
Exercise: Lewis Structures	© R. Paselk 2008
	Supplemental Study Modules
	Use Back Button to Return to Notes

Lewis Structures

Recall that positively charged ions retain the element name in compounds, while negatively charged elemental ions are given the -ide ending. For covalent compounds the more positive (less electronegative) element is usually given the element name, while the more electronegative element is given the -ide ending.

In general we can predict how electrons distribute in compounds using the Hi-Lo-Intermediate electronegativity rules:

Compounds composed of a "Hi" and a "Lo" element will be ionic with the respective ions combining to give a neutral compound.
All other combinations will give covalent compounds following the "octet rule" whenever possible.

A Periodic Table is provided with Hi (red), Lo (gray) and Intermediate (green) Electronegativities indicated to aid in prediting ionic vs. covalent compounds.

Multiple Bonds

For covalent compounds we somethimes get multiple bonds - so now we need a way of determining when such bonds occur. Recall we must show an octet (or duet for Period I) in the outer-most shell (valence electrons). When this does not occur with single electron pairs (bonds) between atoms can sometimes make it happen with multiple bonds. You might find "Clark's Method" useful for determining the bonding patterns of various molecules:

Clark's Method for determining bonding in covalent Lewis Structures
- Add up all of the valence electrons in the structure (remember to add one electron for each negative charge, or subtract one for each positive charge)
  - If S e^- = 6y + 2 where y = # atoms other than H, then octet rule is followed with single bonds only.
  - If S e^- < 6y + 2 then probably have multiple bonding with the number of multiple bonds = D/2 (remember a triple bond is 2 multiple bonds!).
- If you can draw more than one structure, then chose the most symmetrical.

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

From Clark's Method need 6y + 2 = 6 x 2 + 2 = 14 to fullfill octet rule with single bonds;

Comparing with the valence electrons, the molecule has 2 fewer electrons than required for all single bonds, therefore1 double bond

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

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. Strontium nitride

Looking at the Periodic Table, Sr is Lo and N is Hi, therefore ionic;

valence electrons on N = 5, so will add 3electrons to get the nitride ion,

Strontium is in Group 2 so will lose two electrons to give Sr²⁺

LS: to give a neutral compound we will then have to have a ratio of 3 Sr : 2 N, giving:

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. Cyanide ion, CN^-

Looking at the Periodic Table, C is Intermediate while N is Hi, therefore covalent; it is a molecular ion.

valence electrons = 4 + 5 + 1 = 10

Using Clark's Method, 6y + 2 = 6 x 2 + 2 = 14; so molecule has 4 fewer electrons than required for all single bonds, 4/2 = 2 multibonds = 1 triple bond since cannot have multibond with H.

LS: since C needs more electrons and prefers four bonds, make it central giving: [:C:::N:]^-

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. Carbon dioxide, CO₂

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

valence electrons = 4 + 2x6 = 16

Using Clark's Method, 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 = :O::C::O:

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. Chloromethane, CH₃Cl

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

valence electrons = 4 + 3x1 + 7 = 14

Using Clark's Method, 6y + 2 = 6(2) + 2 = 14 = valence electrons, thus 8 electrons in single bonds around central atom.

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

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. 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

Using Clark's Method, 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 the two oxygens bonded to the second giving =

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. Ozone, O₃

All three atoms identical, therefore covalent;

valence electrons = 3x6 = 18

Using Clark's Method, 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:

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. 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 =

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

9. 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 =

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

10. 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=

**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

**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 materials in your text and lab manual.