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.

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. Phosphide ion

The -ide ending indicates phosphide is the elemental anion (negatively charged ion) for phosphorous, so we will have to add electrons.

Looking at the Periodic Table, P is in Group 5, so valence electrons on P = 5, so will add 3 electrons to get the phosphide ion with a -3 charge (alternately, we can determine the charge by subtrating 8 from the group number, 5 - 8 = -3).

Arranging the resulting eight outer (valence) electrons into four pairs, and placing them on the four sides of a square around the symbol for phosphorus, P and "decorating" it, we get:

Note that the ion has a full octet of electrons (four pairs) and that the structure is enclosed in brackets to demostrate that the anion "owns" all of the electrons, and that the charge is distributed over the entire ion.

2. Bromide ion

The -ide ending indicates bromide is the elemental anion (negatively charged ion) for bromine, so we will have to add electrons.

Looking at the Periodic Table, Br is in Group 7, so valence electrons on Br = 7, so will add 1electron to get the bromide ion with a -1 charge,

Note that the ion has a full octet of electrons (four pairs) and that the structure is enclosed in brackets to demostrate that the anion "owns" all of the electrons, and that the charge is distributed over the entire ion

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. Rubidium fluoride

Looking at the Periodic Table, Rb is Lo and F is Hi, therefore ionic;

Rubidium keeps its elemental name so it is the elemental cation (positively charged ion), while fluoride is the elemental anion for fluorine.

valence electrons on F = 7, so will add 1electron to get the fluoride ion,

Rubidium is in Group 1 so will lose one electron to give Rb⁺

LS: to give a neutral compound we will then have to have a ratio of 1 Rb : 1 F, 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

4. Cesium nitride

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

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

Cesium is in Group 1 so will lose two electrons to give Cs ⁺

LS: to give a neutral compound we will then have to have a ratio of 3 Cs : 1 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

5. 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=

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. 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 S, 2 bonds, since only 2 H's

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

**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

Multiple Bonds

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

7. Carbon monoxide, CO

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

valence electrons = 4 + 6 = 10

Using Clark's Method, 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:

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. Barium iodide

Looking at the Periodic Table, Ba is Lo and I is Hi, therefore ionic;

valence electrons on I = 7, so will add 1electron to get the iodide ion,

Barium is in Group 2 so will lose two electrons to give Ba²⁺

LS: to give a neutral compound we will then have to have a ratio of 1 Ba: 2 I, 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

9. 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=

In addition to these exercises you should familiarize yourself with the materials in your text and lab manual.

Lewis Structures

© R A Paselk

Last modified 30 September 2008

Humboldt State University ® Department of Chemistry

Richard A. Paselk

Supplemental Study Modules

Lewis Structures

1. Phosphide ion

2. Bromide ion

3. Rubidium fluoride

4. Cesium nitride

5. Methane, CH4

6. Hydrogen sulfide

Multiple Bonds

7. Carbon monoxide, CO

8. Barium iodide

9. Ammonia, NH3

Lewis Structures

5. Methane, CH₄

9. Ammonia, NH₃