Electronegativity and Lewis Structure Practice
ELECTRONEGATIVITY LEWIS STRUCTURE:
1.Determine total number of valence electrons
2. Any charges? YES – add (-ve charge)/subtract
(+ve charge)
3. Build skeleton structure (incomplete Lewis
Structure)
4. Group 14,15,16 atoms usually “central”
5. Hydrogen and Group 17 atoms “terminal”
6. Make multiple bonds only when necessary
7. Check - Noble gas electronic configuration at
each atom?
CALCULATING FORMAL CHARGE:
1) Draw Lewis Structure
2) Determine neutral valence of each atom (number of valence electrons)
3) Assign each atom half of bonding electrons + lone pairs
4) FC = Neutral Valence – Assigned electrons
Shape of Molecules (Only electron groups 2, 3, and 4)
VSEPR (Valence Shell Electron Pair Repulsion) Theory
MOLECULAR POLARITY
Review
Question 1
A. Draw the Lewis Structure and identify the formal charge on carbon in the bicarbonate ion
(HCO3)-. Show calculation for the formal charge.
1 1
Formal charge on C=valence e- - (lone pair e- +2 bonding e- )= 4-(0+2×8)= 0
B. Draw the Lewis structure for HCN, CH2NH, and CH3NH2. (Note: All contain a Carbon-
Nitrogen bond)
Which molecule to you expect to have the shortest nitrogen-to-carbon bond? Why?
HCN is expected to have the shortest nitrogen-to-carbon bonds. Because the nitrogen-to-carbon
bond in HCN is a triple bond and is the strongest. The stronger the bond, the shorter the bond
length.
Question 2
Arrange the following (Explain your answer – show Lewis structures and geometry where
applicable)
1. From lowest to highest bond angle
PH3, ClO4-, SCl2
SCl2< PH3< ClO4-
PH3 has an electron-group arrangement of AX3E1, the bond angle will be smaller than
ideal(109.5) due to the presence of a lone pair electron.
ClO4- has an electron-group arrangement of AX4, and has the ideal bond angle of
109.5 (even though one of the bonds is a single bond and others double bond – there
are equally contributing resonance structures so overall all the bond angles will be
equivalent)
SCl2 has an electron-group arrangement of AX2E2, the bond angle will be smaller
than that of PH3 due to the presence of two lone pair electrons instead of one. The
more lone pair electrons, the greater the repulsion, and the smaller the bond angle.
2. From lowest to highest formal charge on the atom that is bolded (consider the most
stable Lewis structure only)
ClO-, O3 (central oxygen atom), ClO4-
For ClO- the most stable Lewis structure would have the –ve formal charge on the more
electronegative atom (O in this case)
ClO-< ClO4-< O3
Question 3
a) Draw the Lewis structure(s) for [CH2CHCH2CN]. The molecule has a C-C-C-C-N
skeleton. Include lone pairs in your answer.
b) Calculate formal charge for N in the structure(s) drawn.
c) Indicate electron groups and the molecular geometry around each carbon.
from left to right:
C1: three bonding electron groups, trigonal planar geometry.
C2: three bonding electron groups, trigonal planar geometry.
C3: four bonding electron groups, tetrahedral geometry.
C4: two binding electron groups, linear geometry.
Note: electron group: single bond, double bond, triple bond, lone pair, or even lone electron.
Question 4
a) Draw the Lewis structure(s) for H2SO3 and SO32-. Include lone pairs in your answer.
Indicate all non zero formal charge on the atoms.
(Sulphur can expand its octet – is in the 3rd period. Elements in period 3 and below, can
expand their octet due to available empty d orbitals)
b) Which of the two (H2SO3 or SO32-) has equivalent resonance structures? Show all
equivalent resonance structures for that molecule
Question 5
a. Draw all possible resonance structures for CH3NCO.
a. Give the formal charge on each atom with a non-zero formal charge.
Formal charges given above the structures:
All formal charges 0 in structure 1
Formal charge on N in structure 2 = valence electrons – assigned electrons = 5 – 4 (4 bonded
electrons = 4)= +1
Formal charge on O in structure 2 = valence electrons – assigned electrons = 6 – 7 (3 lone pairs =
3 x 2 = 6; 1 bond =1; 6+ 1 = 7) = +1
Formal charge on N in structure 3 = valence electrons – assigned electrons = 5 – 6 (2 bonded
electrons = 2; + 2 lone pairs = 4; 2 + 4 = 6)= -1
Formal charge on O in structure 3 = valence electrons – assigned electrons = 6 – 5 (1 lone pairs =
= 2; + 3 bond =3; 2+ 3 = 5) = +1
b. Which of the resonance structures is most contributing? Explain why?
Structure 1 is most contributing as the formal charges on all atoms are 0.
Question 6
C2H4 + HBr C2H5Br
In the above reaction, determine the shape of the molecule (around either of the
carbons atom) in the reactant and compare that to the shape of the product.
H H
H H
C C + H Br H C C Br
H H H H
In the reactant:
Both carbons have 3 electron groups and all are bonded: (2 single bonds + 1 double bond) –
trigonal planar
In the product:
Both carbons have 4 electron groups and all are bonded: (4 single bonds) – tetrahedral
Question 7
For each of the following compare the electron geometry (total electron groups) and molecular
geometry (shape of the molecule) around:
a. Central oxygen atom for : H3O+ ; OH-; H2O
b. Central carbon atom for: CH3+ ; CH4; CH3-
