3. Bonding & Chemical Interactions

Terms in this set (31)

Draw out the backbone of the compound—that is, the arrangement of atoms. In general, the least
electronegative atom is the central atom. Hydrogen (always) and the halogens F, Cl, Br, and I
(usually) occupy a terminal position.
In HCN, H must occupy an end position. Of the remaining two atoms, C is the least
electronegative and, therefore, occupies the central position. Therefore, the skeletal structure is
as follows:
H - C - N
Count all the valence electrons of the atoms. The number of valence electrons of the molecule is
The HCN structure above does not satisfy the octet rule for C because C only has four valence
electrons. Therefore, two lone electron pairs from the N atom must be moved to form two more bonds
with C, creating a triple bond between C and N. To make it easier to visualize, bonding electron pairs
are represented as lines. You should be familiar with both dot and line notation for bonds.
H - C ≡ N:
Now, the octet rule is satisfied for all three atoms; C and N have eight valence electrons, and H has
two valence electrons.
Formal Charge
To determine if a Lewis structure is representative of the actual arrangement of atoms in a compound,
one must calculate the formal charge of each atom. In doing so, assume a perfectly equal sharing of all
bonded electron pairs, regardless of actual differences in electronegativity. In other words, assume
the sum of the valence electrons of all atoms present:
H has 1 valence electron
C has 4 valence electrons
N has 5 valence electrons; therefore,
HCN has a total of 10 valence electrons.
Draw single bonds between the central atom and the atoms surrounding it. Each single bond
corresponds to a pair of electrons:
H : C : N
Complete the octets of all atoms bonded to the central atom, using the remaining valence
electrons left be assigned. Recall that H is an exception to the octet rule because it can only have
two valence electrons. In this example, H already has two valence electrons from its bond with
C.
Place any extra electrons on the central atom. If the central atom has less than an octet, try to
write double or triple bonds between the central and surrounding atoms using the lone pairs on
the surrounding atoms.
The HCN structure above does not satisfy the octet rule for C because C only has four valence
electrons. Therefore, two lone electron pairs from the N atom must be moved to form two more bonds
with C, creating a triple bond between C and N. To make it easier to visualize, bonding electron pairs
are represented as lines. You should be familiar with both dot and line notation for bonds.
H - C ≡ N:
Now, the octet rule is satisfied for all three atoms; C and N have eight valence electrons, and H has
two valence electrons.
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