The Foundations of Chemistry

Step 4:

When the octets of the five F atoms have been satisfied, all 40 of the available electrons have

been added. The phosphorus (central) atom has a share of ten electrons.

Calculation of formal charges shows that

for P, FC 5 (50)0 and for F, FC 7 (16) 0

When an atom has a share of more than eight electrons, as does P in PF 5 , we say that

it exhibits an expanded valence shell.The electronic basis of the octet rule is that one sand

three porbitals in the valence shell of an atom can accommodate a maximum of eight

electrons. The valence shell of phosphorus has n3, so it also has 3dorbitals available

that can be involved in bonding. It is for this reason that phosphorus (and many other

representative elements of Period 3 and beyond) can exhibit an expanded valence shell.

By contrast, elements in the second rowof the periodic table can neverexceed eight elec-

trons in their valence shells, because each atom has only one sand three porbitals in that

shell. Thus, we understand why PF 5 can exist but NF 5 cannot.

EXAMPLE 7-8 Limitations of the Octet Rule

Write the Lewis formula for sulfur tetrafluoride, SF 4.

Plan

We apply the usual stepwise procedure. The calculation of SNAin step 2 shows only 6 e

shared, but a minimum of 8 eare required to bond four F atoms to the central S atom.

Limitation type D applies, and we proceed accordingly.

Solution

FF

Step 1: The skeleton is S

FF

Step 2: N 1    8 (S atom) 4   8 (F atoms) 40 eneeded

Step 3: A 1    6 (S atom) 4   7 (F atoms) 34 eavailable

Step 3: SNA 40  34  6 eshared. Four F atoms are bonded to the central

S. This requires a minimum of 8 e, but only 6 ehave been calculated in

step 2. This is therefore an example of limitation type D.

Step 2a: We increase Sfrom 6 eto 8 e.

Step 3:

Step 4: F

F F

F

S

F

F F

F

S

or

F

F

F F

F P

F

F

F F

F P

This is sometimes referred to as
hypervalence.

294 CHAPTER 7: Chemical Bonding

We first complete octets on the atoms around the central atom.