Incomplete octets

• We have seen previously that for elements below atomic number 20 the?octet rule?states that the atoms try to achieve 8 electrons in their valence shells, so they have the same electron configuration as a noble gas
• However, there are some elements that are exceptions to the?octet rule, such a H, Li, Be, B and Al
  • H can achieve a stable arrangement by gaining an electron to become 1s², the same structure as the noble gas helium
  • Li does the same, but losing an electron and going from 1s²2s^1?to 1s²?to become a Li⁺?ion
  • Be from group 2, has two valence electrons and forms stable compounds with just four electrons in the?valence shell
  • B and Al in group 13 have 3 valence electrons and can form stable compounds with only 6 valence electrons

Table showing examples of incomplete octets

Expansion of the octet

• Elements in period 3 and above have the possibility of having more than eight electrons in their valence shell
• This is because there is a d-subshell present which can accommodate additional pairs of electrons
• This is known as the?expansion of the octet
• The concept explains why structures such as PCl₅?and SF₆?exist, which have 10 and 12 bonding pairs of electrons respectively, around the central atom

Five electron pairs

Phosphorus pentachloride, PCl₅

• An example of a molecule with five bonding electron pairs is phosphorus pentachloride, PCl₅
• The total number of valence electrons is = P + 5Cl = 5 + (5 x 7) = 40
• The number of bonding pairs is 5, which accounts for 10 electrons
• The remaining 30 electrons would be 15 lone pairs, so that each Cl has 3 lone pairs
• The completed Lewis diagram looks like this:

Lewis diagram for PCl₅

Sulfur tetrafluoride, SF₄

• The total number of valence electrons is = S + 4F = 6 + (4 x 7) = 34
• The number of bonding pairs is 4, which accounts for 8 electrons
• The remaining 26 electrons would be 13 lone pairs
• Fluorine cannot expand the octet so each fluorine would accommodate 3 lone pairs, accounting for 24 electrons, leaving one lone pair on the sulfur (sulfur has expanded the octet)
• The completed Lewis diagram looks like this:

Lewis diagram for SF₄

Chlorine trifluoride, ClF₃

• The total number of valence electrons is = Cl + 3F = 7 + (3 x 7) = 28
• The number of bonding pairs is 3, which accounts for 6 electrons
• The remaining 22 electrons would be 11 lone pairs
• Fluorine cannot expand the octet so each fluorine would accommodate 3 lone pairs, accounting for 18 electrons, leaving two lone pairs on the chlorine
• The completed Lewis diagram looks like this:

Lewis diagram for ClF₃

Triiodide?ion, l₃^-

• The total number of valence electrons is = 3I?+ the negative charge = (3 x 7) + 1 = 22
• The number of bonding pairs is 2, which accounts for 4 electrons
• The remaining 18 electrons would be 9 lone pairs
• Iodine would accommodate 3 lone pairs, accounting for 12 electrons, leaving three lone pairs on the central iodine
• The completed Lewis diagram looks like this:

Lewis diagram for l₃^-

Six electron pairs

Sulfur hexafluoride, SF₆

• An example of a molecule with six bonding electron pairs is sulfur hexafluoride, SF₆
• The total number of valence electrons is = S + 6F = 6 + (6 x 7) = 48
• The number of bonding pairs is 6, which accounts for 12 electrons
• The remaining 36 electrons would be 18 lone pairs, so that each F has 3 lone pairs, accounting for all electrons and no lone pairs
• The completed Lewis diagram looks like this:

Lewis diagram for SF₆

Bromine pentafluoride, BrF₅

• The total number of valence electrons is = Br + 5F = 7 + (5 x 7) = 42
• The number of bonding pairs is 5, which accounts for 10 electrons
• The remaining 32 electrons would be 16 lone pairs
• Fluorine cannot expand the octet so each fluorine would accommodate 3 lone pairs, accounting for 30 electrons, leaving one lone pairs on the bromine
• The completed Lewis diagram looks like this:

Lewis diagram for BrF₅

Xenon tetrafluoride, XeF₄

• The total number of valence electrons is = Xe + 4F = 8 + (4 x 7) = 36
• The number of bonding pairs is 4, which accounts for 8 electrons
• The remaining 28 electrons would be 14 lone pairs
• Each fluorine would accommodate 3 lone pairs, accounting for 24 electrons, leaving two lone pairs on the xenon
• The completed Lewis diagram looks like this:

Lewis diagram for XeF₄