To draw the Lewis structures for BrF3, ClF5, and IF7, we need to consider the number of valence electrons for each atom and how they connect. Let’s explore each molecule.
BrF3 (Bromine Trifluoride)
Bromine (Br) has 7 valence electrons and each fluorine (F) has 7 valence electrons. In BrF3, bromine is the central atom surrounded by three fluorine atoms.
- Count total valence electrons: 7 (Br) + 3 × 7 (F) = 28 electrons.
- Distributing electrons, Br will bond to three F with single bonds using 6 electrons, leaving 22 electrons.
- Distribute remaining electrons to F atoms to satisfy their octet: each F gets 6 more electrons (3 bonding pairs) which uses up 18 electrons.
- Br will have 2 lone pairs remaining.
The Lewis structure looks like this:
F
|
F--Br--F
|
(lone pairs)
In this case, a total of 12 electrons (3 lone pairs plus 3 bonds) are around Br, which does not obey the octet rule.
ClF5 (Chlorine Pentafluoride)
Chlorine (Cl) has 7 valence electrons, and with five fluorines, we have a total of:
- Valence electrons: 7 (Cl) + 5 × 7 (F) = 42 electrons.
- Cl bonds to 5 F with single bonds using 10 electrons.
- The remaining 32 electrons are used to fulfill the octet for each F.
The Lewis structure is:
F
|
F--Cl--F
|
F
|
F
Here, chlorine has 10 electrons surrounding it (5 bonds), which clearly violates the octet rule.
IF7 (Iodine Heptafluoride)
Iodine (I) has 7 valence electrons, and with seven fluorines:
- Valence electrons: 7 (I) + 7 × 7 (F) = 56 electrons.
- For I, bonding to 7 F uses 14 electrons.
The Lewis structure looks like this:
F
|
I
/ | \
F F F
|
F
Iodine possesses 14 electrons in its outer shell (7 bonds), again not obeying the octet rule.
Summary
The molecules BrF3, ClF5, and IF7 all have central atoms that do not obey the octet rule. Bromine and iodine can have expanded octets, which enables them to accommodate more than 8 electrons. Chlorine, although capable, also has more than 8 electrons due to its multiple bonds with fluorine.