I am asked to draw lewis strcutures of the following molecules and ions
SF6, BrF5, XeF4, PF5 and IF4-.
I have to follow a specific format to draw them, below is an example of SF6, where am i going wrong in the table, since i am not getting six bonds, nor am i successful with the others. Are they an exception to the octet rule and cannot calculate them that way?
Valence Electrons 6e- + 42e- = 48e-
Noble gas Config 8e- + 48e- = 56e-
Total 56e- - 48e- = 8e-
Covalent bonds 8e-/2 = 4
Non-Bonding Pairs 48e- - 8e- = 40e-
Help please!
I believe all are exceptions.
See this link. Go to page 8 for SF6. The others may be there too but I didn't look. http://www.smallscalechemistry.colostate.edu/PowerfulPictures/LewisDiagrams.pdf
Drawing Lewis structures can be a helpful way to visualize the arrangement of atoms and electrons in molecules and ions. However, it is important to note that not all molecules follow the octet rule, which states that atoms tend to gain, lose, or share electrons in order to attain a stable electron configuration with eight valence electrons.
Let's go through each molecule and ion you mentioned and try to determine the correct Lewis structures:
1. SF6:
- Sulfur (S) has 6 valence electrons, and there are 6 fluorine (F) atoms each contributing 7 valence electrons (6*7 = 42).
- In total, the molecule has 48 valence electrons (6+42).
- Starting with sulfur, place one bond between S and each F atom, accounting for 6 electrons as bonding pairs.
- After assigning all the bonding pairs (6 electrons), subtract the valence electrons used so far (48-6 = 42).
- Divide the remaining electrons by 2 to get the number of non-bonding pairs (42/2 = 21).
- Place the remaining 21 electrons around the fluorine atoms as non-bonding pairs.
- Sulfur (S) does not need to follow the octet rule in this case, as it can accommodate an expanded octet, meaning it can have more than eight electrons around it.
- The Lewis structure of SF6 should have one sulfur atom surrounded by six fluorine atoms, with six bonding pairs and zero non-bonding pairs around sulfur.
2. BrF5:
- Bromine (Br) has 7 valence electrons, and there are 5 fluorine (F) atoms each contributing 7 valence electrons (5*7 = 35).
- In total, the molecule has 42 valence electrons (7+35).
- Following the same steps as before, assign bonding and non-bonding pairs to the atoms.
- Start by placing a single bond between Br and each F atom until all the bonding pairs are assigned.
- Determine the number of remaining electrons (42-10 = 32).
- Divide the remaining electrons by 2 to get the number of non-bonding pairs (32/2 = 16).
- Place the 16 electrons around the fluorine atoms as non-bonding pairs.
- Bromine (Br) does not need to follow the octet rule in this case, as it can also accommodate an expanded octet.
- The Lewis structure of BrF5 should have one bromine atom surrounded by five fluorine atoms, with five bonding pairs and three non-bonding pairs around bromine.
3. XeF4:
- Xenon (Xe) has 8 valence electrons, and there are 4 fluorine (F) atoms each contributing 7 valence electrons (4*7 = 28).
- In total, the molecule has 36 valence electrons (8+28).
- Again, follow the same steps as before to assign bonding and non-bonding pairs.
- Start by placing a single bond between Xe and each F atom until all the bonding pairs are assigned.
- Determine the number of remaining electrons (36-8 = 28).
- Divide the remaining electrons by 2 to get the number of non-bonding pairs (28/2 = 14).
- Place the 14 electrons around the fluorine atoms as non-bonding pairs.
- Xenon (Xe) does not need to follow the octet rule here, as it can accommodate an expanded octet.
- The Lewis structure of XeF4 should have one xenon atom surrounded by four fluorine atoms, with four bonding pairs and two non-bonding pairs around xenon.
4. PF5:
- Phosphorus (P) has 5 valence electrons, and there are 5 fluorine (F) atoms each contributing 7 valence electrons (5*7 = 35).
- In total, the molecule has 40 valence electrons (5+35).
- Follow the same steps as before to assign bonding and non-bonding pairs.
- Start by placing a single bond between P and each F atom until all the bonding pairs are assigned.
- Determine the number of remaining electrons (40-10 = 30).
- Divide the remaining electrons by 2 to get the number of non-bonding pairs (30/2 = 15).
- Place the 15 electrons around the fluorine atoms as non-bonding pairs.
- Phosphorus (P) does not need to follow the octet rule here, as it can accommodate an expanded octet.
- The Lewis structure of PF5 should have one phosphorus atom surrounded by five fluorine atoms, with five bonding pairs and three non-bonding pairs around phosphorus.
5. IF4-:
- Iodine (I) has 7 valence electrons, and there are 4 fluorine (F) atoms each contributing 7 valence electrons (4*7 = 28).
- In total, the ion has 36 valence electrons (7+28).
- Follow the same steps as before to assign bonding and non-bonding pairs.
- Start by placing a single bond between I and each F atom until all the bonding pairs are assigned.
- Determine the number of remaining electrons (36-8 = 28).
- Divide the remaining electrons by 2 to get the number of non-bonding pairs (28/2 = 14).
- Place the 14 electrons around the fluorine atoms as non-bonding pairs.
- Iodine (I) needs to follow the octet rule in this case, as it cannot accommodate an expanded octet.
- To satisfy the octet rule for iodine, add an additional lone pair (non-bonding pair) on iodine, giving it a total of three non-bonding pairs around it.
- The Lewis structure of IF4- should have one iodine atom (I-) surrounded by four fluorine atoms, with four bonding pairs and three non-bonding pairs around iodine.
Remember, Lewis structures are a way to represent the arrangement of electrons in a molecule or ion, and they aim to minimize formal charges and maximize stability. Double-checking your calculations and following the steps for assigning pairs of electrons can help in constructing accurate Lewis structures.