For O.
valence electrons = 6 (periodic table)
bonds = 2 (Lewis diagram)
lone pairs = 2 lone pairs or 4 electrons.
6-2-4=0
For C.
valence electrons = 4
bond = 4
lone pairs = 0
4-4-0 = 0
Frankly, I don't like this method of doing it but it works.
i know the formula
FC = # Valence Electron - bonds - lone pairs
but am i supposed to use that formula for all elements on the problems
for instance the teacher did this example
:O:
||
C
/ \
O FC: 6-2-4=0
CFC: 4-4=0
I have no idea where the numbers came from, can someone explain simply how to find the numbers i need to plug in?
4 answers
is there a different way to do this problem?
Yes, but it's hard to explain without a whiteboard. However, I'll try.
Count electrons by themselves. Count electrons in a bond as belonging to each atom equally; that is, if two electrons make a bond then 1 electron belongs to one element and the other element belongs to the other element. Then the formal charge is determined by comparing the electrons the element normally has to the electrons you just counted. Here is how it works.
For the compound you posted, O has 4 non-bonding electrons and 2 in the bond with C (the C=O). [There are 4 electrons in the C=O bond so half belong to C and half to O making O now with 4 + 2 = 6.] How many valence electrons does O have? It has six since O is a group 6 element; therefore, the formal charge is zero. Now for C. The C has 2 electrons from the C=O bond, 1 electrons from 1 C-H bond and another electrons from the other C-H bond, which makes 2 + 2 or 4 total. The number of valence electrons for C is 4; therefore, the formal charge is zero.
Draw N2O
..
:O:
..
N
..
..
..
N
..
Top O. You have 6 electrons not involved in bonding and 2 shared. Electrons we count as belonging to O is 7; O normally has 6, FC is -1
Top N. We count 1e in N-O bond and 3 in the NtriplebondN which makes 4. N is in group V so it normally has 5 which makes it +1. Lower N is 3 for the triple bond and two non-bonding which makes 5; formal charge is zero. Molecule is zero since you have a +1 and a -1.
I think it is far simpler to count the electrons than it is to try and put them into a formula. I can have the electrons counted by the time I can figure out the three numbers to put into the formula we used in the first post. I hope this is of some use to you.
Count electrons by themselves. Count electrons in a bond as belonging to each atom equally; that is, if two electrons make a bond then 1 electron belongs to one element and the other element belongs to the other element. Then the formal charge is determined by comparing the electrons the element normally has to the electrons you just counted. Here is how it works.
For the compound you posted, O has 4 non-bonding electrons and 2 in the bond with C (the C=O). [There are 4 electrons in the C=O bond so half belong to C and half to O making O now with 4 + 2 = 6.] How many valence electrons does O have? It has six since O is a group 6 element; therefore, the formal charge is zero. Now for C. The C has 2 electrons from the C=O bond, 1 electrons from 1 C-H bond and another electrons from the other C-H bond, which makes 2 + 2 or 4 total. The number of valence electrons for C is 4; therefore, the formal charge is zero.
Draw N2O
..
:O:
..
N
..
..
..
N
..
Top O. You have 6 electrons not involved in bonding and 2 shared. Electrons we count as belonging to O is 7; O normally has 6, FC is -1
Top N. We count 1e in N-O bond and 3 in the NtriplebondN which makes 4. N is in group V so it normally has 5 which makes it +1. Lower N is 3 for the triple bond and two non-bonding which makes 5; formal charge is zero. Molecule is zero since you have a +1 and a -1.
I think it is far simpler to count the electrons than it is to try and put them into a formula. I can have the electrons counted by the time I can figure out the three numbers to put into the formula we used in the first post. I hope this is of some use to you.
that makes a lot of sense , thank you i was really confused.
1 answer