Asked by Feather
Formal charges on CH2O and H2SO3
I was taught thus way valence electrons subtracted from addition of dots plus bonds
For CH2O
Carbon 4-(0+3) = + 1
. Oxygen 6-(4+1) = - 1
For H2SO3 I am having issues figuring out the dots and bonds
I was taught thus way valence electrons subtracted from addition of dots plus bonds
For CH2O
Carbon 4-(0+3) = + 1
. Oxygen 6-(4+1) = - 1
For H2SO3 I am having issues figuring out the dots and bonds
Answers
Answered by
DrBob222
I think you're having trouble with the CH2O, too. I think by your formula and the way I do it that the formal charge on C is 0 and the formal charge on O is 0. (and the formal charge on each H is 0).
Here is the dot structure.
http://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/CH2O-lewis-structure.html
The way you've been taught is
For C. valence e = 4
- additional dots = 0
- bonds and there are 4 (1 CH, another CH and 2 for C=O making 4)
so 4-4 = 0
The way I do it.
valence e = 4
1e from the CH bond belongs to C; ( that's 1)
1e from the other CH bond belongs to C(that's 2)
There are 4 electrons shared with the C=O and half of them belong to C.(That's 2 more to make a total of 4)
So C should have 4, it has 4, charge zero.
Here is the dot structure.
http://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/CH2O-lewis-structure.html
The way you've been taught is
For C. valence e = 4
- additional dots = 0
- bonds and there are 4 (1 CH, another CH and 2 for C=O making 4)
so 4-4 = 0
The way I do it.
valence e = 4
1e from the CH bond belongs to C; ( that's 1)
1e from the other CH bond belongs to C(that's 2)
There are 4 electrons shared with the C=O and half of them belong to C.(That's 2 more to make a total of 4)
So C should have 4, it has 4, charge zero.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.