I mark in 3 different colors 3 different arrow.
http://p1cture.me/images/94525455018278966399.jpg
About the green arrow I asked to say what is the corner in H-C-H is not must be exact. So here I answer is need to be less than 120 degrees.

About the red arrow I asked to decide geometry of this both lone pair electrons. So here I think is an a Bent.

About the blue arrow I asked to say what is the corner in O-C-O is not must be exact.
So here I think is 120 degrees.

Do I correct in my answers?

Thanks.

I would say yes for all three. :)

But if you were to give a more precise answer,

1. A closer answer might be a little more than 107 degrees since that is the angle in ammonia, where there is a lone pair of electrons above the H atoms.

2. Right.

3. First taking all the atoms to be the same (O), then it is 120. Since you have C, then the angle O-C-C will be a little smaller, which makes angle O-C-O a little bigger.

:)

Hi. Thank you for answering.

But I have to ask something:
1)about your 1 answer you told "where there is a lone pair of electrons above the H atoms. " but I am not see there any lone pair electrons. (we speak about the green arrow)?

2)where is the blue arrow is a Trigonal planar, so is not a 120
Degrees in the O-C-O angle ?
Again thank you. :)

1. I was referring to Ammonia, which is closer to that structure.

Ammonia has 3 H and 1 lone pair of electrons. Oxygen which is an electronegative element will more or less act like the lone pair of electrons, but weaker.

2. If it was CO3, then it would have been perfectly trigonal planar and 120 degrees. Since you have C instead of a third O, yo don't get exactly 120 degrees, but a little more for the O-C-O bond angle. C is less electronegative, and will less repel the other O.

Oh OK now I understand it correctly.
Thank you very much.

Just a few thoughts about this question.
Since your title concerns the VSEPR theory lets look at these three atoms in this light.
The blue atom (carbon number 1) is sp2 hybridized and will be trigonal planar with 120 o bond angles.
The red atom (hydroxyl oxygen on carbon number 5) is sp3 hybridzed and will be tetrahedral.
The green atom (carbon number 6) will also be sp3 and tetrahedral.
The bond angles at all these atoms will be a little distorted from the geometric bond angles but I wouldn't think enough to be concerned about. The H-O-C angle (red oxygen) will be the most distorted -- but more like water 104 o) than ammonia (107 o).

Thanks for this Information.
But you told " The red atom (hydroxyl oxygen on carbon number 5) is sp3 hybridzed and will be tetrahedral" but if you use the formula AXnEn you get AX2E2 what make it to be bent.

Because there are four clouds of electrons around the O atom they are arranged tetrahedrally. This is the VSEPR theory in action. Because there are just two H atoms bonded to them the arrangement is BENT. Shapes are determined by the atoms in a compound.

I tried to answer about 1 more question :
1)in the top middle arrow how can I know which orbit are overlaps to create the bond H-C?


Thanks.

This is a single bond, so it's a sigma bond.

The 1s orbital of H will overlap the 2s orbital of C.

Many thanks :)

Wait! >.<

I forgot something, C is SP3 hybridised, and hence, you have another orbital in play.

I never did those... maybe you did?

If an 1s and 2SP3 overlap exists... :o

Well I will check it , but I also never did it before.
Maybe Dr bob know.

You are correct. The (ANY) C-H bond is formed by the overlap of a 1s orbital on H and an sp3 hybrid orbial on C.

On further thought, of course the carbon atom can be sp2 or sp hybridized as well.

Today I learned it in the first time.
Thanks to both of you for help

Last thing in this topic, just to make sure I got it.
In the Lower Left arrow the orbit that overlap to create the bond O-C is: from O is will be sp3 and from C is will be sp2
Thanks.

Yes, I would say so :)

Ok so I think I got the point.

Thanks :)

Oh I mean in the Lower right arrow :) but I think you understand what I mean.

Lol, I didn't notice XD

:) is OK.