Question

Can anyone PLEASE help me out with hybridization?
H4I2COC the question is C-sp3
C-sp3
O-sp3

Answer 1

What is the question?

Answer 2

how can I know which hybridization it is? we had multiple choices and that's the answer

Answer 3

why is it not O-sp3
;C-sp2
; C-sp2
.ʡ

Answer 4

Can you help me out with it?

Answer 5

I am still not clear about your question. Can you write down the original question?

Answer 6

H4I2COC

Answer 7

there're multiple choices. which one is correct:

Answer 8

\(\sf H_4I_2COC\) is this the molecular formula you are trying to write ?

Answer 9

yes

Answer 10

ok, what are the choices?

Answer 11

b. C-sp; C-sp3
; P-sp; C-sp

Answer 12

c: C-sp; C-sp; C-sp2

Answer 13

d: C-sp3
; C-sp; C-sp2

Answer 14

and the fist one: C-sp3
C-sp3
O-sp3

Answer 15

Did you understand it? I don't really get how can I tell, only by elimination?

Answer 16

are you sure the molecular formula you are writing is correct?

Answer 17

definitely

Answer 18

I know it was weird for me as well

Answer 19

C-sp3
; C-sp3
; O-sp3

Answer 20

What do you think?

Answer 21

Not sure :/

Answer 22

@abb0t may help you

Answer 23

I hope so... thanks though!

Answer 24

@Kainui

Answer 25

Draw out the molecule.

Answer 26

I've already tried....

Answer 27

Are you sure it's \(\sf H_4I_2COC\)??
The formula makes no sense in the way it's written.

Answer 28

I know

Answer 29

but it was on our test

Answer 30

It's over now. Forget about it.

Answer 31

lol, so how about this one? AsC2l2F3

Answer 32

the answer is sp2 for both crabons

Answer 33

it's from a test I saw from last year

Answer 34

and still, makes no sense to me, I tried eliminating, but all makes sense

Answer 35

@abb0t it's ok, it's not from an ongoing test.

Answer 36

b) C-sp; C-sp2; As-p
C)C-sp; As-p2 ;C-sp3
d)C-sp3; C-sp

Answer 37

all those I should have eliminated if I took the exam last year, WHY?

Answer 38

@abb0t @Abhisar @Kainui

Answer 39

Tough ehe?

Answer 40

The number of atoms bonded to it, should equal to the amount present. See how the -CF\(_3\) group has carbon, and 3 fluorine atoms bonded to it.
Hence, you have 1 s orbital bonding + 3 p orbitals here. 1+3 = 4 bonds

Answer 41

@abb0t thanks!