Hybridization of a molecules atoms?

Question

I am very confused with the idea of hybridization. Can someone explain to me what the hybridization of Oxygen with a double bond to C is?

Answer 1

Do you mean O=C=O? if you mean this then it is sp because their are two regions and no little dots(lone pairs) with sp orbitals there are always two pi bonds, and with sp2 there is one pi bond and with sp3 there are zero pi bonds. If you had say NH3 then you would have the 3 lines going to the hydrogens and then the 2 little dots (equal1 region) above the nitrogen which would give four regions. only worry about the dots above the central atom in hybridization. if you meant O=C then i have no idea. Sorry if this confuses you but at least remember how many pi bonds there are based on the hybridization. Also don't forget to include the little dot regions when figuring out bond angles.

Answer 2

When a molecule becomes hybridized it must have orbitals for all of the valence electrons that are present on the central atom of the molecule. Iodine, I, has 7 valence electrons (e^-) Fluorine, F, has 7 valence e^- Total electrons = 1 x 7 + 5 x 7 = 42 total e^- that must be placed in hybrid orbitals. - Each F atom has 3 pairs of nonbonding; e^- = 30 e^- - The F atoms on IF5 have 5 bonding pairs; of e^- = 10 e^- - One nonbonding pair of e^- resides on the central I atom; = 2 e^- - This gives a total of 42 e^- - 30 of the total number of electrons are present as nonbonding electrons. - That leaves 12 e^-'s or 6 e^- pairs that are associated with the central I atom. - The 6 e^- pairs that reside on the I atom must be in separate orbitals because each orbital can hold only 2 e^-'s. ==>>The only correct choice is e) which is sp^3d^2 <<== - 4 orbitals involve the sp^3 part and 2 orbitals involve the d^2 part. That's 6 orbitals around the central I atom. According to VSEPR theory this molecule has the general form AB5E, where A represents the central atom, B represents the F atoms, and E represents the nonbonding pair of electrons on the Central I atom. The molecular shape is that of a square pyramid. Hope this is helpful.

Answer 3

See the pictures on http://www.towson.edu/~ladon/carbon.html
for the explanation below.


carbon has 4 hybridized orbitals in tetrahedral shape. When one of these 4 orbitals overlap with another one on an adjacent carbon, the overlap becomes the C-C single bond or the sigma bond.

Now to get the C=C double bond or pi bond, you need to picture another orbital on both carbons splitting in 1/2 and then folding over the 1st bond with 1/2 on top and the other 1/2 below.

Another way of thinking about it is you have your 2 carbons bonded with a single C-C or sigma bond. That leaves 3 hybrid orbitals currently not attached to anything. Now suppose of of the 3 unbonded orbitals on each carbon could somehow split into 2 thinner orbitals leaving 2 thinner orbitals, two regular hydbridized orbitals and the last orbital forming the bond on each carbon. Now you have to picture the thinner orbitals folding over the top and bottom of the 1st C-C bond. When the two thinner orbitals overlap beteen the 2 carbons, you now have the C-C bond or sigma bond, the thinner overlapping orbitals making the 2nd C-C pi bond. This leaves you with a C=C bond and two unbonded orbitals on each carbon.

If you repeat the portion for the pi bond again, you can get a 2nd pi bond to give a C≡C triple bond or a sigma+2 pi bonds. This will be C triply bonded to the adjacent carbon leaving only one hybridized orbital on each C cable of forming the 4th and final bond.


To picture hybridization itself, think about the shape of the one S orbital and the shape of the 3 P orbitals(dumbell or figure8 shaped) joining together to form 4 new orbitals that are a blend of the spherical shape and the dumbell shapes. These newly shaped orbitals, arranged in a tetrahedral formation are what we are calling the hybridized orbitals or the result of the blending between the 1 s orbital and 3 p orbitals. You started with 4 orbitals and ended with the same number. Only the shape and orientation in space altered.

The idea is the same for oxygen bonded to carbon except in the case of oxygen, only two of its 4 orbitals are capable of bonding due to the additional electrons it has. In other words, oxygen can only make single and double bonds.

Answer 4

sp2 hybridisation will reult in a bonding pair being used for a sigma bond between the two O atoms, and two lone pairs at approximately 120 degrees to each other (all planar). The unhybridised p orbital, containing one electron and perpendicular to the plane just mentioned, can overlap sideways with its neighbour on the other oxygen atom, forming a pi bond.

Answer 5

let me give u an easy method to find the type of hybridization.
consider,
a molecule,(for ex CO2)
write the electronic configuration.
C-(1s2)(2s2,2p2): no.of valence electrons 4;
O-(1s2)(2s2,2p4):no.of valence electrons:6
total=4+6*2=16(since there r 2 oxygen atoms)
if tot>8
divide total by8 =>2
therefore type of hybridization is :sp
remember 2-sp
3-sp2
4-sp3 or dsp2
5-sp3d
6-sp3d2
7-sp3d3