Those atoms undergo sp hybridization.
In this case, we have two carbon atoms with a triple bond between the two. Since carbon has only four valence electrons, the most likely case is that C2 will gain two electrons (making it C2 2-) so that both carbons will have their 2p orbitals filled. The hybridization that occurs is therefore sp, since each carbon atom has a triple bond and a lone pair. An easy way to count hybridization:
lone pairs count as one
single bonds, double bonds, and triple bonds only count as one
Add up the above... 2=sp, 3=sp2, 4=sp3, 5=sp3d (which doesn't technically exist)
sp hybridisation. Leaving two p orbitals on each C atom available to form pi bonds. There are then three bonds, a sigma bond between the C atoms formed by overlap of the sp hybrid orbitals, and two pi orbitals formed by overlap of the p orbitals
A carbon atom that participates in a triple bond and a single bond has a linear electron geometry, and thus uses sp hybridization.
Those atoms undergo sp hybridization.
The hybridization of the bonding orbitals of carbon in carbon tetrachloride are sp3 hybridized. The hybridization occurs between the s orbital of the hydrogen atom and the px, py, pz orbitals of the carbon atom, hence it is sp3 hybridized.
sp hybridization
sp2. Each carbon is sp2 hybridised leaving a p orbital that can form a pi bond. The ethylene (ethene) molecule is planar.
In this compound, the carbon atoms undergo sp3 hybridization.
There wont be a stable compound with the formula C2Br2. If there is then it will be sp hybridization of carbon. If the question is for CH2Br2, then carbon will be sp3 hybridized.
The hybridization of the bonding orbitals of carbon in carbon tetrachloride are sp3 hybridized. The hybridization occurs between the s orbital of the hydrogen atom and the px, py, pz orbitals of the carbon atom, hence it is sp3 hybridized.
The electronic geometry about the carbon atom is: tetrahedral The orbital hybridization about the carbon atom is: sp^3 The molecular geometry about the carbon atom is: tetrahedral
sp hybridization
The valence shell of carbon is a linear combination of 2s and 2p^3 we typically call the sp^3 hybridized orbital or the sp^2 hybridized orbital depending on the number of sigma bonds to carbon four or three respectively. There is also sp linear hybridization otherwise known as the triple bond motif of carbon.
sp2. Each carbon is sp2 hybridised leaving a p orbital that can form a pi bond. The ethylene (ethene) molecule is planar.
sp hybridization.
Looking at the electron configuration of carbon (at. no. 6) you have 1s2 2s2 2p2. In the 2 p subshell, you have 1 electron in the 2px orbital, and 1 electron in the 2py orbital and no electrons in the 2pz orbital. So, the answer is that there are TWO half filled orbitals in the carbon atom. This is the case BEFORE hybridization. After hybridization, there are FOUR half filled orbitals which are called sp3 hybrids.
In this compound, the carbon atoms undergo sp3 hybridization.
Not quite sure what Hybridization means here, yet No; the only requirement for a compound to be Organic is that it contains Carbon.
There wont be a stable compound with the formula C2Br2. If there is then it will be sp hybridization of carbon. If the question is for CH2Br2, then carbon will be sp3 hybridized.
When carbon atom undergoes sp hybridization there will be two unchanged p orbitals (Px and Py considering z-axis as molecular axis) and two sp orbitals(hybridized orbitals).Mechanism of sp hybridization: The ground state carbon atom has 2 es in the 2s orbital and 2 es in the 2p orbital.On excitation the one of the 2s electron is transferred to an empty 2p orbital( say 2pz).Then the half filled 2s orbital and 2pz orbital undergo hybridization yo form to two equivalent sp orbitals(on the molecular axis).The will be a maximum of 3 bonds that 2 carbon atoms can make: It can be explained by the sp hybridization.When two sp hybridized carbon atoms combine through the z-axis as molecular axis then the unchanged p orbitals undergo lateral overlap forming 2 pi orbitals and one sp orbitals of each carbon atom overlap horizontally forming a sigma orbital.But the other sp orbital of carbon atom cannot overlap since the sp orbitals lie in the same molecular axis.This prevents the formation of the fourth bond.
tetrahidral