Three is the answer expected. Higher valences of phosphorus, in PCl5 for example can be explained by hybridisation although this method is not the only explanation.
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∙ 2014-04-24 10:37:280 bonds
The C atom of HCHO has 3 sigma bonds and a pi bonds. Hence the hybridization of C is sp2.
The hybridization of the carbon atom in the carbonate ion is to have three orbitals on the carbon atom that will be used to form sigma bonds.
Si belong to carbon family when it form four sigma bonds its hybridization is always sp3.
The Hybridization of the 'As' atom in AsF5 is SP3d.
0 bonds
The C atom of HCHO has 3 sigma bonds and a pi bonds. Hence the hybridization of C is sp2.
The hybridization of the carbon atom in the carbonate ion is to have three orbitals on the carbon atom that will be used to form sigma bonds.
It depends on the hybridization of the central atom.
Si belong to carbon family when it form four sigma bonds its hybridization is always sp3.
The rule for hybridization: All sigma bonds (single bonds and only one bond in double or triple bonds) and lone pairs (nonbonding pairs) are found in hybrid orbitals. We understand from this statement that hybrib orbitals cannot form pi bonds. Boron (B) (atomic no:5) has 3 valence electrons and form 3 single bonds and no pi bond. Therefore it is impossible for B to make bonds without hybridization. Nitrogen (N) (atomic no: 7) has 5 valence electrons. Normally it can form 3 sigma bonds to complete its octet. Note that although B also have 3 sigma bonds it can not complete its octet. In such a case nitrogen can not form any bond without hybridization like in the case of ammonia NH3. Together with the non bonding pair of N , the type of hybridization of N is sp3 when all the bonds of N are single (sigma) bonds. On the other hand , N can be form 1 double bond like in the case of nitrate ion, NO3^-. 2 of the O atoms are bound to N by single bonds and one O atom is bound to N atom by a double bond (one sigma and one pi). In such a case since pi bonds are formed without hybridization, 2 of the bonds of N are formed by hybridization and 1 of them is formed without hybridization. Together with the non bonding pair of N , the type of hybridization of N is sp2 when it forms one double bond. In some cases N can form a triple bond, like in the case of cyanate ion CN^-. There is a triple bond between C and N atoms. One of them is sigma the others are pi bonds. This time 2 bonds are formed without hybridization. Together with the non bonding pair of N , the type of hybridization of N is sp when it forms one triple bond. Oxygen (O) (atomic no: 8) has 6 valence electrons. Normally it can form 2 sigma bonds to complete its octet. In such cases oxygen can not form any bond without hybridization like in the case of ammonia H2O. Together with the two non bonding pair of O , the type of hybridization of O is sp3 when all the bonds of O are single (sigma) bonds. However , O can be form 1 double bond like in the case of ozone molecule, O3 . 1 of the O atoms are bound to the central O atom by a single bond and the other O atom is bound to Central O atom by a double bond (one sigma and one pi). Since pi bonds are formed without hybridization, 2 of the bonds of O are formed by hybridization and 1 of them is formed without hybridization. Together with the non bonding pair of O , the type of hybridization of O is sp2 when it forms one double bond. Remark: As you may notice, in H2O, O has two nonbonding pairs, but in O3, the central O has 1 nonbonding pair. One of them is used to make a coordinate covalent (dative) bond with the O atom to form a single bond by completing its octet vacancy. As a summary; When N forms a double bond, 1 bond (1 pi) When N forms a triple bond, 2 bonds (2 pi) When O forms a double bond, 1 bond (1 pi) cannot be made by by hybridization. O cannot form a triple bond.
The Hybridization of the 'As' atom in AsF5 is SP3d.
If the given compound exists, there'll be 4 sigma bonds and a single remaining electron in the valence shell of the central atom. With 5 activation groups, the hybridization would be sp3d.
There are 4 carbon atoms, which each individually act as a central atom since they are surrounded entirely by the hydrogen atoms. Each carbon forms 4 sigma bonds, therefore, each carbon atom has a hybridization state of sp^3.
sp3 hybridised el.orbitals forming into four sigma bonds with hydrogen atoms
atom that has four bonds forming a tetrahedral shape, mixture of an s and 3 p orbitals
Sp3 d