How many bonds can Si3s23p2 make without hybridization?

Answer 1

wo. A strange question! if you hybridise the 3s and 3 p orbitals you end up with sp3 and still get the same answer. Perhaps the hybridisation involves d orbitals, if that is what you are being taught.

Answer 2

Boron, B, has an electronic configuration of 1s2 2s2 2p1. With 3 electrons boron can form 3 single covalent bonds. This explains why boron has a valency of 3. It forms trihalides such as BCl3 which is planar and boron is sp2 hybridised.

Answer 3

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.

Answer 4

The answer if you use valence bond theory is none as electrons need to be promoted to hybrid orbitals. If you are using molecular orbital theory then this is not a restriction.

Answer 5

Silicon (Si) in its ground state can make a total of 4 bonds without hybridization, since it has 4 valence electrons in its 3s and 3p orbitals.

Answer 6

one

Answer 7

2 bonds

Answer 8

zero(0)

Answer 9

2