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How many bonds can Calcium make without hybridization?
0 bonds
How many bonds can Be2S2 make without hybridization?
one
How many bonds can Si make without hybridization?
2
How many bonds can Al make without hybridization?
1
How many bonds can Mg 3s2 atom make without hybridization?
0 bonds
How many bonds can Be 2s2 make without hybridization?
zero(0)
How many bonds can oxygen make?
Without hybridization, oxygen has a valence electron configuration of 2s22p4. Which means it has 2 unpaired electrons; therefore it can form 2 bonds.
How many bonds can each atom make without hybridization?
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.
How many bonds can each atom make without hybridization P3s23p3?
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.
How many bonds would a carbon atom make when it is most stable?
The strongest and most stable bonds involve carbon (C) to carbon bonds. C in sp, sp2, and sp3 hybridization, that is single, double and triple bonds, are the most stable.
How many bonds can S3s23p4 make without hybridization?
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.
How many bonds can Be2s2 make with out hybridization?
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.
