1.Which of the following does not obey octet rule and why?

NO , PCl3 , SF4, SO2.

2.Calculate the bond order of B2+.

3.Compare realtive stability of N2 and N2 +.

4. What is the hybridisation of central atom in PCl5 , NH4+ AND BF3 ?

1) The octet rule will be violated in all those cases where there are more or less than eight  electrons in the valence shell of the central atom. Now consider NO. It is an odd electron species with an incomplete octet for both oxygen and nitrogen. As a result, it does not obey the octet rule. In case of SF4 also, there are 10 electrons in the valence shell of sulphur. Hence this too does not obey the octet rule. Now we are  left with PCl3 and SO2. In PCl3, the octet for both phosphorus and Chlorine atoms is complete. Therefore, it follows the octet rule. In case of SO2 as well the octet rule is being violated.

Thus we find that the octet rule is followed in case of only PCl3

2) The atomic number of boron is 5. Thus the total number of electrons that have to be arranged in the molecular orbitals are 10. Therefore in case of B2 molecule, the electronic configuration will be 


In case of B2+ molecule, the electron is removed from a ∏2p orbital. This leaves us with 5 electrons in bonding molecular orbital and 4 electrons in antibonding molecular orbital. Thus the bond order in will be

 = (number of electrons in bonding molecular orbital- number of electrons in antibonding molecular orbital)/2

 = (5-4) / 2 = 0.5

3) There are 14 electrons in N2 molecule, therefore we have the following arrangement of electrons

(σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (Π2px 2 = Π2py 2)(σ2p z )2

here bond order is = (Nb-Na) / 2 = (10-4) / 2 = 3

Now because in N2 +, there is one electron less in bonding orbital as compared to electrons in N2, therefore, there will be 1 electron less in the σ2p z orbital. Thus the bond order in N2 molecule will be

= (9-4) / 2 = 2.5

Hence the bond order in N2 molecule is more than N2+. Therefore N2 is more stable than N2+.

4) The hybridisation of the central atom in the said compounds is as follows

 PCl5 = Sp3d

NH4+ = Sp3

BF3 = Sp2

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