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

NO , PCl3 , SF_{4}, SO_{2.}

2.Calculate the bond order of B_{2}+.

3.Compare realtive stability of N2 and N_{2} ^{ +. }

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 PCl_{3} and SO_{2}. In PCl_{3}, the octet for both phosphorus and Chlorine atoms is complete. Therefore, it follows the octet rule. In case of SO_{2} as well the octet rule is being violated.

Thus we find that the octet rule is followed in case of only PCl_{3}.

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 B_{2} molecule, the electronic configuration will be

σ1S^{2}σ*1S^{2}σ2S^{2}σ*2S^{2}∏2p_{x}^{1}=∏2p_{y}^{1}

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 N_{2} molecule, therefore we have the following arrangement of electrons

(σ1*s*)^{2} (σ*1*s*)^{2 }(σ2*s*)^{2} (σ*2*s*)^{2} (Π2p_{x} ^{2} = Π2p_{y} ^{2})(σ2*p* _{ z })^{2}

here bond order is = (N_{b}-N_{a}) / 2 = (10-4) / 2 = 3

Now because in N_{2} ^{+}, there is one electron less in bonding orbital as compared to electrons in N_{2}, therefore, there will be 1 electron less in the σ2*p* _{ z }* *orbital. Thus the bond order in N_{2} molecule will be

= (9-4) / 2 = 2.5

Hence the bond order in N_{2} molecule is more than N_{2}^{+}. Therefore N_{2} is more stable than N_{2}^{+}.

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

PCl_{5} = Sp^{3}d

NH_{4}^{+ }= Sp^{3}

BF_{3} = Sp^{2}

**
**