​In the formation of NH3, if nitrogen uses its pure p-orbital then choose, the false statement  
(1) All the three bond in NH3 will be of 90°
(2) All the three bond in NH3 will have to same bond strength
(3) L.P. will be present in 2s orbital
(4) The geometry of the molecule will be trigonal bipyramidal
​Kindly explain the answer and what does the question mean by pure p orbital ?

Dear Student,

According to the concept of hybridisation, pure orbital mix together to form hybrid orbitals which participate in bonding. In NH3 1 s and 3 p orbitals hybridise together to form 4 hybrid sp​3 orbitals of which three orbitals form bond with H and one orbital holds lone pair of electron. 
But, in the question it has been given that pure p-orbitals are used for bonding i.e. hybridisation donot occur. 
Now, valence shell electronic configuration of N is =2s2 2p3
Since pure p-orbitals are used, so 3 H will form bond by sharing 3 unpaired electrons in 3 p orbitals respectively i.e. px, py and pz
The electron pair in 2s acts as a lone pair. Now, the 3 p orbitals are perpendicular to each other. Therefore, all three bonds in NH3 will be 90º. 
All the bonds will also be equivalent. So, all N-H bond strength will also be equal. 
Therefore, options (1), (2) and (3) are correct. 
But, its shape cannot be triagonal bipyramidal. Because, even if we donot use the concept of hybridisation but according to VSEPR theory, 
Total no. of electrons pair=No. of bond pair+No. of lone pair=3+1=4
If no. of electron pair is 4 then its shape can be tetrahedral. 
For TBP geometry, no. of electron pairs must be 5. 

Hence, the incorrect answer is (4)


  • 7
i think answer should be (4)

Acc to me, formation of NH3 involves hybridised SP3 orbital, pure p orbital = involving only p orbital in bonding

As no orbitals are hybridised no hybrid orbital shape can be given to NH3
  • 0
What are you looking for?