Question

Bond angle decreases from NH3 (107.3°)to NF3(102.1°)and from H2O (104.5°)to OF2(103.2°).explain the decrease?​

Answer 1

Explanation:

In H2O

Sigma bond is 2

Lone pair is 2

So, steric number= 2+2= 4

Thus hybridization is sp3. Since it has 2 lone pair so, both the lone pair will repel each other and the bond angle reduces to 104.5°

In NH3

Sigma bond is 3

Lone pair is 1

So, steric no. is (3+1)= 4

Thus, hybridization is sp3. Since it has only 1 lone pair so due to replusion between lone pair and bond pair the bond angle also reduces (107°)

But, if we compare both of them together so in case of H20 there are 2 lone pair and the lone pair-lone pair repulsion is greater than the bond pair-bond pair repulsion so the bond angle reduces much more in this case than NH3. And in NH3 there is only one lone pair. So due to less repulsion bond angle reduces less.

And so, the bond angles of H20 and NH3 are 104.5° and 107° respectively

Answer 2

Answer:

Explanation:

The bond angle in H20 is 104.5 degress due to the presence of two lone pair of electrons. This fact can be explained with the help of valence shell electron pair repulsion theory (VSEPR).  

Both NH3 and NF3 are pyramidal in shape with one lone pair on N. However F has larger electronegatively than H, The electron pair is more towards F in NF3 . Hence repulsion between bond pairs in NF3, is less than NH  3

​As a result the bond angle decreases to 102.4°, whereas in NH 3 , it decreased to 107.3° only.

The actual bond angle for H2O is 104.5˚ while the angle for OF2 is 103.1˚. Oxygen is more electronegative than hydrogen. This causes the bonding pair electrons to move towards O, which results in electron repulsion and a larger bond angle.