Question

Question 11.7 Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.

Class XI The p-Block Elements Page 323

Answer 1

concept :- In boron only 3 electrons are present in valance shell. it has one empty p-orbital , in which it can accept electrons .

for finding structure , count the number of bonds ( or bond pair ) and then find hybridisation . because if number of bond pair are 3, hybridisation will be sp², and structure will be trigonal planar and if the number of bond pair will be 4 , hybridisation will be sp³ and structure will be tetrahedral .


solution : in BF3 , Boron is sp²- hybridisation as 8t contains three bond pairs and therefore BF3 molecule is trigonal planar in shape . on the other hand , In [BF4]- is sp³ - hybridisation because of the presence of 4 bond pairs and hence, [BF4]- species is tetrahedral in shape.

Answer 2

Answer:

In BF3, boron is SP2 hybridized.

∴ shape of BF3 = planar.

In [BH4]–, boron is sp3 hybridized, thus the shape is tetrahedral.

Explanation:

Hope this helps you !!