Question

7.
Identify the molecule having equal percentage of sand p orbitals
b) CO2
c) BCI3
a) CCI4
d)so2​

Answer 1

Answer:

CO2 has equal percentage of s and p orbitals as its hybridization is sp....

hope this may help you....

Explanation:

Answer 2

Answer:

CO2

Explanation:

no.of hybridising orbital=1/2*(valence electron in central atom+no.of monovalent atom+negative charge-positive charge)

For:

CO2=1/2*4=2,sp

BCl3=1/2*(3+3)=3,sp2

CCl4=1/2*(4+4)=4,sp3

SO2=1/2*6=3,sp2