TS EAMCET 20-July-2022 Shift 2 Solved Paper

On finding number of bond pairs in each
(A) H2CO3‌

2×1+4+3×6

8

=‌

24

8

=3
⇒sp2 hybridised
(B) SiF4‌

4+4×7

8

=‌

32

8

=4⇒sp3 hybridised
(C) BF3‌

3+3×7

8

=‌

24

8

=3⇒sp2 hybridised
(D) HClO2‌

1+7+2×6

8

=‌

20

8

(2 lone pair +4 bond pair sp3 hybridised
A and C only are having sp2 hybridised central atom.