TS EAMCET 2015 Solved Paper

No of e− pairs =

G+M−C+A

2

G= Group No. of central atom
M= No. of Monovalent atom attached to central atom
C= Cationic charge
A= Anionic charge
For ClF3=

7+3

2

=5 in which 3B.P+ 2LP
SF4=

6+4

2

=5 in which 4 B.P + 1LP
BrF5=

7+5

2

=6 in which 4 B.P + 1LP