#MCQ ON VSEPR THEORY#SHAPE /GEOMETRY OF MOLECULES#IIT JAM GATE NET PYQ SOLUTIONS WITH TRICKS BY ASHIS SIR. As you know a couple of questions always asked in IIT JAM GATE NET examinations regarding the Geometry of molecules. In this video I am not going to discuss much about the theory but only I will discuss how to predict the shape of molecules, the hybridizations and no. of lone pairs present in the given molecules. In this video I have given you a trick how to predict the shape and geometry of all the molecules easily. I am sure after watching this video you can solve all this questions regarding VSEPR Theory.

2. JAM 2020

3. GATE 2021

4. NET 2017 DEC

5. N= No of Surrounding atoms
M= Outermost electrons of Central Atom
X=Valency of Each S.As
Total no. of pair of electrons
= N (n1+ n2 + …….) + ½ [ M – (N × X)]
B.P L.P
FOR NEUTRAL MOLECULE

6. N= No of Surrounding atoms
M= Outermost electrons of Central Atom
X =Valency of Each S.As
Y= Charge of ions( for +ve charge –Y and –ve charge +Y
Total no. of pair electrons
= N (n1+ n2 + …….) + ½ [ M – (N × X ± Y)]
B.P L.P
FOR CHARGE SPECIES

9. Total no. of pair of electrons
= 2+ ½ [ 2 – (2 × 1)] = 2 + 0 = 2
B.P L.P
B.P L.P
Total no. of pair of electrons
= 2+ ½ [ 4 – (2 × 2)] = 2 + 0 = 2
B.P L.P
B.P L.P
Total no. of pair of electrons
= 3+ ½ [ 3 – (3 × 1)] = 3 + 0 = 3
B.P L.P
B.P L.P
Total no. of pair of electrons
= 4+ ½ [ 4 – (4 × 1)] = 4 + 0 = 3
B.P L.P
B.P L.P
Linear
Linear
Trigonal planar
Tetrahedral
BeCl2
CO2
BF3
CH4

10. Total no. of pair of electrons
= 3+ ½ [ 5 – (3 × 1)] = 3 + 1 = 4
B.P L.P
B.P L.P
Trigonal
pyramidal
Total no. of pair of electrons
= 2+ ½ [ 6 – (2 × 1)] = 2 + 2 = 4
B.P L.P B.P L.P
Bent
or
V Shape
Total no. of pair of electrons
= 2+ ½ [ 6 – (2 × 1)] = 2 + 2 = 4
B.P L.P
B.P L.P
Bent
or
V Shape
Total no. of pair of electrons
= 2+ ½ [ 6 – (2 × 1)] = 2 + 2 = 4
B.P L.P
B.P L.P
Bent
or
V Shape
NH3
OF2
H2O
H2S

11. Total no. of pair of electrons
= 5+ ½ [ 5 – (5 × 1)] = 5 + 0 = 5
B.P L.P
B.P L.P
Total no. of pair of electrons
= 3+ ½ [ 5 – (3 × 1)] = 3 + 1 = 4
B.P L.P
B.P L.P
Trigonal
pyramidal
Trigonal
bipyramidal
Total no. of pair of electrons
= 3+ ½ [ 7 – (3 × 1)] = 3 + 2 = 5
B.P L.P
B.P L.P
Trigonal
planar
Total no. of pair of electrons
= 7+ ½ [ 7 – (7 × 1)] = 7 + 0 = 7
B.P L.P B.P L.P
Pentagonal
bipyramidal
PCl5
PF3
ClF3
IF7

12. Total no. of pair of electrons
= 6+ ½ [ 6 – (6 × 1)] = 6 + 0 = 6
B.P L.P
B.P L.P
Octahedral
Total no. of pair of electrons
= 4+ ½ [ 6 – (4 × 1)] = 4 + 1 = 5
B.P L.P
B.P L.P
See Saw
Total no. of pair of electrons
= 2+ ½ [ 6 – (2 × 2)] = 2 + 1 = 3
B.P L.P
B.P L.P
Angular
SF6
SF4
SO2

13. Total no. of pair of electrons
= 2+ ½ [ 8 – (2 × 1)] = 2 + 3 = 5
B.P L.P B.P L.P Linear
Total no. of pair of electrons
= 4+ ½ [ 8 – (4 × 1)] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
Total no. of pair of electrons
= 6+ ½ [ 8 – (6 × 1)] = 6 + 1 = 7
B.P L.P B.P L.P
Pentagonal
pyramidal
Total no. of pair of electrons
= 3+ ½ [ 8 – (3 × 2)] = 3 + 1 = 4
B.P L.P B.P L.P
Trigonal
bpyramidal
XeF2
XeF4
XeF6
XeO3

14. Total no. of pair of electrons
= (6+1)+½ [ 8 –(6 × 1)-(1 × 2)] = 7+0= 7
B.P L.P B.P L.P Pentagonal
bipyramidal
Total no. of pair of electrons
= (4+1)+½ [ 8 –(4 × 1)-(1 × 2)] = 5+1= 6
B.P L.P B.P L.P Square
pyramidal
Total no. of pair of electrons
= (2+2)+½ [ 8 –(2 × 1)-(2 × 2)] = 4+1= 5
B.P L.P B.P L.P See Saw
XeOF6
XeOF4
XeO2F2

15. Total no. of pair of electrons
= 3+½ [ 4 –(3 × 2)+2] = 3 + 0 = 3
B.P L.P B.P L.P
Trigonal
planar
Total no. of pair of electrons
= 3+½ [ 5 –(3 × 2)+1] = 3 + 0= 3
B.P L.P B.P
Trigonal
planar
Total no. of pair of electrons
= 3+½ [ 7 –(3 × 2)+1] = 3 + 1= 3
B.P L.P B.P
Trigonal
pyramidal
L.P
L.P
CO3
2-
NO3
-
ClO3
-

16. IF4
+ Total no. of pair of electrons
= 4+½ [ 7 –(4 × 1)-1] = 4 + 1 = 5
B.P L.P B.P L.P
See saw
Total no. of pair of electrons
= 4+½ [ 6 –(4 × 2)+2] = 4 + 0 = 4
B.P L.P B.P L.P Tetrahedral
Total no. of pair of electrons
= 4+½ [ 7 –(4 × 2)+1] = 4 + 0= 4
B.P L.P B.P
Tetrahedral
L.P
Total no. of pair of electrons
= 4+½ [ 5 –(4 × 1)-1] = 4 + 0 = 4
B.P L.P B.P L.P
Tetrahedral
SO4
2-
ClO4
-
NH4
+

17. JAM 2020
Ans C
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)-1] = 2 + 2 = 4
B.P L.P B.P L.P
Tetrahedral
Br3
+ = BrBr2
+
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
Trigonal bipyramidal
Br3
- = BrBr2
-
Total no. of pair of electrons
= 3+ ½ [ 7– (3 × 1)] = 3 + 2 = 5
B.P L.P B.P L.P
Trigonal bipyramidal
BrF3

18. JAM 2018
Ans D

19. XeF2
XeF4
XeF6
XeO3
Total no. of pair of electrons
= 2+ ½ [ 8 – (2 × 1)] = 2 + 3 = 5
B.P L.P B.P L.P Linear
Total no. of pair of electrons
= 4+ ½ [ 8 – (4 × 1)] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
Total no. of pair of electrons
= 6+ ½ [ 8 – (6 × 1)] = 6 + 1 = 7
B.P L.P B.P L.P
Pentagonal
pyramidal
Total no. of pair of electrons
= 3+ ½ [ 8 – (3 × 2)] = 3 + 1 = 4
B.P L.P B.P L.P
Trigonal
bpyramidal

20. JAM 2017
Ans A,C
Total no. of pair of electrons
= 4 + ½ [ 8 – (4 × 1)] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
XeF4
Total no. of pair of electrons
= 4 + ½ [ 6 – (4 × 1)] = 4 + 1 = 6
B.P L.P B.P L.P
See saw
SF4
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)-1] = 4 + 1 = 6
B.P L.P B.P L.P
See saw
ClF4
+
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
Square planar
ClF4
-

21. JAM 2015
Ans D
CO3
2-
Total no. of pair of electrons
= 3 +½ [ 4 –(3 × 2)+2] = 3 + 0 = 3
B.P L.P B.P L.P
Trigonal
planar
Total no. of pair of electrons
= 3 +½ [ 5 –(3 × 2)+1] = 3 + 0= 3 Trigonal
planar
B.P L.P B.P L.P
Total no. of pair of electrons
= 3 +½ [ 6 –(3 × 2)+2] = 3 + 1= 4 Trigonal
pyramidal
B.P L.P B.P L.P
Total no. of pair of electrons
= 3+ ½ [ 7– (3 × 1)] = 3 + 2 = 5
B.P L.P B.P L.P
BrF3
SO3
2-
T Shape
NO3
-

22. JAM 2014
Ans D
Total no. of pair of electrons
= 4 + ½ [ 6 – (4 × 1)] = 4 + 1 = 5
B.P L.P
B.P L.P
Trigonal
bipyramidal
TeO4

23. JAM 2012
Ans B
Total no. of pair of electrons
= (2+1)+½ [ 6 –(2 × 1)-(1 × 2)] = 3 + 1= 4
B.P L.P B.P L.P
Trigonal
pyramidal
SOCl2

24. JAM 2010
Ans D
Total no. of pair of electrons
= 4 + ½ [ 6 – (4 × 1)] = 4 + 1 = 5
B.P L.P B.P L.P
See saw
SF4
Total no. of pair of electrons
= 4 + ½ [ 8 – (4 × 1)] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
XeF4
Total no. of pair of electrons
= (2+2) +½ [ 8 –(2 × 1)-(2 × 2)] = 4 + 1 = 5
B.P L.P B.P L.P See saw
XeO2F2
Total no. of pair of electrons
= 4 + ½ [ 5 – (4 × 1)+1] = 4 + 1 = 5
B.P L.P B.P L.P
See saw
BF4
-

25. JAM 2008
Ans C
SO2
Total no. of pair of electrons
= 2+ ½ [ 6 – (2 × 2)] = 2 + 1 = 3
B.P L.P
B.P L.P
Angular

26. JAM 2007
Ans C
CH3
-
Total no. of pair of electrons
= 3+½ [ 4 –(3 × 1)+1] = 3 + 1= 3
B.P B.P
Trigonal
pyramidal
L.P
L.P

27. GATE 2021
Ans A
Total no. of pair of electrons
= 3+ ½ [ 7– (3 × 1)] = 3 + 2 = 5
B.P L.P B.P L.P
ClF3 T Shape
Total no. of pair of electrons
= (2+1) +½ [ 8 –(2 × 1)-(1 × 2)] = 3 + 2 = 5
B.P L.P B.P L.P
XeOF2
T Shape
Total no. of pair of electrons
= 2 + ½ [ 5– (2 × 3)+1] = 2 + 0 = 2
B.P L.P B.P L.P
Linear
N3
- =NN2
-
Total no. of pair of electrons
= (2+3) +½ [ 8 –(2 × 1)-(3× 2)] = 5 + 0 = 5
B.P L.P B.P L.P
Trigonal bipyramidal
XeO3F2

28. GATE 2020
Ans D
XeOF4
Total no. of pair of electrons
= (4+1)+½ [ 8 –(4 × 1)-(1 × 2)] = 5 +1 = 6
B.P L.P B.P L.P
Square
pyramidal
Total no. of pair of electrons
= 5 + ½ [ 7 – (5 × 1)] = 5 + 1 = 6
B.P L.P B.P L.P
Square
pyramidal
IF5
Total no. of pair of electrons
= 5 + ½ [ 6 – (5 × 1)+1] = 5 + 1 = 5
B.P L.P B.P L.P
SF5
- Square
Pyramidal
Total no. of pair of electrons
= 5 + ½ [ 8 – (5 × 1)+1] = 5 + 2 = 7
B.P L.P B.P L.P
XeF5
- Pentagonal
planar

29. GATE 2016
Ans D
Total no. of pair of electrons
= 5 + ½ [ 8 – (5 × 1)+1] = 5 + 2 = 7
B.P L.P B.P L.P
XeF5
- Pentagonal
planar
Total no. of pair of electrons
= 5 + ½ [ 8 – (5 × 1)-1] = 5 + 1 = 6
B.P L.P B.P L.P
XeF5
+ Square
pyramidal

30. GATE 2011
Ans D
Total no. of pair of electrons
= 3 + ½ [ 6 – (2 × 1)-(1 x 1) -1] = 3 + 1 = 4
B.P L.P B.P L.P
SF2Cl+ Trigonal
pyramidal
Trigonal
pyramidal

31. GATE 2010
Ans D
XeOF5
-
Total no. of pair of electrons
= (5+1) +½ [ 8 –(5× 1)-(1 × 2)+1] = 6 +1 = 7
B.P L.P B.P L.P
Pentagonal
monopyramidal

32. GATE 2009
Ans D
Total no. of pair of electrons
= 2 + ½ [ 4 – (2 × 2)] = 2 + 0 = 2
B.P L.P B.P L.P Linear
CO2
Total no. of pair of electrons
= 2 + ½ [ 6 – (2 × 2)] = 2 + 1 = 3
B.P L.P B.P L.P Angular
SO2
Total no. of pair of electrons
= 3 + ½ [ 6 – (3 × 2)] = 3 + 0 = 3
B.P L.P B.P L.P
Trigonal
planar
SO3
Total no. of pair of electrons
= 3 + ½ [ 6 – (3 × 2)] = 3 + 0 = 3
B.P L.P B.P L.P
Trigonal
planar
SeO3
Total no. of pair of electrons
= 2 + ½ [ 6 – (2 × 2)] = 2 + 1 = 3
B.P L.P
Angular
TeO2
Total no. of pair of electrons
= 2 + ½ [ 5 – (2 × 2)-1] = 2 + 0 = 3
B.P L.P
NO2
+ Linear
Total no. of pair of electrons
=4+½[4–(4 × 2)+4] =4+0=4
B.P L.P
SiO4
4-
Tetrahedral
Total no. of pair of electrons
=4+½[5–(4 × 2)+3] =4+0=4
B.P L.P
PO4
3-
Tetrahedral
SO3 = 16+24 = 40
SeO3 = 34+24 = 58
SiO4
4- = (14+32)-4 = 42
PO4
3- = (15+32)-3 = 42

33. GATE 2007
Ans A
Total no. of pair of electrons
=6+½[4–(6 × 1)+2] =6+0=6 (sp3d2)
B.P L.P
SiF6
2-
Total no. of pair of electrons
= 4 + ½ [ 8 – (4 × 1)] = 4 + 2 = 6 (sp3d2)
B.P L.P B.P L.P
XeF4
[NiCl4]2- (sp3) VBT
[PtCl4]2- (dsp2) VBT
NiCO4 (sp3) VBT
XeO2F2
Total no. of pair of electrons
= (2+2)+½ [ 8 –(2 × 1)-(2 × 2)] = 4+1= 5 (sp3d)
B.P L.P
[Co(NH3)6]3+ (d2sp3) VBT
[Co(H2O)6]3+ (sp3d2) VBT

34. NET 2022 FEB NTA
Ans 3
Total no. of pair of electrons
= 3+ ½ [ 7– (3 × 1)] = 3 + 2 = 5
B.P L.P B.P L.P
Trigonal
planar
BrF3
Total no. of pair of electrons
XeO3 = 3 + ½ [ 8 – (3 × 2)] = 3 + 1 = 4
B.P L.P B.P L.P
Trigonal
pyramidal
Total no. of pair of electrons
= 5 + ½ [ 8 – (5 × 1)+1] = 5 + 2 = 7
B.P L.P B.P L.P
XeF5
-
Pentagonal
planar
Total no. of pair of electrons
= (1+2) +½ [ 7 –(1 × 1)-(2 × 2)] = 3 + 1 = 5
B.P L.P B.P L.P
FClO2
Trigonal
pyramidal
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
ClF4
-

35. NET 2020 NOV 1
Ans 3
CO3
2-
Total no. of pair of electrons
= 3 +½ [ 4 –(3 × 2)+2] = 3 + 0 = 3
B.P L.P B.P L.P
Trigonal
planar
Total no. of pair of electrons
= 3 +½ [ 6 –(3 × 2)+2] = 3 + 1= 4 Trigonal
pyramidal
B.P L.P
SO3
2-
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
BrF4
-

36. NET 2019 JUNE
Ans 1
Total no. of pair of electrons
= (2+1) +½ [ 8 –(2 × 1)-(1 × 2)] = 3 + 2 = 5
B.P L.P B.P L.P
XeOF2
T Shape
Total no. of pair of electrons
XeO3 = 3 + ½ [ 8 – (3 × 2)] = 3 + 1 = 4
B.P L.P B.P L.P
Trigonal
pyramidal

37. NET 2018 DEC
Ans 1

38. NET 2017 JUNE
Ans 1
Total no. of pair of electrons
= 5 + ½ [ 8 – (5 × 1)+1] = 5 + 2 = 7
B.P L.P B.P L.P
XeF5
-
Pentagonal
planar
Total no. of pair of electrons
= 5 + ½ [ 7 – (5 × 1)] = 5 + 1 = 6
B.P L.P B.P L.P
Square
pyramidal
BrF5

39. NET 2017 DEC
Ans 1
Total no. of pair of electrons
XeO3 = 3 + ½ [ 8 – (3 × 2)] = 3 + 1 = 4
B.P L.P B.P L.P
Trigonal
pyramidal
Total no. of pair of electrons
= 3 + ½ [ 6 – (3 × 2)] = 3 + 0 = 3
SO3
Trigonal
planar
B.P L.P B.PL.P
ClO3
-
Total no. of pair of electrons
= 3 +½ [ 7 –(3 × 2)+1] = 3 + 1= 3
B.P L.P B.P
Trigonal
pyramidal
L.P

40. NET 2017 DEC
Ans 4
XeO2F2
Total no. of pair of electrons
= (2+2)+½ [ 8 –(2 × 1)-(2 × 2)] = 4+1= 5 (Trigonal bipyramidal)
B.P L.P
SNF3
Total no. of pair of electrons
= (3+1)+½ [ 6 –(3 × 1)-(1 × 3)] = 4+0= 4 (Tetrahedral)
B.P L.P

41. NET 2017 DEC
Ans 1

42. NET 2016 JUNE
Ans 2
Total no. of pair of electrons
= 5 + ½ [ 6 – (5 × 1)+1] = 5 + 1 = 6
B.P L.P B.P L.P
TeF5
-
Square pyramidal

43. NET 2015 JUNE
Ans 2
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)-1] = 2 + 2 = 4
B.P L.P B.P L.P
Tetrahedral
Br3
+ = BrBr2
+
Total no. of pair of electrons
= 4 + ½ [ 7– (4 × 1)-1] = 4 + 1 = 5
B.P L.P B.P L.P
Trigonal bipyramidal
I5
+ = II4
+

44. NET 2015 DEC
Ans 3
XeO2F2
Total no. of pair of electrons
= (2+2)+½ [ 8 –(2 × 1)-(2 × 2)] = 4+1= 5 (See saw)
B.P L.P
Total no. of pair of electrons
= 2 + ½ [ 8 – (2 × 1)] = 2 + 3 = 5
B.P L.P B.P L.P
Linear
XeF2

45. NET 2015 DEC
Ans 1
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
Square
planar
BrF4
-
Total no. of pair of electrons
= 6 + ½ [ 8 – (6 × 1)] = 6 + 1 = 7
B.P L.P B.P L.P
Pentagonal
pyramidal
XeF6
Total no. of pair of electrons
= 6 + ½ [ 5 – (6 × 1)+3] = 6 + 1 = 7
B.P L.P B.P L.P
SbCl6
3-
Pentagonal
pyramidal

46. NET 2014 JUNE
Ans B
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)-1] = 2 + 2 = 4
B.P L.P B.P L.P
Bent
I3
+ = II2
+
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
Linear
I3
- = II2
-
Total no. of pair of electrons
= 2 + ½ [ 6– (2 × 1)] = 2 + 2 = 5
B.P L.P B.P L.P
SCl2
Total no. of pair of electrons
= 2 + ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
Linear
ClF2
-
Bent

47. NET 2013 JUNE
Ans A
Total no. of pair of electrons
= 4 + ½ [ 8 – (4 × 1)] = 4 + 2 = 6
B.P L.P B.P L.P
XeF4
Total no. of pair of electrons
= 3+ ½ [ 7– (3 × 1)] = 3 + 2 = 5
B.P L.P B.P L.P
ClF3
XeO2F2
Total no. of pair of electrons
= (2+2)+½ [ 8 –(2 × 1)-(2 × 2)] = 4 + 1= 5
B.P L.P
Total no. of pair of electrons
XeO4 = 4 + ½ [ 8 – (4 × 2)] = 4 + 0 = 4
B.P L.P B.P L.P
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
lCl4
-

48. NET 2013 DEC
Ans C
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)-1] = 2 + 2 = 4
B.P L.P B.P L.P
Tetrahedral
I3
+ = II2
+
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
Trigonal bipyramidal
I3
- = II2
-

49. NET 2012 JUNE
Ans B
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
I3
- = II2
-

50. NET 2012 DEC
Ans D
Total no. of pair of electrons
XeO3 = 3 + ½ [ 8 – (3 × 2)] = 3 + 1 = 4
B.P L.P B.P L.P
Trigonal
pyramidal
Total no. of pair of electrons
= 3 + ½ [ 6 – (3 × 2)] = 3 + 0 = 3
SO3
Trigonal
planar
B.P L.P B.PL.P
CO3
2-
NO3
-
Total no. of pair of electrons
= 3 +½ [ 4 –(3 × 2)+2] = 3 + 0 = 3
B.P L.P B.P L.P
Trigonal
planar
Total no. of pair of electrons
= 3 +½ [ 5 –(3 × 2)+1] = 3 + 0= 3 Trigonal
planar
B.P L.P B.P L.P

51. NET 2011 JUNE
Ans A
Total no. of pair of electrons
= 4 + ½ [ 8 – (4 × 1)] = 4 + 2 = 6
B.P L.P B.P L.P
XeF4
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
lCl4
-
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
BrF4
-
Total no. of pair of electrons
= 4 + ½ [ 6 – (4 × 1)] = 4 + 1 = 5
B.P L.P B.P L.P
SF4

52. NET 2011 JUNE
Ans B
Total no. of pair of electrons
= 4 + ½ [ 6 – (4 × 1)] = 4 + 1 = 5
B.P L.P B.P L.P
SF4
SO4
2-
Total no. of pair of electrons
= 4 +½ [ 6 –(4 × 2)+2] = 4 + 0 = 4
B.P L.P B.P L.P
Tetrahedral
Total no. of pair of electrons
= (2+2) +½ [ 6 –(2 × 1)-(2 × 2)] = 4 + 0= 4
B.P L.P B.P L.P
SO2Cl2
T Shape
Tetrahedral

53. NET 2011 JUNE
Ans B
Total no. of pair of electrons
= 2+ ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
I3
- = II2
-
Total no. of pair of electrons
= 2 + ½ [ 4 – (2 × 2)] = 2 + 0 = 2
B.P L.P B.P L.P
CO2
NO2
-
Total no. of pair of electrons
= 2 +½ [ 5 –(2 × 2)+1] = 2 + 1 = 3
B.P L.P B.P L.P
NO2
+
Total no. of pair of electrons
= 2 +½ [ 5 –(2 × 2)-1] = 2 + 0 = 3
L.P B.P L.P

54. NET 2011 DEC
Ans A
Total no. of pair of electrons
= 2 + ½ [ 5– (2 × 3)+1] = 2 + 0 = 2
B.P L.P B.P L.P
Linear
N3
- =NN2
-
Total no. of pair of electrons
= 2 + ½ [ 8 – (2 × 1)] = 2 + 3 = 5
B.P L.P B.P L.P
Linear
XeF2
Total no. of pair of electrons
= 4 + ½ [ 7 – (4 × 1)+1] = 4 + 2 = 6
B.P L.P B.P L.P
lCl4
- Square
planar
[PtCl4]2- (dsp2) VBT Square
planar
Total no. of pair of electrons
= 2 + ½ [ 7– (2 × 1)-1] = 2 + 2 = 4
B.P L.P B.P L.P
Bent
ClF2
+
Total no. of pair of electrons
= 2 + ½ [ 7– (2 × 1)+1] = 2 + 3 = 5
B.P L.P B.P L.P
Linear
lCl2
-
XeO3 = 3 + ½ [ 8 – (3 × 2)]
= 3 + 1 = 4
Trigonal
pyramidal
= 3 + ½ [ 6 – (3 × 2)]
= 3 + 0 = 3
SO3
Trigonal
planar