1: Predict the shapes and bond angles in:

i) sulphur dichloride (SCl₂)     non linear    104.5°

ii) silane (SiH₄)  tetrahedral   109.5°

iii) carbon disulphide (CS₂)  linear   180°

iv) sulphur trioxide  trigonal planar   120°

v) ozone  non-linear   117.5°

 

2: State and explain the shape of the ammonia molecule, NH₃

The central N atom in NH₃ has 3 bonding pairs and 1 lone pair around it. The electron pairs all repel to get as far apart as possible, and the lone pair repels more strongly than the bonding pairs. As a result the bond angles are 107º and the shape is PYRAMIDAL.

3:  Compare the shapes of the H₂O molecule and the H₃O⁺ ion, and use this to explain why the bond angle in H₂O is 104.5º while the bond angle in H₃O⁺ is 107º.

The central O atom in H₂O has 2 bonding pairs and 2 lone pairs around it giving it a NON-LINEAR shape, while the central O atom in H₃O⁺ has 3 bonding pairs and 1 lone pair around it giving it a PYRAMIDAL shape. In both cases the four electron pairs repel to get as far apart as possible, but because lone pairs repel more than bonding pairs the bond angle in H₂O closes to 109.5 – (2 x 2.5) = 104.5º while the bond angleH₃O⁺ in only closes to 109.5 – 2.5 = 107º.

4:  Draw the 3D shape of the following molecules

i) chloromethane, CH₃Cl

ii) phosphine, PH₃

iii) phosphorus pentachloride, PCl5

5: State the shape around each of the three carbon atoms in propene, CH₂CHCH₃

The 3D shape can be drawn as: 

The carbon on the left has 3 bonding pairs and no lone pairs (remember a double bond counts as 1 bonding pair for the purposes of VSEPR Theory). It thus has TRIGONAL PLANAR shape.

The carbon in the middle is the same as the one on the left, with 3 bonding pairs and no lone pairs, and therefore TRIGONAL PLANAR shape.

The carbon on the right has 4 bonding pairs and no lone pairs, so it has TETRAHEDRAL shape.