Assuming that all the four valencies of carbon atom in propane pointing towards the corners of a regular tetrahedron, If the distance between the terminal carbon atoms in propane is $X$ $˚ A$ . Find $100 X$ . Given, $C - C$ bond length is $1.54 ˚ A$ ,
(sin $109.5$ is $0.94$ )

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Solution

The distance $X$ between the terminal $C$ atoms is given by the expression $2 \times C - C_{b o n d l e n g t h} \times s i n 109.5 = 2 \times 1.54 \times 0.94 = 2.9 A^{0}$ .

Hence, $100 X = 100 \times 2.9 A^{0} = 290 A^{0}$ .

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˙ A

C - C

1.54

˙ A

On catalytic hydrogenation ethylene gives ethane. During this reaction

hybridization of carbon atoms changes from SP² to SP³

bond angle decreases from 120 to 109.5

C-C Bond length increases from 1.34 A° to 1.54 A°

all of these

In propane, the number of Carbon-Carbon single bond is/are:

^{'} A^{'}

1.54 A^{\circ}

1.34 A^{\circ}

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