Calculate the I-I distance in each of the isomeric compounds `C_(2)H_(2)I_(2)` as shown below
Give Bond length of `C -I implies 210` pm
Bond length of `C =C implies 133` pm


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Explanation

The I -I distance `bar((AC)) =bar((AD))+bar((DC))=bar((2AD))`
`C-I` bond length `bar((AB)) = 210` pm
In the `60^(@)` right angle triangle `(ABD)`
`2bar((AD)) =2 (210 sin 60^(@)) = 364` pm
The I-I distance ` =bar((AH)) =bar((AF))+bar((GH))+bar((FG))`
Now `bar(AF) =bar(GH)=bar(BD) = 210 cos 60^(@) = 105` pm
`bar(FG)` is the `C =C` double bond distance `= 133` pm
Thus, `bar(AH) = 105 + 133 + 105 = 343` pm
The I-I distance `=AK` the hypotenuse or right angle triangle `ACK`
The legs of the triangle `ACK` are `AC` are `AC` and `CK`
`AC= 364` pm (as calculated in part (a) above)
`bar(CK) = bar(AH)` (as calculated in part (b) above)
` =343` pm
Then, `barAK = sqrtbar((CK)^(2) + (bar(AC))^(2))`
`= sqrtbar((343)^(2)+ (364)^(2))`
` = 500` pm


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