Question

The platinum-chlorine distance has been found to be 2.32$\mathring{A}$ in several crystalline compounds. If this value applies to both of the compounds shown in figure, what is the chlorine -chlorine distance in (a) structure I and (b) structure II?
structure I and II are geometrical isomers and based on a position of $Cl-$atoms, distance between them can be determined.

• A. (a) $4.64\mathring{A}$ (b) $3.28\mathring{A}$
• B. (a) $4.24\mathring{A}$ (b) $4.58\mathring{A}$
• C. (a) $4.44\mathring{A}$ (b) $8.18\mathring{A}$
• D. None of these

Answer 1

Answer: (A)

(a) $4.64\mathring{A}$ (b) $3.28\mathring{A}$

(a) Given $(Pt-Cl)$ distance = 2.32 $\mathring{A}$
Since both $Cl$ and $Pt$ atoms are in straight line hence in (i) $Cl-Cl$$distance$=2 \times 2.32 \mathring{A} = 4.64\mathring{A}$\left(b\right)In\left(II\right),$Cl-Cl$distance$= \sqrt{(2.32)^2 + (2.32)^2}= 2.32\sqrt{2}= 2.32 \times 1.414= 3.28\mathring{A}$