Question

Arrange the following complexes in increasing order of conductivity of their solutions.
(i) $\left[Co\right(N{H}_3{)}_6\left]C{l}_3\right]$
(ii) $\left[Co\right(N{H}_3{)}_{4}C{l}_2]Cl$
(iii) $\left[Co\right(N{H}_3{)}_6]C{l}_3$
(iv) $\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2$

• A. $\left(i\right)<\left(ii\right)<\left(iv\right)<\left(iii\right)$
• B. $\left(ii\right)<\left(i\right)<\left(iii\right)<\left(iv\right)$
• C. $\left(i\right)<\left(iii\right)<\left(ii\right)<\left(iv\right)$
• D. $\left(iv\right)<\left(i\right)<\left(ii\right)<\left(iii\right)$

Answer 1

Answer: (A)

$\left(i\right)<\left(ii\right)<\left(iv\right)<\left(iii\right)$

Higher the number of ions in the solution, higher is the conductivity.

No. of ions in

$\left[Co\right(N{H}_3{)}_{3}C{l}_3]=0;$

$\left[Co\right(N{H}_3{)}_{4}C{l}_2]Cl=2;$
$\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2=3;$

$\left[Co\right(N{H}_3{)}_6]C{l}_3=4$

Hence, Option "A" is the correct answer.