Question

Which of the following 0.1 M complex compound solutions will have the minimum electrical conductivity?

• A. Hexammine platinum (IV) chloride
• B. Chloropentaammine platinum (IV) chloride
• C. Dichiaro tetrammine platinum (IV) chloride
• D. Trichloro triammine platinum (IV) chloride

Answer 1

Answer: (D)

The formula of given compiex are as follows :
(a) Hexammine platinum (IV) chioride $[Pt(N{H}_3{)}_6]C{l}_4$
(b) Chioropentammine piatinum (IV) chloride
$[Pt(N{H}_3{)}_{5}Cl]C{l}_3$
(c) Dichlorotetrammine platinum (JV) chioride
$[Pt(N{H}_3{)}_{4}C{l}_2]C{l}_2$
(d) Trichlorotriammine piatinum (IV) chioride
$[Pt(N{H}_3{)}_{3}C{l}_3]Cl$
In aqueous soiution, the complexes ionise as
$[Pt(N{H}_3{)}_6]C{l}_4\leftrightharpoons \underset{5ions}{\underbrace{[Pt(N{H}_3{)}_6{]}^{4+}+4C{l}^{-}}}$
$[Pt(N{H}_3{)}_{5}Cl]C{l}_3\leftrightharpoons \underset{4ions}{\underbrace{[Pt(N{H}_3{)}_{5}Cl{]}^{3+}+3C{l}^{-}}}$
$[Pt(N{H}_3{)}_{4}C{l}_2]C{l}_2\leftrightharpoons \underset{2ions}{\underbrace{[Pt(N{H}_3{)}_{4}Cl{]}^{2+}+C{l}^{-}}}$
$[Pt(N{H}_3{)}_{3}C{l}_3]Cl\leftrightharpoons \underset{2ions}{\underbrace{[Pt(N{H}_3{)}_{3}C{l}_3{]}^{+}+C{l}^{-}}}$
Trichlorotriammine platinum (IV) chioride gives the minimum number of ions in the soiution. Hence, it has the minimum electricaI conductivity.