Question

The total number of possible isomers of the complex compound $|C{u}^{II}{(N{H}_3)}_4||P{t}^{II}C{l}_4|$ is

• A. 3
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (D)

$|C{u}^{II}(N{H}_3)||P{t}^{II}C{l}_4|$ shows $|Cu(N{H}_3)Cl||PtC{l}_{3}N{H}_3|,$ $|Cu(N{H}_3)C{l}_3||Pt{(NH)}_{3}Cl|$ and $|CuC{l}_4||Pt{(N{H}_3)}_4|$ .Other isomers such as $|Cu{(N{H}_3)}_{2}C{l}_2||Pt{(N{H}_3)}_{2}C{l}_2|$ does not exist due to both parts being neutral.