MCQ
The complex $[Pt (py) (NH_3) (Br) (Cl)]$ will have how many geometrical isomers ?
- A$0$
- B$2$
- ✓$3$
- D$4$
✓
Answer
Correct option: C.
$3$
c
The given complex is a type of [Mabcd] compound where the M is metal and a,b,c,d are 4 different ligands attached to it. This type of complexes forms 3 geometrical isomers as depicted in the image.
The given complex is a type of [Mabcd] compound where the M is metal and a,b,c,d are 4 different ligands attached to it. This type of complexes forms 3 geometrical isomers as depicted in the image.
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$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{C{H_3\,\,}} \\
{\,|\,\,\,\,\,\,\,\,\,\,}
\end{array}\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow[{{H_2}{O_2}|\mathop {OH}\limits^\Theta }]{{{B_2}{H_6}|THF}}\,(Z)$ Major product
→{\begin{array}{*{20}{c}}
{C{H_3\,\,}} \\
{\,|\,\,\,\,\,\,\,\,\,\,}
\end{array}\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C - CH = C{H_2}\,\,} \\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow[{{H_2}{O_2}|\mathop {OH}\limits^\Theta }]{{{B_2}{H_6}|THF}}\,(Z)$ Major product
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