Which from following complexes is an example of {MA}₄ {BC} type distereoisomers ?

Correct Answer is : [Pt(NH3)4ClBr]2+≤ft[{Pt}≤ft({NH}₃)₄ {ClBr}]²+}[Pt(NH3)4ClBr]2+

≤ft[{Pt}≤ft({NH}₃)≤ft({H}₂ {O}) {Cl}₂]

≤ft[{Pt}≤ft({NH}₃)₄ {ClBr}]²+}

≤ft[{Co}≤ft({NH}₃)₄ {Cl}₂]^{+}

Correct Answer is : [Pt(NH3)4ClBr]2+≤ft[{Pt}≤ft({NH}₃)₄ {ClBr}]²+}[Pt(NH3)4ClBr]2+

Test Index

MHTCET 19 Apr 2026 Shift 2 Paper

Question : 91 of 150

Marks: +1, -0

\mathrm{MA}_{4} \mathrm{BC}

type distereoisomers ?

$\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{2}\right]$
$\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{ClBr}\right]^{2+}$
$\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}$
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$

Solution:

Concept:

Octahedral complexes of the type

\mathrm{MA}_4\mathrm{BC}

can show cis-trans geometrical isomerism, and these isomers are diastereomers.

Explanation:

In the general formula

\mathrm{MA}_4\mathrm{BC}

, four ligands are identical (A), while B and C are two different monodentate ligands.

The complex

\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\mathrm{ClBr}\right]^{2+}

exactly matches this form.

Here,

\mathrm{M}=\mathrm{Pt}

\mathrm{A}=\mathrm{NH}_3

\mathrm{B}=\mathrm{Cl}^-

, and

\mathrm{C}=\mathrm{Br}^-

This is an octahedral complex with coordination number 6.

Because

\mathrm{Cl}^-

and

\mathrm{Br}^-

can be placed adjacent (cis) or opposite (trans), two geometrical isomers are formed.

These cis and trans forms are not mirror images of each other, so they are diastereomers.

Other complexes given in the options do not have the

\mathrm{MA}_4\mathrm{BC}

general formula.

Answer:

Option B:

\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4\mathrm{ClBr}\right]^{2+}

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