In square planar complexes (also in octahedral complexes), some ligands (A) weaken the bond of the other ligand trans to them (B). As a result, when substitution by another ligand takes place, the ligand with weaker bond (B) gets replaced. This is called transeffect of A. Given the following order of transeffect $[C O > C_{2} H_{4} > B r^{-} > C l^{-} > N H_{3}]$ . Predict the product of the following reaction: $[P t C l_{3} (C O)]^{-} + N H_{3} \to$

T r a n s - [P t (C O) C l_{2} (N H_{3})]

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C i s - [P t (C O) C l_{2} (N H_{3})]

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Both A and B in equal amount

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Neither (A) or (B)

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Solution

The correct option is A $T r a n s - [P t (C O) C l_{2} (N H_{3})]$
As transeffect of $C O$ is more, the $C l^{-}$ trans to it gets substituted by $N H_{3}$ .

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Similar questions

Cis-trans isomerism is found in square planar complexes of the molecular formula: (a and b are monodentate ligands)

Q. Which of the following square planar complexes will show cis-trans isomerism?

[M is the metal while a and b are monodentate ligands.]

Q. The magnetic moments of following, arranged in increasing order will be:

[atomic number of Co = 27]

(1)

C o^{3 +}

(octahedral complex with a strong field ligand)

(2)

C o^{3 +}

(octahedral complex with a weak field ligand)

(3)

C o^{2 +}

(tetrahedral complex)

(4)

C o^{2 +}

(square planar complex)

Which of the following statements is correct regarding the chirality (optical isomerism) of the cis and trans isomers of the type $M (a a)_{2} b_{2}$ (M stands for a metal, a and b are achiral ligands and aa is a bidentate ligand) ?

Q. Which of the following square planar complex does not show geometrical isomerism?

AA is symmetrical bidentate ligand
AB is unsymmetrical bidentate ligand
a and b are monodentate ligands

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The correct option is A Trans−[Pt(CO)Cl2(NH3)]As transeffect of CO is more, the Cl– trans to it gets substituted by NH3.

The correct option is A $T r a n s - [P t (C O) C l_{2} (N H_{3})]$
As transeffect of $C O$ is more, the $C l^{-}$ trans to it gets substituted by $N H_{3}$ .