A coordination complex of type MX_2Y_2 (M = m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a tetrahedral complex of type $MX_2Y_2$,all positions are equivalent and adjacent to one another,so it does not show geometrical isomerism. Th

Vedclass · Accepted Answer

$1$ and $2$

Vedclass · Answer

(A) For a tetrahedral complex of type $MX_2Y_2$,all positions are equivalent and adjacent to one another,so it does not show geometrical isomerism. Thus,only $1$ isomer is possible.For a square planar complex of type $MX_2Y_2$,it can exist in two geometrical isomeric forms: $cis$ and $trans$. Thus,$2$ isomers are possible.Therefore,the number of isomers for tetrahedral and square planar geometries are $1$ and $2$ respectively.

$A$ coordination complex of type $MX_2Y_2$ ($M$ = metal ion; $X, Y$ = monodentate ligands) can have either a tetrahedral or a square planar geometry. The maximum number of possible isomers in these two cases are respectively:

Similar Questions

Identify the type of isomerism observed when an ambidentate ligand binds to a central metal atom and provide two examples of ambidentate ligands.

The diamagnetic species is

Discuss the magnetic properties of coordination compounds.

The pair of complex compounds $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ are an example of

$A$ solution of $[Ni(H_{2}O)_{6}]^{2+}$ is green but a solution of $[Ni(CN)_{4}]^{2-}$ is colourless. Explain.

Vedclass Test Series

Exam Paper Generator

Online Exam Module