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

$[Co(en)_3]^{3+}$ ion is expected to show

The value of the 'spin only' magnetic moment for one of the following configurations is $2.84 \ BM$. The correct one is:

Identify the types of isomers shown in the structures $(i)$ and $(ii)$:

In nitroprusside ion,the iron and $NO$ exist as $Fe^{II}$ and $NO^+$ rather than $Fe^{III}$ and $NO$. These forms can be differentiated by

Square planar complex of the type $MAXBL$ (where $A, B, X$ and $L$ are unidentate ligands) shows which of the following sets of isomers?

Vedclass Test Series

Exam Paper Generator

Online Exam Module