The number of possible geometric isomers for | vedclass

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

Question

(C) For a square planar complex of the type $[Mabcd]$,where all four ligands are different,the number of geometric isomers is $3$.These isomers are fo

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For a square planar complex of the type $[Mabcd]$,where all four ligands are different,the number of geometric isomers is $3$.These isomers are formed by fixing one ligand (e.g.,$a$) and arranging the other three ligands $(b, c, d)$ in different positions relative to it.The three possible arrangements are:$1$. $a$ trans to $b$ (with $c$ trans to $d$)$2$. $a$ trans to $c$ (with $b$ trans to $d$)$3$. $a$ trans to $d$ (with $b$ trans to $c$)Thus,the total number of geometric isomers is $3$.

The number of possible geometric isomers for a square planar complex $[Mabcd]$ is ............. ($M =$ central metal,$a, b, c, d =$ monodentate ligands).

Similar Questions

The pentaamminenitrocobalt $(III)$ ion exhibits which of the following properties?

Amongst $[Ni(H_{2}O)_{6}]^{2+}$, $[Ni(PPh_{3})_{2}Cl_{2}]$, $[Ni(CO)_{4}]$ and $[Ni(CN)_{4}]^{2-}$, the paramagnetic species are

Which of the following exhibit ionization isomerism?
$I) [Cr(NH_3)_4Cl_2]Cl$
$II) [Ti(H_2O)_5Cl](NO_3)_2$
$III) [Pt(en)(NH_3)Cl]NO_3$
$IV) [Co(NH_3)_4(NO_3)_2]NO_3$

The species that can have a $trans$-isomer is: ($en = \text{ethane-}1,2\text{-diamine}$,$ox = \text{oxalate}$)

The number of isomers possible for the complex $K_2[PtClBr_2(SCN)]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module