Total number of geometrical isomers possible | vedclass

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

Question

(C) $1$. $[NiCl_4]^{2-}$: This is a tetrahedral complex ($sp^3$ hybridization). Tetrahedral complexes do not exhibit geometrical isomerism. Number of

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) $1$. $[NiCl_4]^{2-}$: This is a tetrahedral complex ($sp^3$ hybridization). Tetrahedral complexes do not exhibit geometrical isomerism. Number of isomers = $0$.$2$. $[CoCl_2(NH_3)_4]^{+}$: This is an octahedral complex of the type $[MA_4B_2]$. It shows $2$ geometrical isomers ($cis$ and $trans$).$3$. $[Co(NH_3)_3(NO_2)_3]$: This is an octahedral complex of the type $[MA_3B_3]$. It shows $2$ geometrical isomers ($fac$ and $mer$).$4$. $[Co(NH_3)_5Cl]^{2+}$: This is an octahedral complex of the type $[MA_5B]$. It does not show geometrical isomerism. Number of isomers = $0$.Total number of geometrical isomers = $0 + 2 + 2 + 0 = 4$.

Total number of geometrical isomers possible for the complexes $[NiCl_4]^{2-}$,$[CoCl_2(NH_3)_4]^{+}$,$[Co(NH_3)_3(NO_2)_3]$ and $[Co(NH_3)_5Cl]^{2+}$ is

Similar Questions

The correct order of the spin-only magnetic moments of the following complexes is
$(I) [Cr(H_2O)_6]Br_2$
$(II) Na_4[Fe(CN)_6]$
$(III) Na_3[Fe(C_2O_4)_3] (\Delta_0 > P)$
$(IV) (Et_4N)_2[CoCl_4]$

Which of the following complex compounds has the highest number of geometrical isomers?

Atomic numbers of $Cr$,$Fe$,and $Co$ are $24$,$26$,and $27$ respectively. Which of the following inner orbital octahedral complexes are paramagnetic?

Which of the following compounds does not show optical activity?

Which of the following complexes will exhibit optical isomerism?

Vedclass Test Series

Exam Paper Generator

Online Exam Module