On treatment of [Ni(NH_3)_4]^{2+} with concen | vedclass

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Question

(B) The complex $[Ni(NH_3)_4]^{2+}$ has a square planar geometry. When treated with $HCl,$ it forms two isomers of $[Ni(NH_3)_2Cl_2].$ $I$ is the $cis

Vedclass · Accepted Answer

$I$ cis,$II$ trans; both square planar

Vedclass · Answer

(B) The complex $[Ni(NH_3)_4]^{2+}$ has a square planar geometry. When treated with $HCl,$ it forms two isomers of $[Ni(NH_3)_2Cl_2].$ $I$ is the $cis$-isomer and $II$ is the $trans$-isomer. Both isomers possess a square planar geometry because $Ni^{2+}$ in this coordination environment typically undergoes $dsp^2$ hybridization. Oxalic acid is a bidentate ligand that can form a chelate ring by replacing two $cis$ ligands. Since $I$ reacts with oxalic acid,it must be the $cis$-isomer,while $II$ is the $trans$-isomer which does not react because the $Cl$ atoms are not adjacent.

Compound	Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$	$P$. $2$
$B$. $[Co(en)_3]Cl_3$	$Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$	$R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$	$S$. $5$

Similar Questions

$[Fe(NO_2)_3Cl_3]$ and $[Fe(ONO)_3Cl_3]$ show:

The total number of isomers for square planar complex $[Pd(H_2O)(Br)(NH_3)(CN)]$ is/are

Compound Total stereoisomers
$A$. $[Co(en)_2Cl_2]^+$ $P$. $2$
$B$. $[Co(en)_3]Cl_3$ $Q$. $3$
$C$. $[Pt(NH_3)_2Cl_2]$ $R$. $4$
$D$. $[Pt(NH_3)_3Cl_3]^+$ $S$. $5$

Match the compounds with their total number of stereoisomers and select the correct code:

Which of the following is paramagnetic?

Which from the following ligands is able to form linkage isomers?

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