The number of geometrical isomers of the comp | vedclass

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(A) The given complex is $[Co(NO_2)_2(NH_3)_2]$. Assuming a square planar geometry (as it is a $4$-coordinate complex),it follows the form $[MA_2B_2]$

Vedclass · Accepted Answer

$2$

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(A) The given complex is $[Co(NO_2)_2(NH_3)_2]$. Assuming a square planar geometry (as it is a $4$-coordinate complex),it follows the form $[MA_2B_2]$.For a square planar complex of the type $[MA_2B_2]$,there are $2$ possible geometrical isomers: $cis$ and $trans$.In the $cis$ isomer,the identical ligands are adjacent to each other.In the $trans$ isomer,the identical ligands are opposite to each other.Therefore,the total number of geometrical isomers is $2$.

List-$I$ (Complex ion)	List-$II$ (Calculated spin only magnetic moment $(BM)$)
$A$. $[Cr(H_2O)_6]^{2+}$	$I$. $3.87$
$B$. $[Co(H_2O)_6]^{2+}$	$II$. $5.92$
$C$. $[Cu(H_2O)_6]^{2+}$	$III$. $4.90$
$D$. $[Mn(H_2O)_6]^{2+}$	$IV$. $1.73$

The number of geometrical isomers of the complex $[Co(NO_2)_2(NH_3)_2]$ is

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