The total number of possible isomers for squa | vedclass

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(B) The square planar complex is of the type $[Mabcd]^{n\pm}$,where $M = Pt^{2+}$,$a = Cl^-$,$b = NO_2^-$,$c = NO_3^-$,and $d = SCN^-$.For a square pl

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$12$

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(B) The square planar complex is of the type $[Mabcd]^{n\pm}$,where $M = Pt^{2+}$,$a = Cl^-$,$b = NO_2^-$,$c = NO_3^-$,and $d = SCN^-$.For a square planar complex $[Mabcd]$,there are $3$ geometrical isomers.In this complex,$NO_2^-$ and $SCN^-$ are ambidentate ligands.An ambidentate ligand can coordinate through two different donor atoms.Since there are $2$ ambidentate ligands,each geometrical isomer can exist in $2 	imes 2 = 4$ linkage isomeric forms.Therefore,the total number of isomers = (Number of geometrical isomers) $	imes$ (Number of linkage isomers per geometrical isomer) = $3 	imes 4 = 12$.

List-$I$ (Complex)	List-$II$ (Spin only Magnetic moment)
$A$. $[CoF_6]^{3-}$	$I$. $0$
$B$. $[Co(C_2O_4)_3]^{3-}$	$II$. $\sqrt{24}$
$C$. $[FeF_6]^{3-}$	$III$. $\sqrt{8}$
$D$. $[Mn(CN)_6]^{3-}$	$IV$. $\sqrt{35}$
	$V$. $\sqrt{15}$

The total number of possible isomers for square planar $[Pt(Cl)(NO_2)(NO_3)(SCN)]^{2-}$ is:

Similar Questions

The correct order regarding the number of unpaired electrons present in the following complexes is
$I. [Fe(CN)_6]^{4-} \quad II. [Fe(H_2O)_6]^{2+} \quad III. [Co(H_2O)_6]^{2+}$

Which of the following is diamagnetic?

Match the following:

List-$I$ (Complex) List-$II$ (Spin only Magnetic moment)

$A$. $[CoF_6]^{3-}$ $I$. $0$

$B$. $[Co(C_2O_4)_3]^{3-}$ $II$. $\sqrt{24}$

$C$. $[FeF_6]^{3-}$ $III$. $\sqrt{8}$

$D$. $[Mn(CN)_6]^{3-}$ $IV$. $\sqrt{35}$

$V$. $\sqrt{15}$

Which of the following is expected to be a paramagnetic complex?

Paramagnetic coordination compounds contain ........ electrons.

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