Which of the following is not a square planar | vedclass

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(D) $1$. $[AgF_4]^-$: $Ag(III)$ is a $d^8$ metal ion,which typically forms square planar complexes.$2$. $[AuCl_4]^-$: $Au(III)$ is a $d^8$ metal ion,w

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$[NiCl_2(PMe_3)_2]$

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(D) $1$. $[AgF_4]^-$: $Ag(III)$ is a $d^8$ metal ion,which typically forms square planar complexes.$2$. $[AuCl_4]^-$: $Au(III)$ is a $d^8$ metal ion,which forms square planar complexes.$3$. $[RhCl(PPh_3)_3]$: This is a $16$-electron $Rh(I)$ complex,which is square planar (Wilkinson's catalyst).$4$. $[NiCl_2(PMe_3)_2]$: $Ni(II)$ is a $d^8$ ion. While many $Ni(II)$ complexes are square planar,the geometry depends on the ligand field strength. For $[NiCl_2(PMe_3)_2]$,the complex is tetrahedral due to the steric bulk of the $PMe_3$ ligands and the weak field nature of $Cl^-$ ligands.

$A. sp^3$	$(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$	$(ii). [Ni(CO)_4]$
$C. sp^3d^2$	$(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$	$(iv). [CoF_6]^{3-}$
	$(v). [Fe(CO)_5]$

Which of the following is not a square planar complex?

Similar Questions

The magnetic moment for two complexes of empirical formula $Ni(NH_3)_4(NO_3)_2 \cdot 2H_2O$ is $0$ and $2.84 \ BM$ respectively. The second complex is not a neutral complex. The correct formula and geometry of the first complex is:

Match the following hybridization in Column $I$ with the corresponding coordination complexes in Column $II$.
$A. sp^3$ $(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$ $(ii). [Ni(CO)_4]$
$C. sp^3d^2$ $(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$ $(iv). [CoF_6]^{3-}$
$(v). [Fe(CO)_5]$

For $K_3[Co(CN)_6]$,which of the following statements is not correct?

What is true for $[Fe(CN)_6]^{3-}$ and $[FeF_6]^{3-}$?

The complex ion $[Cu(NH_3)_4]^{2+}$ has which of the following geometries and electronic configurations?

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