Which of the following pairs differ in geomet | vedclass

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(A) $1$. $[Ni(CO)_4]$: $Ni(0)$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. Hybridization is $sp^3$,geometry is tetrahedral,and it i

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$[Ni(CO)_4]$ & $[Ni(CN)_4]^{2-}$

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(A) $1$. $[Ni(CO)_4]$: $Ni(0)$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. Hybridization is $sp^3$,geometry is tetrahedral,and it is diamagnetic $(n=0)$.$2$. $[Ni(CN)_4]^{2-}$: $Ni(II)$ is $3d^8$. $CN^-$ is a strong field ligand. Hybridization is $dsp^2$,geometry is square planar,and it is diamagnetic $(n=0)$.$3$. Since both are diamagnetic but have different geometries (tetrahedral vs square planar),option $A$ is correct.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$	$N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$	$P$. $d^2sp^3$,diamagnetic

Which of the following pairs differ in geometry but resemble in magnetic behavior?

Similar Questions

Which of the following complexes has a square planar geometry and $dsp^2$ hybridization?

Match the correct option.
$A$. $[CoF_6]^{3-}$ $M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$ $N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$ $O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$ $P$. $d^2sp^3$,diamagnetic

Which one of the following complexes is an outer orbital complex?

The number of unpaired electrons in $[NiCl_{4}]^{2-}$, $Ni(CO)_{4}$ and $[Cu(NH_{3})_{4}]^{2+}$ respectively are

Which of the following statements is $NOT$ correct regarding the complex $[Co(NH_3)_6]^{3+}$?

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