The geometry,hybridization,and magnetic momen | vedclass

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(D) $CN^-$ is a strong field ligand,while $Br^-$ and $F^-$ are weak field ligands.For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $CN^-$ caus

Vedclass · Accepted Answer

Square planar,tetrahedral,octahedral : $dsp^2, sp^3, sp^3d^2 : 0, 5.9, 4.9$

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(D) $CN^-$ is a strong field ligand,while $Br^-$ and $F^-$ are weak field ligands.For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $CN^-$ causes pairing,leading to $dsp^2$ hybridization (square planar) with $0$ unpaired electrons and $\mu = 0 \ BM$.For $[MnBr_4]^{2-}$,$Mn^{2+}$ has $d^5$ configuration. $Br^-$ is a weak ligand,leading to $sp^3$ hybridization (tetrahedral) with $5$ unpaired electrons and $\mu = \sqrt{5(5+2)} = 5.9 \ BM$.For $[FeF_6]^{4-}$,$Fe^{2+}$ has $d^6$ configuration. $F^-$ is a weak ligand,leading to $sp^3d^2$ hybridization (octahedral) with $4$ unpaired electrons and $\mu = \sqrt{4(4+2)} = 4.9 \ BM$.

The geometry,hybridization,and magnetic moment of the ions $[Ni(CN)_4]^{2-}$,$[MnBr_4]^{2-}$,and $[FeF_6]^{4-}$ respectively are:

Similar Questions

Assertion $:$ Complex ion $[Co(NH_3)_6]^{2+}$ is readily oxidized to $[Co(NH_3)_6]^{3+}$.
Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.

If $Ni^{2+}$ is replaced by $Pt^{2+}$ in the complex $[NiCl_2Br_2]^{2-}$,which of the following properties are expected to change?
$A.$ Geometry
$B.$ Geometrical isomerism
$C.$ Optical isomerism
$D.$ Magnetic properties

Nickel combines with a uninegative monodentate ligand $(X^{-})$ to form a paramagnetic complex $[NiX_4]^{2-}$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

In which of the following complex ions is the value of the magnetic moment (spin only) $\sqrt{3} \ BM$ and outer $d$-orbitals are used in hybridization?

Which of the following is true for the $[NiCl_4]^{2-}$ complex?

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