Identify the number of unpaired electrons pre | vedclass

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(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.In the complex $[CoF_6]^{3-}$,the oxidation state of $

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$4$,Octahedral

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(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.The $3d$ orbitals have $4$ unpaired electrons.The complex undergoes $sp^3d^2$ hybridization,which corresponds to an octahedral geometry.

List-$I$ (Complex)	List-$II$ (Hybridization)
$A. Ni(CO)_4$	$I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$	$II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$	$III. dsp^2$
$D. [CoF_6]^{3-}$	$IV. sp^3$

Identify the number of unpaired electrons present and the geometry,respectively,of the complex $[CoF_6]^{3-}$ (given atomic number of $Co = 27$).

Similar Questions

Which among the following statements about $[Ni(CN)_4]^{2-}$ is $NOT$ true?

Match the complexes in List-$I$ with their hybridization in List-$II$.
List-$I$ (Complex) List-$II$ (Hybridization)
$A. Ni(CO)_4$ $I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$ $II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$ $III. dsp^2$
$D. [CoF_6]^{3-}$ $IV. sp^3$

Predict the number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion.

The magnetic moment of $[NiX_4]^{2-}$ ion is found to be zero. Then the ion is : ($X =$ monodentate anionic ligand)

Which one of the following is an inner orbital complex as well as diamagnetic in behaviour? (Atomic number: $Zn = 30$,$Cr = 24$,$Co = 27$,$Ni = 28$)

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