The hexafluoroferrate(III) ion is an outer or | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The chemical formula for the hexafluoroferrate$(III)$ ion is $[FeF_6]^{3-}$.In this complex,the oxidation state of $Fe$ is $x + 6(-1) = -3$,so $x

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The chemical formula for the hexafluoroferrate$(III)$ ion is $[FeF_6]^{3-}$.In this complex,the oxidation state of $Fe$ is $x + 6(-1) = -3$,so $x = +3$.The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.Since $F^-$ is a weak field ligand,it forms an outer orbital complex ($sp^3d^2$ hybridization).In the $3d$ orbitals,the $5$ electrons remain unpaired according to Hund's rule.Therefore,the number of unpaired electrons is $5$.

$LIST-I$	$LIST-II$
$P. dsp^2$	$1. [FeF_6]^{4-}$
$Q. sp^3$	$2. [Ti(H_2O)_3Cl_3]$
$R. sp^3d^2$	$3. [Cr(NH_3)_6]^{3+}$
$S. d^2sp^3$	$4. [FeCl_4]^{2-}$
	$5. Ni(CO)_4$
	$6. [Ni(CN)_4]^{2-}$

The hexafluoroferrate$(III)$ ion is an outer orbital complex. The number of unpaired electrons in it is.....

Similar Questions

The hybridisation and magnetic behavior of the complex $[Ni(CO)_4]$ are

Square-planar geometry is shown by

Among the following,the species having square planar geometry for the central atom are:
$(i)$ $XeF_4$
$(ii)$ $SF_4$
$(iii)$ $[NiCl_4]^{2-}$
$(iv)$ $[PtCl_4]^{2-}$

Match each set of hybrid orbitals from $LIST-I$ with complex(es) given in $LIST-II$. The correct option is

$LIST-I$ $LIST-II$

$P. dsp^2$ $1. [FeF_6]^{4-}$

$Q. sp^3$ $2. [Ti(H_2O)_3Cl_3]$

$R. sp^3d^2$ $3. [Cr(NH_3)_6]^{3+}$

$S. d^2sp^3$ $4. [FeCl_4]^{2-}$

$5. Ni(CO)_4$

$6. [Ni(CN)_4]^{2-}$

What is the geometry of $[Ni(CO)_4]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module