Which among the following is most paramagneti | vedclass

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

Question

(B) Paramagnetic character depends on the number of unpaired electrons in the $d$-subshell.In these complex ions,$H_2O$ is a weak field ligand,so the

Vedclass · Accepted Answer

$[Fe(H_2O)_6]^{2+}$

Vedclass · Answer

(B) Paramagnetic character depends on the number of unpaired electrons in the $d$-subshell.In these complex ions,$H_2O$ is a weak field ligand,so the electrons remain unpaired according to Hund's rule:$(a)$ In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ has a $3d^3$ configuration. Number of unpaired electrons $= 3$.$(b)$ In $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ has a $3d^6$ configuration. In an octahedral field,this corresponds to $t_{2g}^4 e_g^2$,resulting in $4$ unpaired electrons.$(c)$ In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ has a $3d^9$ configuration. Number of unpaired electrons $= 1$.$(d)$ In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ has a $3d^{10}$ configuration. Number of unpaired electrons $= 0$.Since paramagnetic character is directly proportional to the number of unpaired electrons,$[Fe(H_2O)_6]^{2+}$ is the most paramagnetic.

Which among the following is most paramagnetic?

Similar Questions

Consider the following isomers of $[Cr(NH_3)_2 Cl_4]^-$ and identify the correct classification of these isomers as $cis$ or $trans$.

The coordination number,oxidation number,number of electrons in $d$-orbital,and number of unpaired electrons respectively in the complex $[Co(NH_3)_4CO_3]ClO_4$ are . . . . . . .

Identify the diamagnetic complexes among the following complexes $K$ to $P$:
$(K)\ K_3[Fe(CN)_6]$
$(L)\ [Co(NH_3)_6]Cl_3$
$(M)\ Na_3[Co(ox)_3]$
$(N)\ [Ni(H_2O)_6]Cl_2$
$(O)\ K_2[Pt(CN)_4]$
$(P)\ [Zn(H_2O)_6](NO_3)_2$

The correct statement about the magnetic properties of $[Fe(CN)_6]^{3-}$ and $[FeF_6]^{3-}$ is $(Z = 26)$.

The $[Cr(NH_3)_6]^{3+}$ ion is ...

Vedclass Test Series

Exam Paper Generator

Online Exam Module