The spin-only magnetic moments of [Mn(CN)_6]^ | vedclass

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

Question

(C) The oxidation state of $Mn$ in $[Mn(CN)_6]^{2-}$ is $+4$. The electronic configuration of $Mn^{4+}$ is $[Ar] 3d^3$.Since $CN^-$ is a strong field

Vedclass · Accepted Answer

$1.73$ and $5.92$

Vedclass · Answer

(C) The oxidation state of $Mn$ in $[Mn(CN)_6]^{2-}$ is $+4$. The electronic configuration of $Mn^{4+}$ is $[Ar] 3d^3$.Since $CN^-$ is a strong field ligand,the $3d$ electrons are arranged as follows: $t_{2g}^3 e_g^0$. The number of unpaired electrons $(n)$ is $1$.The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.The oxidation state of $Mn$ in $[MnBr_4]^{2-}$ is $+2$. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$.Since $Br^-$ is a weak field ligand,no pairing occurs. The number of unpaired electrons $(n)$ is $5$.The spin-only magnetic moment is $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Thus,the values are $1.73 \ BM$ and $5.92 \ BM$.

$A$. Tetrahedral $NiCl_2Br_2^{2-}$	$B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$	$D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Complexes	No. of unpaired electrons
$A. [CrF_6]^{4-}$	$P. 5$
$B. [MnF_6]^{4-}$	$Q. 2$
$C. [Cr(CN)_6]^{4-}$	$R. 1$
$D. [Mn(CN)_6]^{4-}$	$S. 4$

The spin-only magnetic moments of $[Mn(CN)_6]^{2-}$ and $[MnBr_4]^{2-}$ in Bohr magnetons,respectively are

Similar Questions

Which of the following molecules/ions can exhibit isomerism:
$A$. Tetrahedral $NiCl_2Br_2^{2-}$ $B$. Square planar $Pt(NH_3)_2Cl_2$
$C$. Octahedral $Co(NH_3)_3Cl_3$ $D$. Square planar $Pd(NH_3)_3Br^{+}$
$E$. Octahedral $Co(en)_3^{3+}$

Here,$en = 1,2-\text{diaminoethane}$.

The correct match is:

Complexes No. of unpaired electrons

$A. [CrF_6]^{4-}$ $P. 5$

$B. [MnF_6]^{4-}$ $Q. 2$

$C. [Cr(CN)_6]^{4-}$ $R. 1$

$D. [Mn(CN)_6]^{4-}$ $S. 4$

Among the complex ions,$[Co(en)_2Cl_2]^{+}$,$[CrCl_2(C_2O_4)_2]^{3-}$,$[Fe(H_2O)_4(OH)_2]^{+}$,$[Fe(NH_3)_2(CN)_4]^{-}$,$[Co(en)_2(NH_3)Cl]^{2+}$ and $[Co(NH_3)_4(H_2O)Cl]^{2+}$,the number of complex ion$(s)$ that show$(s)$ cis-trans isomerism is

Consider the following complex ions,$P$,$Q$ and $R$.
$P = [FeF_6]^{3-}$,$Q = [V(H_2O)_6]^{2+}$ and $R = [Fe(H_2O)_6]^{2+}$.
The correct order of the complex ions,according to their spin-only magnetic moment values (in $B.M.$) is

Number of paramagnetic complexes among the following is . . . . . . . $[MnBr_4]^{2-}$,$[NiCl_4]^{2-}$,$[Ni(CN)_4]^{2-}$,$[Ni(CO)_4]$,$[CoF_6]^{3-}$,$[Fe(CN)_6]^{4-}$,$[Mn(CN)_6]^{3-}$,$[Ti(CN)_6]^{3-}$,$[Cu(H_2O)_6]^{2+}$,$[Co(C_2O_4)_3]^{3-}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module