Observe the following complexes: [Mn(CN)_6]^{ | vedclass

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

Question

(A) Let us analyze each complex:$1$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong-field ligand,causing pairing. The configuration become

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let us analyze each complex:$1$. $[Mn(CN)_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $CN^-$ is a strong-field ligand,causing pairing. The configuration becomes $t_{2g}^4 e_g^0$. It is an inner-orbital complex $(d^2sp^3)$ and is paramagnetic ($2$ unpaired electrons).$2$. $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong-field ligand. The configuration becomes $t_{2g}^5 e_g^0$. It is an inner-orbital complex $(d^2sp^3)$ and is paramagnetic ($1$ unpaired electron).$3$. $[Co(C_2O_4)_3]^{3-}$: $Co^{3+}$ is $3d^6$. $C_2O_4^{2-}$ is a weak-field ligand. It forms an outer-orbital complex $(sp^3d^2)$ and is diamagnetic.$4$. $[MnCl_6]^{3-}$: $Mn^{3+}$ is $3d^4$. $Cl^-$ is a weak-field ligand. It forms an outer-orbital complex $(sp^3d^2)$ and is paramagnetic.$5$. $[Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong-field ligand. The configuration becomes $t_{2g}^6 e_g^0$. It is an inner-orbital complex $(d^2sp^3)$ and is diamagnetic.$6$. $[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak-field ligand. It forms an outer-orbital complex $(sp^3d^2)$ and is paramagnetic.Inner-orbital complexes are those using $(n-1)d$ orbitals. These are $[Mn(CN)_6]^{3-}, [Fe(CN)_6]^{3-}, [Fe(CN)_6]^{4-}$.Among these,the paramagnetic ones are $[Mn(CN)_6]^{3-}$ and $[Fe(CN)_6]^{3-}$.Thus,the total number is $2$.

Observe the following complexes: $[Mn(CN)_6]^{3-}, [Fe(CN)_6]^{3-}, [Co(C_2O_4)_3]^{3-}, [MnCl_6]^{3-}, [Fe(CN)_6]^{4-}, [CoF_6]^{3-}$. From the above,the number of inner orbital complexes with paramagnetic nature is

Similar Questions

What is the number of unpaired electrons present in $[CoF_6]^{3-}$?

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

The spin-only magnetic moment of $[MnBr_4]^{x-}$ is $5.9 \ BM$. The geometry of the complex and $x$ respectively are

Hybridization in $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are respectively . . . . . . .

$K_3[CoF_6]$ is a high-spin complex. The hybridization state of $Co$ in this complex is .....................

Vedclass Test Series

Exam Paper Generator

Online Exam Module