A magnetic moment of 1.73 BM will be shown b | vedclass

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

Question

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu = 1.73 \ BM$,we have

Vedclass · Accepted Answer

$[Cu(NH_3)_4]^{2+}$

Vedclass · Answer

(C) The magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu = 1.73 \ BM$,we have $\sqrt{n(n+2)} = 1.73$,which implies $n = 1$.$1$. $TiCl_4$: $Ti$ is in $+4$ oxidation state $(d^0)$,so $n = 0$.$2$. $[CoCl_6]^{4-}$: $Co$ is in $+2$ oxidation state $(d^7)$. $Cl^-$ is a weak field ligand,so electrons remain unpaired,$n = 3$.$3$. $[Cu(NH_3)_4]^{2+}$: $Cu$ is in $+2$ oxidation state $(d^9)$. In the square planar complex,there is $1$ unpaired electron in the $3d$ orbital,$n = 1$.$4$. $[Ni(CN)_4]^{2-}$: $Ni$ is in $+2$ oxidation state $(d^8)$. $CN^-$ is a strong field ligand,causing pairing,$n = 0$.Thus,$[Cu(NH_3)_4]^{2+}$ has $1$ unpaired electron and a magnetic moment of $1.73 \ BM$.

$A$ magnetic moment of $1.73 \ BM$ will be shown by which of the following?

Similar Questions

The number of possible isomers for the square planar complex $K_2[PdClBr_2SCN]$ is .................

Which of the following polymerisation isomers of the compound having empirical formula $Cr(NH_3)_3(NO_2)_3$ has the lowest molecular mass?

The calculated spin-only magnetic moments of $[Cr(NH_3)_6]^{3+}$ and $[CuF_6]^{3-}$ in $BM$,respectively,are (Atomic numbers of $Cr$ and $Cu$ are $24$ and $29$,respectively).

In the nitroprusside ion,the iron exists as $Fe^{2+}$ and $NO$ as $NO^{+}$ rather than $Fe^{3+}$ and $NO$ respectively. These forms of ions are established with the help of

Which one of the following complex ions is diamagnetic in nature?

Vedclass Test Series

Exam Paper Generator

Online Exam Module