Which of the following ions has the lowest ma | vedclass

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

Question

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Ti^

Vedclass · Accepted Answer

$Ti^{3+}$

Vedclass · Answer

(A) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$1$. For $Ti^{3+}$ $(3d^1)$: $n = 1$,$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$2$. For $Co^{3+}$ $(3d^6)$: In the presence of weak field ligands,$n = 4$,$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \ BM$.$3$. For $Cr^{3+}$ $(3d^3)$: $n = 3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.$4$. For $Fe^{3+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Comparing the values,$Ti^{3+}$ has the lowest number of unpaired electrons $(n=1)$,therefore it has the lowest magnetic moment.

Which of the following ions has the lowest magnetic moment?

Similar Questions

If the magnetic moment of $[Ag(CN)_2]^{-1}$ is zero,then the number of unpaired electrons will be

Which of the following complexes exhibits the highest paramagnetic behaviour? (Where $gly = \text{glycine}$,$en = \text{ethylenediamine}$ and $bipy = \text{bipyridyl}$) (At. nos. $Ti = 22, V = 23, Cr = 24, Fe = 26, Co = 27$)

Which of the following ions is paramagnetic?

What is the ionization isomer of $[Cr(H_2O)_4Cl(NO_2)]Cl$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module