The magnetic moment of X^{n+} is sqrt{24} B. | vedclass

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

Question

(C) The magnetic moment formula is $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \ B.M.$,we have $

Vedclass · Accepted Answer

$4, 2$

Vedclass · Answer

(C) The magnetic moment formula is $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons.Given $\mu = \sqrt{24} \ B.M.$,we have $\sqrt{n(n+2)} = \sqrt{24}$,which implies $n(n+2) = 24$.Solving for $n$,we get $n^2 + 2n - 24 = 0$,so $(n+6)(n-4) = 0$. Since $n$ must be positive,the number of unpaired electrons is $4$.The atomic number of $X$ is $26$,which corresponds to $Fe$. The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.For $Fe^{2+}$,the configuration is $[Ar] 3d^6$,which has $4$ unpaired electrons.Thus,the number of unpaired electrons is $4$ and the value of '$n$' is $2$.

The magnetic moment of $X^{n+}$ is $\sqrt{24} \ B.M.$ Find the number of unpaired electrons and the value of '$n$' respectively. (Atomic number of $X = 26$)

Similar Questions

Which of the following complexes is an example of the $MA_2BC$ type of stereoisomers?

How many geometrical isomers are shown by the complex $[Pt(NH_3)(NH_2OH)(H_2O)(Py)]^+$?

$A$ compound absorbs light in the wavelength region $490-500 \,nm$. Its complementary colour is

Which of the following is paramagnetic?

Identify the metal ions among $Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+}$ and $Ti^{2+}$ having a spin-only magnetic moment value more than $3.0 \ BM$. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module