Question

Magnetic moment of ${X^{n + }}$ $(Z = 26)$ is $\sqrt {24} BM$. Hence, number of unpaired electrons and value of $n$ respectively are:
A. $4,2$
B. $2,4$
C. $3,1$
D. $0,2$

Answer 1

We know that $Fe$ has the atomic number of $26$. The magnetic moment is given by the formula $\sqrt {(x)(x + 2)} BM$. Here $x$ is the number of unpaired electrons of the ion. The full form of $BM$ is Bohr Magneton.

Complete step by step solution:
As in the question new are given that the magnetic moment is $\sqrt {26} BM$. We will directly solve using the formula of magnetic moment. So,
$\Rightarrow \sqrt {(x)(x + 2)} = \sqrt {24} \\\ \Rightarrow {x^2} + 2x - 24 = 0 \\\ \Rightarrow {x^2} + 6x - 4x - 24 = 0 \\\ \Rightarrow x(x + 6) - 4(x + 6) = 0 \\\ \Rightarrow (x - 4)(x + 6) = 0 \\\ \Rightarrow x = 4, - 6 \\\$
Only $x = 4$ is acceptable, so the number of unpaired electrons will be four, $x = 4$.
The electronic configuration of $Fe(26) = [Ar]3{d^6}4{s^2}$, now as we have found that there are four unpaired electrons in total so we will remove two electrons from the outermost shell that is from $4s$. So after removing the two electrons we will get $F{e^{ + 2}}$. The electronic configuration of $F{e^{ + 2}}$is $[Ar]3{d^6}$. In $3{d^6}$ there are four unpaired electrons and two paired electrons.
So from the above explanation and calculation it is clear to us that the number of unpaired electrons will be four and the value of $n$ will be two$(F{e^{ + 2}})$. The correct answer of the given question is option: A. $4,2$

Hence, the correct answer is option A.

Additional information:
Iron is a very important metal. It has an atomic weight of $55.85gmo{l^{ - 1}}$. Iron is used in the manufacturing of steel. Iron is used to make many alloys with nickel, chromium, manganese, vanadium. Iron is a transition element.

Note: Always remember that the magnetic moment is given by $\sqrt {(x)(x + 2)} BM$, where $x$ is the number of unpaired electrons of the ion. The electronic configuration of $Fe$ is $Fe(26) = [Ar]3{d^6}4{s^2}$. The colour of $F{e^{ + 2}}$ is very pale green. Always solve the numerical carefully and avoid calculation errors.