Animated Solution for Chemistry - d and f-Block Elements: In the ground state of atomic $\text{Fe}(Z=26)$, the spin-only magnetic moment is ........... $\times 10^{-1}\text{ BM}$. (Round off to the nearest integer). [Given : $\sqrt{3}=1.73,\sqrt{2}=1.41$]

Whenever we are asked to find the magnetic moment of a transition metal, our very first instinct should be to write down its electronic configuration. Iron ($\text{Fe}$) is a classic $3d$-series element with an atomic number of $26$.

Its ground state electronic configuration is $[\text{Ar}]3d^{6}4s^2$.

To visualize this, imagine the five $3d$ orbitals. According to Hund's Rule of Maximum Multiplicity, we first fill each of the five orbitals with one electron (all with parallel spins). That accounts for five electrons. The sixth electron must then pair up in the first orbital. This leaves us with exactly four unpaired electrons ($n=4$).

The spin-only magnetic moment ($\mu$) arises purely from the intrinsic spin of these unpaired electrons. The formula is beautifully simple:

Here, $\text{BM}$ stands for Bohr Magneton, the standard unit for magnetic moment.

Let's substitute $n=4$ into our master equation:

Now, we need to evaluate $\sqrt{24}$. The problem kindly provides us with the values of $\sqrt{2}=1.41$ and $\sqrt{3}=1.73$. We can break down $\sqrt{24}$ using basic surd properties:

Substituting the given values:

The question asks for the answer in the format of $x\times 10^{-1}\text{ BM}$. Let's adjust our decimal point:

Finally, we round off $48.786$ to the nearest integer, which gives us $49$.

This is a fantastic example of how a solid grasp of electronic configuration combined with careful arithmetic leads straight to the correct answer!