Question

In compound $FeC{l}_2$, the orbital angular momentum of the last electron in its cation and magnetic moment (in Bohr Magneton) of this compound are:
(A) ${\displaystyle \frac{h}{4\pi }}\sqrt{6}$, $\sqrt{35}$
(B) ${\displaystyle \frac{h}{2\pi }}\sqrt{6}$, $\sqrt{24}$
(C) 0, $\sqrt{35}$
(D) None of these

Answer 1

Hint: Think about the concept of orbital angular momentum. Find out the oxidation state of iron in $FeC{l}_2$ and write its electronic configuration. Find the azimuthal quantum number, l of the last electron. Then, use the formula of orbital angular momentum, $L={\displaystyle \frac{h}{2\pi }}\sqrt{l(l+1)}$. Substitute the value of l in the equation to get the answer. Then find a magnetic moment.

Complete step by step solution:
- Iron has the atomic number 26
- Its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$
- In $FeC{l}_2$, iron is in +2 oxidation state. So its electronic configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^0$
- So the last electron enters in 3d orbital which is shown as $\underset{+2}{}\underset{+1}{}\underset{0}{}\underset{-1}{}\underset{-2}{}$
- Therefore, azimuthal quantum number of last electron is l = +2
- Now, we know the formula of orbital angular momentum, $L={\displaystyle \frac{h}{2\pi }}\sqrt{l(l+1)}$
- Substituting the value l=+2 in the formula we get, $L={\displaystyle \frac{h}{2\pi }}\sqrt{2(2+1)}$
$$L={\displaystyle \frac{h}{2\pi }}\sqrt{2(2+1)}={\displaystyle \frac{h}{2\pi }}\sqrt{6}$$
- Therefore, orbital angular momentum is ${\displaystyle \frac{h}{2\pi }}\sqrt{6}$
- Magnetic moment is calculated using the formula, $\mu =\sqrt{n(n+2)}$ B.M.
- Therefore, for ferrous ion, number of unpaired electrons, n=4
- Therefore, magnetic moment, $\mu =\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}B.M.$
- Therefore, the magnetic moment is $\sqrt{24}$B.M.

- Therefore, the answer is option (B).

Note: Remember the formula for calculating orbital angular momentum is $L={\displaystyle \frac{h}{2\pi }}\sqrt{l(l+1)}$ where h is the Planck’s constant and l is azimuthal quantum number and the formula for calculating magnetic moment for s-block, p-block, d-block elements is $\mu =\sqrt{n(n+2)}$ where n is the number of unpaired electrons.