Ionization & Magnetic Moment
Order of Electron Removal | Spin-Only Formula | Class 11
1. Order of Removing Electrons (Ion Formation)
When an atom turns into a positive ion (cation), it loses electrons. A very common mistake is assuming that electrons are removed in the reverse order of the Aufbau filling principle.
Electrons are always removed from the outermost principal shell first (i.e., the shell with the highest principal quantum number, $n$).
The Transition Metal Trap
During the filling of orbitals (Aufbau), the $4s$ orbital fills before the $3d$ orbital because it has lower energy. However, once the $3d$ orbital contains electrons, it shields the $4s$ electrons, pushing the $4s$ shell further out. Therefore, when ionizing, electrons are removed from $4s$ BEFORE $3d$.
Example: Iron ($Fe$, $Z=26$)
- Neutral $Fe$: $[Ar] \ 3d^6 \ 4s^2$ (Notice $4s$ is the highest principal shell, $n=4$)
- $Fe^{2+}$ ion: Loses two electrons from the $4s$ shell $\rightarrow$ $[Ar] \ 3d^6$
- $Fe^{3+}$ ion: Loses two from $4s$ and one from $3d$ $\rightarrow$ $[Ar] \ 3d^5$
2. Paramagnetism and Diamagnetism
The magnetic properties of an atom or ion depend on whether its electrons are paired or unpaired.
- Diamagnetic: All electrons are paired (e.g., $\uparrow\downarrow$). These substances are weakly repelled by a magnetic field.
- Paramagnetic: Contains one or more unpaired electrons (e.g., $\uparrow$). These substances are attracted to a magnetic field.
3. Spin-Only Magnetic Moment Formula
The magnetic moment due strictly to the spin of unpaired electrons is called the Spin-Only Magnetic Moment. It is denoted by $\mu$ (mu).
Where:
- $\mu$ = Magnetic moment
- $n$ = Number of unpaired electrons
- B.M. = Bohr Magneton (the standard unit of magnetic moment)
4. Examples & Calculations
| Ion | Configuration | Unpaired Electrons ($n$) | Magnetic Moment ($\mu$) |
|---|---|---|---|
| $Sc^{3+}$ | $[Ar] \ 3d^0$ | 0 | $\sqrt{0(2)} = 0 \text{ B.M.}$ |
| $Ti^{3+}$ | $[Ar] \ 3d^1$ | 1 | $\sqrt{1(3)} = 1.73 \text{ B.M.}$ |
| $Ni^{2+}$ | $[Ar] \ 3d^8$ | 2 | $\sqrt{2(4)} = 2.83 \text{ B.M.}$ |
| $Mn^{2+}$ | $[Ar] \ 3d^5$ | 5 | $\sqrt{5(7)} = 5.92 \text{ B.M.}$ |
Pro-Tip: The calculated value of $\mu$ is always a decimal starting with the number of unpaired electrons (e.g., if $n=4$, $\mu \approx 4.90$ B.M.).
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