A compound of a metal ion $\mathrm{M}^{x+}(\mathrm{Z}=24)$ has a spin only magnetic moment of $\sqrt{15}$…

A compound of a metal ion $\mathrm{M}^{x+}(\mathrm{Z}=24)$ has a spin only magnetic moment of $\sqrt{15}$ Bohr Magnetons. The number of unpaired electrons in the compound are
  1. 2
  2. 4
  3. 5
  4. 3

Solution

(d) Magnetic moment $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ where $\mathrm{n}=$ number of unpaired electrons $\sqrt{15}=\sqrt{\mathrm{n}(\mathrm{n}+2)} \therefore \mathrm{n}=3$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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