QJEEM24Chf669: JEE Main 2024 Chemistry - Class 12 JEEM24Chf669 Solution: Which of the following electronic configuration would be…

The magnetic moment of an atom or ion with unpaired electrons is given by the formula $$\mu =\sqrt{n(n+2)}$$ Bohr magnetons (BM), where $$n$$ is the number of unpaired electrons. The magnetic moment depends on the number of unpaired electrons: more unpaired electrons lead to a higher magnetic moment.

Looking at the given options with respect to their electron configurations and calculating their respective magnetic moments based on their unpaired electrons:

• $$[\mathrm{Ar}]3{\mathrm{d}}^7$$ has 3 unpaired electrons, yielding a magnetic moment of $$\sqrt{15}$$ BM.
• $$[\mathrm{Ar}]3{\mathrm{d}}^8$$ has 2 unpaired electrons, for a magnetic moment of $$\sqrt{8}$$ BM.
• $$[\mathrm{Ar}]3{\mathrm{d}}^3$$ also has 3 unpaired electrons, similar to $$3{\mathrm{d}}^7$$, so its magnetic moment is likewise $$\sqrt{15}$$ BM.
• Finally, $$[\mathrm{Ar}]3{\mathrm{d}}^6$$ has 4 unpaired electrons, resulting in the highest magnetic moment among the options, $$\sqrt{24}$$ BM.

Hence, the electron configuration associated with the highest magnetic moment is $$[\mathrm{Ar}]3{\mathrm{d}}^6$$.