Search any question & find its solution
Question:
Answered & Verified by Expert
The "spin only" magnetic moment of $\mathrm{Ni}^{2+}$ in aqueous solution would be [At. no. of $\mathrm{Ni}=28$ ]
Options:
Solution:
2379 Upvotes
Verified Answer
The correct answer is:
$\sqrt{8} \mathrm{BM}$
$\mathrm{Ni}^{2+}$ in aqueous solution refers to
${\left.\left[\mathrm{Ni}_{2} \mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot \mathrm{Ni}^{2+}$ has following electronic $}
$ orientation.
$\therefore$ Number of unpaired electrons $=2$
Spin only magnetic moment $=\sqrt{n(n+2)}$
$$
=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM}
$$
${\left.\left[\mathrm{Ni}_{2} \mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot \mathrm{Ni}^{2+}$ has following electronic $}
$ orientation.
$\therefore$ Number of unpaired electrons $=2$
Spin only magnetic moment $=\sqrt{n(n+2)}$
$$
=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.