Search any question & find its solution
Question:
Answered & Verified by Expert
The magnetic moment of $\mathrm{Ni}^{2+}$ ion (atomic number of $\mathrm{Ni}=28$ ) in BM unit is
Options:
Solution:
1961 Upvotes
Verified Answer
The correct answer is:
2.84
$\mathrm{Ni}_{28}=[\mathrm{Ar}]_{18} 3 \mathrm{~d}^8 4 \mathrm{~s}^2, \mathrm{Ni}^{2+}=[\mathrm{Ar}]_{18} 3 \mathrm{~d}^8 4 \mathrm{~s}^2$
Number of unpaired electrons $=2$
Magnetic moment,
$\mu=\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}=2.84 \mathrm{BM}$
Number of unpaired electrons $=2$
Magnetic moment,
$\mu=\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}=2.84 \mathrm{BM}$
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.