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Observe the following ions
$$
\mathrm{V}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Ti}^{3+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}, \mathrm{Ni}^{3+}, \mathrm{Co}^{3+}, \mathrm{Cu}^{+}
$$
How many ions in the above list have zero magnetic moment?
Options:
$$
\mathrm{V}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Ti}^{3+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}, \mathrm{Ni}^{3+}, \mathrm{Co}^{3+}, \mathrm{Cu}^{+}
$$
How many ions in the above list have zero magnetic moment?
Solution:
2067 Upvotes
Verified Answer
The correct answer is:
4
$\mathrm{V}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^3 ; \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^9 ; \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ $\mathrm{Zn}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{10} ; \mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5 ; \mathrm{Ti}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^1$ $\mathrm{Ni}^{+}=[\mathrm{Ar}] 3 \mathrm{~d}^7 ; \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6 ; \mathrm{Cu}^{+}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}$ Thus, $\mathrm{Zn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Co}^{3+}$ and $\mathrm{Cu}^{+}$have zero magnetic moment, due to even number of electrons that can get paired in a complex.
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