Arrange the following metal complex/ compounds in the increasing order of spin only magnetic moment.
Question:

Arrange the following metal complex/ compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system. (Atomic numbers $\mathrm{Ce}=58, \mathrm{Gd}=64$ and $\mathrm{Eu}=63 .)$

(a) $\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{6}\right]$

(b) $\mathrm{Gd}\left(\mathrm{NO}_{3}\right)_{3}$ and

(c) $\mathrm{Eu}\left(\mathrm{NO}_{3}\right)_{3}$

  1. $(b)<(a)<(c)$

  2. (c) $<(a)<(b)$

  3. $(a)<(b)<(c)$

  4. $(a)<(c)<(b)$


Correct Option: , 4

Solution:

(a) $\mathrm{Ce} \rightarrow[\mathrm{Xe}] 4 \mathrm{f}^{2} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{2}$

In complex $\mathrm{Ce}^{4+} \rightarrow[\mathrm{Xe}] 4 \mathrm{f}^{0} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{0}$

there is no unpaired electron so $\mu_{\mathrm{m}}=0$

(b) ${ }_{64} \mathrm{Gd}^{3+} \rightarrow[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{0}$

contain seven unpaired electrons so, $\mu_{\mathrm{m}}=\sqrt{7(7+2)}=\sqrt{63}$ B.M.

(c) ${ }_{63} \mathrm{Eu}^{3+} \rightarrow\left[{ }_{54} \mathrm{Xe}\right] 4 \mathrm{f}^{6} 5 \mathrm{~d}^{0} 6 \mathrm{~s}^{0}$

contain six unpaired electron

so, $\mu_{\mathrm{m}}=\sqrt{6(6+2)}=\sqrt{48}$ B. M

Hence, order of spin only magnetic movement

$\mathrm{b}>\mathrm{c}>\mathrm{a}$

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