$(a)$ $\left( NH _{4}\right)_{2}\left[ Ce \left( NO _{3}\right)_{6}\right]$
$(b)$ $Gd \left( NO _{3}\right)_{3}$ અને
$(c)$ $Eu \left( NO _{3}\right)_{3}$
In complex \(Ce ^{4+} \rightarrow[ Xe ] 4 f ^{0} 5 d ^{0} 6 s ^{0}\)
there is no unpaired electron so \(\mu_{ m }=0\)
\((b)\) \({ }_{64} Gd ^{3+} \rightarrow[ Xe ] 4 f ^{7} 5 d ^{0} 6 s ^{\circ}\)
contain seven unpaired electrons so, \(\mu_{ m }=\sqrt{7(7+2)}=\sqrt{63} B.M\)
\((c)\) \({ }_{63} Eu ^{3+} \rightarrow\left[{ }_{54} Xe \right] 4 f ^{6} 5 d ^{0} 6 s ^{\circ}\)
contain six unpaired electron
so, \(\mu_{ m }=\sqrt{6(6+2)}=\sqrt{48} B . M\)
Hence, order of spin only magnetic movement
\(b > c > a\)
$\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right]^{2+}, \quad\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-} \text {, }$
$(A)$ $(B)$
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{H}_2 \mathrm{O}\right)\right]^{3+}, \quad\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{2+}$
$(C)$ $(D)$
અવશોષિત પ્રકાશ ની તરંગ સંખ્યા ના સંદર્ભ માં $A, B, C$ અને $D$ નો સાયો ક્રમ શોધો.
(પરમાણુ ક્રમાંક : $Ti = 22, Cr = 24, Co = 27, Zn = 30$)