Coordination Compounds - Result Question 41
####43. Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code : [2018]
Column I
a. $Co^{3+}$
b. $Cr^{3+}$
c. $Fe^{3+}$
d. $Ni^{2+}$ Column II i. $\sqrt{8} BM$
ii. $\sqrt{35} BM$
iii. $\sqrt{3} BM$
iv. $\sqrt{24} BM$ v. $\sqrt{15} BM$
\begin{tabular}{lllll} & a & b & c & d \\
(a) & iv & v & ii & i \\
(b) & i & ii & iii & iv \\
(c) & iii & v & i & ii \\
(d) & iv & i & ii & iii \end{tabular}
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Solution:
- (a) $Co^{3+}=[Ar] 3 d^{6}$, unpaired $e^{-}(n)=4$
Spin magnetic moment
$=\sqrt{4(4+2)}=\sqrt{24}$ B.M.
$Cr^{3+}=[Ar] 3 d^{3}$, unpaired $e^{-}(n)=3$
Spin magnetic moment
$=\sqrt{3(3+2)}=\sqrt{15}$ B.M.
$Fe^{3+}=[Ar] 3 d^{5}$, unpaired $e^{-}(n)=5$
Spin magnetic moment
$=\sqrt{5(5+2)}=\sqrt{35}$ B.M.
$Ni^{2+}=[Ar] 3 d^{8}$, unpaired $e^{-}(n)=2$
Spin magnetic moment
$=\sqrt{2(2+2)}=\sqrt{8}$ B.M.
44 (b) In a solution containing $HgCl_2, I_2$ and $I^{-}$, both $HgCl_2$ and $I_2$ compete for $I^{-}$.
Since formation constant of $[HgI_4]^{2-}$ is very large $(1.9 \times 10^{30})$ as compared with $\Gamma_3$ $(K_f=700)$.
$\therefore \quad I^{-}$will preferentially combine with $HgCl_2$.
$ \begin{gathered} HgCl_2+2 I^{-} \to HgI_2 \downarrow+2 Cl^{-} \\ Red ppt^{HgI_2+2 I^{-} \to[HgI_4]^{2-}} \\ \text{ soluble } \end{gathered} $
