Coordination Compounds 2 Question 5
####5. The incorrect statement is
(2019 Main, 10 April II)
(a) the gemstone, ruby, has $Cr^{3+}$ ions occupying the octahedral sites of beryl
(b) the color of $\left[CoCl\left(NH _3\right) _5\right]^{2+}$ is violet as it absorbs the yellow light
(c) the spin only magnetic moments of $\left.Fe\left(H _2 O\right) _6\right]^{2+}$ and $\left[Cr\left(H _2 O\right) _6\right]^{2+}$ are nearly similar
(d) the spin only magnetic moment of $\left[Ni\left(NH _3\right) _4\left(H _2 O\right) _2\right]^{2+}$ is $2.83 BM$
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Solution:
- The explanation of given statements are as follows :
(a) Ruby, a pink or blood-red coloured gemstone belongs to corundum $\left(Al _2 O _3\right.$, alumina) system which has trigonal crystalline lattice containing the repeating unit of $Al _2 O _3-Cr^{3+}$. So, ruby does not belong to beryl lattice $\left(Be _3 Al _2 Si _6 O _{18}\right)$.
Thus, statement (a) is incorrect. (b) $\left[Co(Cl)\left(NH _3\right) _5\right]^{2+}$ is a low spin octahedral complex of $Co^{3+}$. It absorbs low energy yellow light and high energy complementary violet light will be shown off. Thus, statement (b) is correct.
(c) $\left[Fe\left(H _2 O\right) _6\right]^{2+}$ and $\left[Cr\left(H _2 O\right) _6\right]^{2+}$ are the high-spin octahedral complexes of $Fe^{2+}\left(3 d^{6}, n=4\right)$ and $Cr^{2+}\left(3 d^{5}, n=5\right)$ ions and weak field ligand, $H _2 O$ respectively. So, spin-only magnetic moment $=\sqrt{n(n+2)}$ of the complexes.
$$ \begin{aligned} {\left[Fe\left(H _2 O\right) _6\right]^{2+}, } & \mu _1 & =\sqrt{4(4+2)} \ \quad(n=4), & & =\sqrt{24}=4.89 BM \ {\left[Cr\left(H _2 O\right) _6\right]^{2+}, } & \mu _2 & =\sqrt{5(5+2)} \ \quad(n=5), & & =\sqrt{35}=5.92 BM \end{aligned} $$
So, $\mu _1 \approx \mu _2$. Thus, statement (c) is correct.
(d) $\left[Ni\left(NH _3\right) _4\left(H _2 O\right) _2\right]^{2+}$ is also a high-spin octahedral complex of $Ni^{2+}\left(3 d^{8}, n=2\right)$
$$ \mu=\sqrt{2(2+2)}=\sqrt{8}=2.83 BM $$
Thus, statement $(d)$ is correct.
Key Idea The wavelength $(\lambda)$ of light absorbed by the complexes is inversely proportional to its $\Delta _0$ CFSE
(magnitude). $\Delta _0$ (CFSE) $\propto 1 / \lambda$
The complexes can be written as: I. $\left.\left[CoCl\left(NH _3\right) _5\right]^{2+} \equiv\left[Co\left(NH _3\right) _5(Cl)\right]^{2+}\right]$
II. $\left[Co\left[NH _3\right] _5 H _2 O\right]^{3+} \equiv\left[Co\left(NH _3\right) _5\left(H _2 O\right)\right]^{3+}$
III. $\left[Co\left(NH _3\right) _5\right]^{3+} \equiv\left[Co\left(NH _3\right) _5\left(NH _3\right)\right]^{3+}$
So, the differentiating ligands in the octahedral complexes of $Co$ (III) in I, II and III are $Cl^{\ominus}, H _2 O$ and $NH _3$ respectively. In the spectrochemical series, the order of this power for crystal field splitting is $Cl^{-}<H _2 O<NH _3$.
So, the crystal field splitting energy (magnitude) order will be
$$ \Delta _0^{CFSE}(I)<\Delta _0^{CFSE} \text { (II) }<\Delta _0^{CFSE} \text { (III) } $$
and the order of wavelength $(\lambda)$ of light absorbed by the complexes will be
$$ \lambda(I)>\lambda(II)>\lambda(III) $$
$$ \left[\because \operatorname{Energy}\left(\Delta _0^{CFSE}\right) \propto \frac{1}{\lambda}\right] $$
