SATHEE: Coordination Compounds 2 Question 5

Coordination Compounds 2 Question 5

####5. The incorrect statement is

(2019 Main, 10 April II)

(a) the gemstone, ruby, has $C r^{3 +}$ ions occupying the octahedral sites of beryl

(b) the color of ${[C o C l {(N H_{3})}_{5}]}^{2 +}$ is violet as it absorbs the yellow light

(c) the spin only magnetic moments of ${F e {(H_{2} O)}_{6}]}^{2 +}$ and ${[C r {(H_{2} O)}_{6}]}^{2 +}$ are nearly similar

(d) the spin only magnetic moment of ${[N i {(N H_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}$ is $2.83 B M$

Show Answer

Solution:

The explanation of given statements are as follows :

(a) Ruby, a pink or blood-red coloured gemstone belongs to corundum $(A l_{2} O_{3}$ , alumina) system which has trigonal crystalline lattice containing the repeating unit of $A l_{2} O_{3} - C r^{3 +}$ . So, ruby does not belong to beryl lattice $(B e_{3} A l_{2} S i_{6} O_{18})$ .

Thus, statement (a) is incorrect. (b) ${[C o (C l) {(N H_{3})}_{5}]}^{2 +}$ is a low spin octahedral complex of $C o^{3 +}$ . It absorbs low energy yellow light and high energy complementary violet light will be shown off. Thus, statement (b) is correct.

(c) ${[F e {(H_{2} O)}_{6}]}^{2 +}$ and ${[C r {(H_{2} O)}_{6}]}^{2 +}$ are the high-spin octahedral complexes of $F e^{2 +} (3 d^{6}, n = 4)$ and $C r^{2 +} (3 d^{5}, n = 5)$ ions and weak field ligand, $H_{2} O$ respectively. So, spin-only magnetic moment $= \sqrt{n (n + 2)}$ of the complexes.

$\begin{aligned} {[F e {(H_{2} O)}_{6}]}^{2 +}, & μ_{1} & = \sqrt{4 (4 + 2)} (n = 4), & = \sqrt{24} = 4.89 B M {[C r {(H_{2} O)}_{6}]}^{2 +}, & μ_{2} & = \sqrt{5 (5 + 2)} (n = 5), & = \sqrt{35} = 5.92 B M \end{aligned}$

So, $μ_{1} \approx μ_{2}$ . Thus, statement (c) is correct.

(d) ${[N i {(N H_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}$ is also a high-spin octahedral complex of $N i^{2 +} (3 d^{8}, n = 2)$

$μ = \sqrt{2 (2 + 2)} = \sqrt{8} = 2.83 B M$

Thus, statement $(d)$ is correct.

Key Idea The wavelength $(λ)$ of light absorbed by the complexes is inversely proportional to its $Δ_{0}$ CFSE

(magnitude). $Δ_{0}$ (CFSE) $\propto 1 / λ$

The complexes can be written as: I. ${[C o C l {(N H_{3})}_{5}]}^{2 +} \equiv {[C o {(N H_{3})}_{5} (C l)]}^{2 +}]$

II. ${[C o {[N H_{3}]}_{5} H_{2} O]}^{3 +} \equiv {[C o {(N H_{3})}_{5} (H_{2} O)]}^{3 +}$

III. ${[C o {(N H_{3})}_{5}]}^{3 +} \equiv {[C o {(N H_{3})}_{5} (N H_{3})]}^{3 +}$

So, the differentiating ligands in the octahedral complexes of $C o$ (III) in I, II and III are $C l^{⊖}, H_{2} O$ and $N H_{3}$ respectively. In the spectrochemical series, the order of this power for crystal field splitting is $C l^{-} < H_{2} O < N H_{3}$ .