Complex A has a composition of $\mathrm{H}_{12} \mathrm{O}_{6} \mathrm{Cl}_{3} \mathrm{Cr}$. If the complex on treatment with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ loses $13.5 \%$ of its original mass, the correct molecular formula of A is: [Given : atomic mass of $\mathrm{Cr}=52 \mathrm{amu}$ and $\mathrm{Cl}=35 \mathrm{amu}$]

(A) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .2 \mathrm{H}_{2} \mathrm{O}$

(B) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$

(C) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

(D) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Coordination Compounds

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Complex A has a composition of

H_{12} O_{6} Cl_{3} Cr

. If the complex on treatment with conc.

H_{2} SO_{4}

loses

13.5%

of its original mass, the correct molecular formula of A is:
[Given : atomic mass of

Cr = 52 amu

and

Cl = 35 amu

]

(A) $[Cr (H_{2} O)_{4} Cl_{2}] Cl .2 H_{2} O$ ✓ Correct
(B) $[Cr (H_{2} O)_{6}] Cl_{3}$
(C) $[Cr (H_{2} O)_{5} Cl] Cl_{2} \cdot H_{2} O$
(D) $[Cr (H_{2} O)_{3} Cl_{3}] \cdot 3 H_{2} O$

Answer: (A)

Solution

With atomic masses

C r = 52

and

C l = 35

, the formula unit

H_{12} O_{6} C l_{3} C r

has mass

12 + 96 + 105 + 52 = 265

. Concentrated sulfuric acid is a dehydrating agent that strips off only the water held outside the coordination sphere, the water of crystallisation. The mass lost is

13.5%

265 \approx 36

, which is

36/18 = 2

molecules of water. With two water molecules lying outside, the complex must be

[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O

, carrying four waters and two chlorides inside the coordination sphere.

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Coordination Compounds

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