A complex is composed of one chromium ion, three bromides and six water molecules. Upon addition of excess… · ABC26IN0398

Subject: Inorganic Chemistry · Chapter: Coordination Chemistry · Topic: Coordination – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

A complex is composed of one chromium ion, three bromides and six water molecules. Upon addition of excess $\mathrm{AgNO}_{3}$, 1.0 g aqueous solution of the complex gave 0.94 g of AgBr. The molecular formula of the complex is (Atomic weight: $\mathrm{Cr}=52$, $\mathrm{Br}=80$, $\mathrm{Ag}=108$, $\mathrm{O}=16$ and $\mathrm{H}=1$)
(a)$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{3}$
(b)$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Br}\right] \mathrm{Br}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$
(c)$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br} \cdot 2 \mathrm{H}_{2} \mathrm{O}$
(d)$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Br}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$
Answer
Answer: B ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
Complex M = 52 + 3×80 + 6×18 = 400 g mol⁻¹, so 1.0 g = 2.5×10⁻³ mol. AgBr formed = 0.94/188 = 5.0×10⁻³ mol = 2 Br⁻ per complex, so two ionic bromides: [Cr(H₂O)₅Br]Br₂·H₂O.

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