JEE Main · 2010 · 11 Apr 2010 · Offline · Coordination Compounds

A solution containing 2.675 g of

CoCl_{3} .6 NH_{3}

(molar mass

= 267.5 g mol^{- 1}

) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of

AgNO_{3}

to give 4.78 g of AgCl (molar mass

= 143.5 g mol^{- 1}

). The formula of the complex is (At. Mass of

Ag = 108 u

)

(A) $[Co (NH_{3})_{6}] Cl_{3}$ ✓ Correct
(B) $[CoCl_{2} (NH_{3})_{4}] Cl$
(C) $[CoCl_{3} (NH_{3})_{3}]$
(D) $[CoCl (NH_{3})_{5}] Cl_{2}$

Answer: (A)

Solution

The complex supplies

\frac{2.675}{267.5} = 0.01

mol. After the cation exchanger only the chlorides that were free in solution, those outside the coordination sphere, are left to precipitate with silver, and they give

\frac{4.78}{143.5} = 0.0333

mol of AgCl. That is three chlorides for every formula unit, so all three are ionic and none is bonded to cobalt. The six ammonia molecules fill the coordination sphere, so the complex is

[Co (N H_{3})_{6}] C l_{3}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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JEE Main · 2010 · 11 Apr 2010 · Offline · Coordination Compounds

Solution

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