The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by \(\mathrm{NiCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}\) to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of \(\mathrm{NiCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}\) are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is ____. (Atomic weights in \(\mathrm{g} \mathrm{mol}^{-1}: \mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{Cl}=35.5, \mathrm{Ca}=40, \mathrm{Ni}=59\))

d and f Block Elements — JEE Advanced · 2018 · 20 May 2018 · Paper 1 Question with Solution

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by

NiCl_{2} .6 H_{2} O

to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of

NiCl_{2} .6 H_{2} O

are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is ____.

(Atomic weights in

g mol^{- 1} : H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59

)

Answer: 2992

Solution

Step 1: Reaction of ammonium sulphate with calcium hydroxide:

(NH_{4})_{2} SO_{4} + Ca (OH)_{2} \to 2 NH_{3} + CaSO_{4} \cdot 2 H_{2} O

Molar mass of

(NH_{4})_{2} SO_{4}

= 132 g/mol
Moles = 1584/132 = 12 mol

This produces 24 mol of

NH_{3}

and 12 mol of gypsum

CaSO_{4} \cdot 2 H_{2} O

.

Mass of gypsum = 12 × 172 = 2064 g

Step 2: Formation of nickel-ammonia complex:

NiCl_{2} \cdot 6 H_{2} O + 6 NH_{3} \to [Ni (NH_{3})_{6}] Cl_{2} + 6 H_{2} O

Molar mass of

NiCl_{2} \cdot 6 H_{2} O

= 238 g/mol
Moles = 952/238 = 4 mol

This requires 24 mol of

NH_{3}

(exactly what we have).

Molar mass of

[Ni (NH_{3})_{6}] Cl_{2}

= 59 + 6(17) + 2(35.5) = 232 g/mol
Mass = 4 × 232 = 928 g

Total mass = 2064 + 928 = 2992 g

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer

LIST-I	LIST-II
(I) $[Cr (CN)_{6}]^{4 -}$	(P) $t_{2 g}$ orbitals contain 4 electrons

LIST-I

LIST-II

(I)

[Cr (CN)_{6}]^{4 -}

(P)

t_{2 g}

orbitals contain 4 electrons

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · d and f Block Elements

Solution

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