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Coordination and Oxidation Number

F eSO 4 solut...

Question

$F e S O_{4}$ solution mixed with ${(N H_{4})}_{2} S O_{4}$ solution in 1:1 molar ratio gives the test of $F e^{2 +}$ ion but $C u S O_{4}$ solution mixed wtih aqueous ammonia in 1:4 molar ratio does not give the test of $C u^{2 +}$ ion. Explain why?

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Solution

Let the oxidation number of metal be x.

When $F e S O_{4}$ and $(N H_{4})_{2} S O_{4}$ solution are mixed in 1:1 molar ratio, Mohr's salt (a double salt) is formed.
$F e S O_{4} (a q) + (N H_{4})_{2} S O_{4} (a q) \to F e S O_{4} (N H_{4})_{2} S O_{4} M o h r s s a l t .6 H_{2} O$
$F e S O_{4} (N H_{4})_{2} S O_{4} 6 H_{2} O_{(a q)} ⇌ F e_{(a q)}^{2 +} + 2 N H_{4 (a q)}^{+} + S O_{4 (a q)}^{2 -} + 6 H_{2} O_{(l)}$
Because $F e^{2 +}$ ions are formed on dissolution of mohr's salt, its aqueous solution gives the test of $F e^{2 +}$ ions.
But when $C u S O_{4}$ (aq) is mixed with $N H_{3}$ (aq), complex formation takes place.
$C u S O_{4 (a q)} + 4 N H_{3 (a q)} \to [C u {(N H_{3})}_{4}] S O_{4 (a q)} T e t r a a m i n e c o p p e r (I I) s u l p h a t e$
This complex does not produce $C u 2 +$ ions on dissolution as Cu has now become a part of ${[C u {(N H_{3})}_{4}]}_{4}^{2 +}$ complex ion, so, the solution of $C u S O_{4}$ and $N H_{3}$ does not give the test of $C u^{2 +}$ ion.

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