JEE MAIN - Chemistry (2014 (Offline) - No. 17)
The equation which is balanced and represents the correct product(s) is :
[Mg (H2O)6 ]2+ + (EDTA)4- $$\buildrel {excess\,NaOH} \over
\longrightarrow $$ [Mg (EDTA) ]2+ + 6H2O
CuSO4 + KCN $$\to$$ K2 [Cu (CN)4] + K2SO4
Li2O + 2KCl $$\to$$ 2LiCl + K2O
[CoCl(NH3)5]+ + 5H+ $$\to$$ Co2+ + $$5NH_4^+$$ + Cl-
Explanation
(A) [Mg (H2O)6 ]2+ + (EDTA)4- $$\buildrel {excess\,NaOH} \over \longrightarrow $$ [Mg (EDTA) ]2+ + 6H2O
The above equation is incorrect because the product formed would be [Mg (EDTA)]2− .
(B) CuSO4 + KCN $$\to$$ K2[Cu (CN)4] + K2SO4
The above equation is incorrect. Thus, the correct equation is
2KCN + CuSO4 $$\to$$ K2SO4 + Cu + (CN)2(Cyanogen Gas)
(C) Li2O + 2KCl $$\to$$ 2LiCl + K2O
The above equation is incorrect because K2O, a stronger base cannot be generated by a weaker base, Li2O.
(D) [CoCl(NH3)5]+ + 5H+ $$\to$$ Co2+ + $$5NH_4^+$$ + Cl-
The above equation is correct because amine complexes decomposes under acidic medium. Thus, the complex [CoCl(NH3)5] decomposes under acidic conditions to give ammonium ions.
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