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Question
A green chromium compound (A) on fusion with alkali gives a yellow compound (B) which on acidification gives an orange coloured compound (C). ‘C’ on treatment with NH4Cl gives an orange coloured product (D), which on heating decomposes to give back (A). Identify A, B, C and D. Write equations for reactions.
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Solution
The green chromium compound (A) is chromium(III) oxide, Cr2O3.
On fusion with alkali like sodium hydroxide in the presence of air or oxygen, it gives a yellow coloured compound (B), sodium chromate, Na2CrO4.
Acidification of sodium chromate gives an orange coloured compound (C), sodium dichromate, Na2Cr2O7.
This reacts with ammonium chloride to give ammonium dichromate, (NH4)2Cr2O7, which is orange in colour and is the compound (D).
On heating, D decomposes to form green Cr2O3 again, hence regenerating compound A.
A = Cr2O3 (Chromium(III) oxide, green)
B = Na2CrO4 (Sodium chromate, yellow)
C = Na2Cr2O7 (Sodium dichromate, orange)
D = (NH4)2Cr2O7 (Ammonium dichromate, orange)
Formation of B from A (fusion with alkali in the presence of oxygen):
\[\ce{Cr2O3 + 4NaOH + 3O2 -> 2Na2CrO4 + 2H2O}\]
Formation of C from B (acidification of sodium chromate):
\[\ce{2Na2CrO4 + H2SO4 -> Na2Cr2O7 + Na2SO4 + H2O}\]
Formation of D from C (reaction with ammonium chloride):
\[\ce{Na2Cr2O7 + 2NH4Cl -> (NH4)2Cr2O7 + 2NaCl}\]
Decomposition of D to regenerate A:
\[\ce{(NH4)2Cr2O7 ->[Heat] Cr2O3 + N2 + 4H2O}\]
