Consider the following reactions A+NaCl+H2SO4→CrO2Cl2+ Side Products Little amount CrO2Cl2( Vapour )+NaOH→B+NaCl+H2O B+H+→C+H2O The number of terminal ' O ' present in the compound ' C ' is ____.
Answer: 6
Solution
The sequence is K2Cr2O7(A) reacting with NaCl and H2SO4 to give chromyl chloride CrO2Cl2; its vapour with NaOH gives sodium chromate Na2CrO4(B); acidification then gives sodium dichromate, whose anion is Cr2O72−(C). In the dichromate ion two CrO4 tetrahedra share a single bridging oxygen, leaving each chromium with three terminal oxygens. The number of terminal oxygen atoms is 6.Structure of dichromate ion showing terminal oxygen atoms