Consider the following reactions A + Little amountNaCl + H2SO4 CrO2Cl2 + Side Products CrO2Cl2(Vapour) + NaOH B + NaCl + H2O B + H+ C + H2O The number of terminal 'O' present in the compound 'C' is
Answer
6
Solution
Chromyl chloride test: A is K2Cr2O7.
K2Cr2O7 + 4NaCl + 6H2SO4 2CrO2Cl2 + 2KHSO4 + 4NaHSO4 + 3H2O
CrO2Cl2 + 4NaOH Na2CrO4 + 2NaCl + 2H2O ⇒ B is Na2CrO4 (yellow)
2Na2CrO4 + 2H+ Na2Cr2O7 + 2Na+ + H2O ⇒ C is Na2Cr2O7 (orange)
Structure of the dichromate ion Cr2O72-:
[figure]
Two tetrahedra share one bridging O (Cr–O–Cr); each Cr carries 3 terminal O.
∴ Number of terminal O = 2 3 = 6
Answer: 6