Light green (Compound 'A') White Residue (B) C + D + E (i) 'D' and 'E' are two acidic gas. (ii) 'D' is passed through HgCl2 solution to give yellow ppt. (iii) 'E' is passed through water first and the

Question

Light green (Compound 'A') $\xrightarrow{\Delta}$ White Residue (B) 
$\xrightarrow{	ext{High Temp.}}$ C + D + E
(i) 'D' and 'E' are two acidic gas.
(ii) 'D' is passed through HgCl2 solution to give yellow ppt.
(iii) 'E' is passed through water first and then H2S is passed, white turbidity 
is obtained.
(iv) A is water soluble and addition of HgCl2 in it, yellow ppt is obtained but 
white ppt does not turn into grey on addition of excess solution of 'A'. [1]

Prashant Kotian · Accepted Answer

Correct answer: B. Step 1: The conversion between chromate (CrO4^2-) and dichromate (Cr2O7^2-) is pH-dependent. Chromate is stable in basic/neutral conditions, while dichromate is stable in acidic conditions. Step 2: Increasing acidity (lowering pH to an acidic value, x < 7, e.g., 6) converts yellow chromate to orange dichromate: 2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O. Conversely, adding base (raising pH to a basic value, y > 7, e.g., 8) shifts the equilibrium back to chromate. Step 3: Thus, x must be acidic (6) and y must be basic (8), which corresponds to option (a).

💡 Solution & Explanation

Practice 22,000+ questions like this