Question

Consider the following reaction:

\[ A + NaCl + H_2SO_4 \xrightarrow[\text{Little amount}]{} CrO_2Cl_2 + \text{Side products} \] \[ CrO_2Cl_2(\text{Vapour}) + NaOH \rightarrow B + NaCl + H_2O \] \[ B + H^+ \rightarrow C + H_2O \]

The number of terminal 'O' present in the compound 'C' is ________.

Hint:

Identify the chromium-containing species formed in each reaction. Chromyl chloride (\( CrO_2Cl_2 \)) reacts with NaOH to form chromate ions (\( CrO_4^{2-} \)), which in acidic solution convert to dichromate ions (\( Cr_2O_7^{2-} \)). Draw the structure of the dichromate ion to count the terminal oxygen atoms.

Answer 1

Correct Answer: 6

The sequence converts chromyl chloride into a chromate in base and then to dichromate in acid. We are asked to find the number of terminal oxygen atoms in the final species \( C \).

Concept Used:

Chromyl chloride, \( \mathrm{CrO_2Cl_2} \), is formed from a dichromate and concentrated acid in presence of a chloride salt. In aqueous base, \( \mathrm{CrO_2Cl_2} \) gives chromate \( \mathrm{CrO_4^{2-}} \); acidification of chromate yields dichromate \( \mathrm{Cr_2O_7^{2-}} \) via condensation with one bridging oxygen between two \( \mathrm{CrO_4} \) tetrahedra:

\[ 2\,\mathrm{CrO_4^{2-}} + 2\,\mathrm{H^+} \longrightarrow \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O} \]

In the dichromate ion, two \( \mathrm{CrO_4} \) tetrahedra share a single bridging oxygen (\(\mathrm{Cr{-}O{-}Cr}\)), leaving the remaining oxygens as terminal (=O) oxygens.

Step-by-Step Solution:

Step 1: Identify reagent \( A \) that produces chromyl chloride in the presence of \( \mathrm{NaCl} \) and \( \mathrm{H_2SO_4} \). Typically, \( A \) is a dichromate such as \( \mathrm{K_2Cr_2O_7} \) or \( \mathrm{Na_2Cr_2O_7} \):

\[ \mathrm{K_2Cr_2O_7} + 4\,\mathrm{NaCl} + 6\,\mathrm{H_2SO_4} \;\xrightarrow[]{}\; 2\,\mathrm{CrO_2Cl_2}\,(\text{vapour}) + \cdots \]

Step 2: React \( \mathrm{CrO_2Cl_2} \) with aqueous \( \mathrm{NaOH} \) to form chromate (\( B \)) and chloride:

\[ \mathrm{CrO_2Cl_2} + 4\,\mathrm{NaOH} \longrightarrow \mathrm{Na_2CrO_4}\;(B) + 2\,\mathrm{NaCl} + 2\,\mathrm{H_2O} \]

Thus, \( B \equiv \mathrm{CrO_4^{2-}} \) (as \( \mathrm{Na_2CrO_4} \)).

Step 3: Acidify chromate to obtain dichromate (\( C \)) via condensation with loss of water:

\[ 2\,\mathrm{CrO_4^{2-}} + 2\,\mathrm{H^+} \longrightarrow \mathrm{Cr_2O_7^{2-}}\;(C) + \mathrm{H_2O} \]

Step 4: Analyze the structure of \( \mathrm{Cr_2O_7^{2-}} \) to count terminal oxygens. Dichromate consists of two \( \mathrm{CrO_4} \) tetrahedra sharing one bridging oxygen (\(\mathrm{Cr{-}O{-}Cr}\)). Out of the 7 oxygens, 1 is bridging; the remaining 6 are terminal (=O) oxygens (three terminal oxygens on each Cr center):

\[ \text{Terminal O in } \mathrm{Cr_2O_7^{2-}} = 7 - 1 = 6 \]

Final Computation & Result

The final species \( C \) is \( \mathrm{Cr_2O_7^{2-}} \). It contains one bridging oxygen and hence 6 terminal oxygen atoms.

Answer 2

\( Cr_2O_7^{2-} + HCl + H_2SO_4 \rightarrow CrO_2Cl_2 \) 

\( CrO_2Cl_2(vapour) + NaOH \rightarrow Na_2CrO_4 + NaCl + H_2O \)

\( Na_2CrO_4 + H^+ \rightarrow Na_2Cr_2O_7 + H_2O \) 

\( 2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O \) 

\( CrO_4^{2-} \xrightarrow{H^+} Cr_2O_7^{2-} \)

No of terminal "O" = 6