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 reaction sequence transforms chromyl chloride into a chromate in a basic environment, followed by conversion to dichromate in an acidic environment. The objective is to determine the count of terminal oxygen atoms within the final species, designated as \( C \).

Concept Used:

Chromyl chloride, represented by the formula \( \mathrm{CrO_2Cl_2} \), is synthesized from a dichromate and concentrated acid in the presence of a chloride salt. When subjected to an aqueous basic solution, \( \mathrm{CrO_2Cl_2} \) yields chromate (\( \mathrm{CrO_4^{2-}} \)). Subsequent acidification of chromate leads to the formation of dichromate (\( \mathrm{Cr_2O_7^{2-}} \)) through a condensation process involving one bridging oxygen atom that links two \( \mathrm{CrO_4} \) tetrahedra:

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

Within the dichromate ion, two \( \mathrm{CrO_4} \) tetrahedra are joined by a single bridging oxygen (\(\mathrm{Cr{-}O{-}Cr}\)). The remaining oxygen atoms are classified as terminal oxygens, indicated by double bonds (=O).

Step-by-Step Solution:

Step 1: Identify the reagent \( A \) responsible for producing chromyl chloride when \( \mathrm{NaCl} \) and \( \mathrm{H_2SO_4} \) are present. This reagent is typically 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 produce chromate (\( B \)) and chloride:

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

Consequently, \( B \) corresponds to \( \mathrm{CrO_4^{2-}} \) (in the form of \( \mathrm{Na_2CrO_4} \)).

Step 3: Acidify the chromate to obtain dichromate (\( C \)) through a condensation reaction with the elimination of water:

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

Step 4: Examine the structure of \( \mathrm{Cr_2O_7^{2-}} \) to count the terminal oxygen atoms. The dichromate ion comprises two \( \mathrm{CrO_4} \) tetrahedra linked by a single bridging oxygen (\(\mathrm{Cr{-}O{-}Cr}\)). Of the total seven oxygen atoms, one is bridging; the remaining six are terminal (=O) oxygens, with three terminal oxygens attached to each chromium center:

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

Final Computation & Result

The terminal oxygen count for the final species \( C \), which is \( \mathrm{Cr_2O_7^{2-}} \), is 6, given that it contains one bridging oxygen atom.