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 ________.
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 ________.
Show Hint
Correct Answer: 6
Approach Solution - 1
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.
Approach Solution -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
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Inorganic chemistry Questions
- Identity the incorrect pair from the following:
- Among the following halogens \(F_2, \, Cl_2, \, Br_2, \, \text{and} \, I_2\)
Which can undergo disproportionation reaction? - Iron (III) catalyses the reaction between iodide and persulphate ions, in which
A. \(\text{Fe}^{3+}\) oxidises the iodide ion
B. \(\text{Fe}^{3+}\) oxidises the persulphate ion
C. \(\text{Fe}^{2+}\) reduces the iodide ion
D. \(\text{Fe}^{2+}\) reduces the persulphate ion
Choose the most appropriate answer from the options given below: - The difference in the 'spin-only' magnetic moment values of \( \text{KMnO}_4 \) and the manganese product formed during titration of \( \text{KMnO}_4 \) against oxalic acid in acidic medium is ______ BM (nearest integer).
- Identify the inorganic sulphides that are yellow in colour: (A) \( ({NH}_4)_2{S} \)
(B) \( {PbS} \)
(C) \( {CuS} \)
(D) \( {As}_2{S}_3 \)
(E) \( {As}_2{S}_5 \)
Choose the correct answer from the options given below:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,