Question:
Consider the following species:
(i) $\text{ClO}^-$
(ii) $\text{ClO}_2^-$
(iii) $\text{ClO}_3^-$
(iv) $\text{ClO}_4^-$
Among them, the species that undergoes(undergo) disproportionation reactions is(are):
Consider the following species:
(i) $\text{ClO}^-$
(ii) $\text{ClO}_2^-$
(iii) $\text{ClO}_3^-$
(iv) $\text{ClO}_4^-$
Among them, the species that undergoes(undergo) disproportionation reactions is(are):
(i) $\text{ClO}^-$
(ii) $\text{ClO}_2^-$
(iii) $\text{ClO}_3^-$
(iv) $\text{ClO}_4^-$
Among them, the species that undergoes(undergo) disproportionation reactions is(are):
Show Hint
Disproportionation is thermodynamically favorable for intermediate oxidation states. If a species contains an element in its absolute maximum oxidation state, it is exclusively an oxidizing agent and cannot disproportionate.
Updated On: Jun 16, 2026
- (i), (ii), and (iii) only
- (iv) only
- (i) and (ii) only
- (iii) and (iv) only
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks us to identify which of the given chlorine oxyanion species are capable of undergoing disproportionation (a redox reaction where a single species is simultaneously oxidized and reduced).
Step 2: Key Formula or Approach:
For an element in a specific species to undergo disproportionation, it must exist in an intermediate oxidation state.
1. If the element is in its maximum possible oxidation state, it can only be reduced, so it cannot undergo disproportionation.
2. If the element is in its minimum possible oxidation state, it can only be oxidized, so it cannot undergo disproportionation.
Step 3: Detailed Explanation:
• Let us calculate the oxidation state of chlorine (Cl) in each of the given species:
itemize
• (i) In $\text{ClO}^-$: Let the oxidation state of Cl be $x$.
\[ x + (-2) = -1 \implies x = +1 \]
• (ii) In $\text{ClO}_2^-$: Let the oxidation state of Cl be $y$.
\[ y + 2(-2) = -1 \implies y = +3 \]
• (iii) In $\text{ClO}_3^-$: Let the oxidation state of Cl be $z$.
\[ z + 3(-2) = -1 \implies z = +5 \]
• (iv) In $\text{ClO}_4^-$: Let the oxidation state of Cl be $w$.
\[ w + 4(-2) = -1 \implies w = +7 \]
Chlorine is a Group 17 halogen, so its possible oxidation states range from $-1$ to $+7$.
In the perchlorate ion, $\text{ClO}_4^-$, the oxidation state of chlorine is $+7$, which is its maximum possible oxidation state. Since it cannot be oxidized any further, it cannot undergo disproportionation.
In $\text{ClO}^-$, $\text{ClO}_2^-$, and $\text{ClO}_3^-$, the oxidation states ($+1$, $+3$, and $+5$) are intermediate.
Therefore, these species can be both oxidized to higher states (like $+5$ or $+7$) and reduced to lower states (like $-1$), meaning they readily undergo disproportionation reactions.
itemize
Step 4: Final Answer:
Therefore, only species (i), (ii), and (iii) can undergo disproportionation (Option A).
The question asks us to identify which of the given chlorine oxyanion species are capable of undergoing disproportionation (a redox reaction where a single species is simultaneously oxidized and reduced).
Step 2: Key Formula or Approach:
For an element in a specific species to undergo disproportionation, it must exist in an intermediate oxidation state.
1. If the element is in its maximum possible oxidation state, it can only be reduced, so it cannot undergo disproportionation.
2. If the element is in its minimum possible oxidation state, it can only be oxidized, so it cannot undergo disproportionation.
Step 3: Detailed Explanation:
• Let us calculate the oxidation state of chlorine (Cl) in each of the given species:
itemize
• (i) In $\text{ClO}^-$: Let the oxidation state of Cl be $x$.
\[ x + (-2) = -1 \implies x = +1 \]
• (ii) In $\text{ClO}_2^-$: Let the oxidation state of Cl be $y$.
\[ y + 2(-2) = -1 \implies y = +3 \]
• (iii) In $\text{ClO}_3^-$: Let the oxidation state of Cl be $z$.
\[ z + 3(-2) = -1 \implies z = +5 \]
• (iv) In $\text{ClO}_4^-$: Let the oxidation state of Cl be $w$.
\[ w + 4(-2) = -1 \implies w = +7 \]
Chlorine is a Group 17 halogen, so its possible oxidation states range from $-1$ to $+7$.
In the perchlorate ion, $\text{ClO}_4^-$, the oxidation state of chlorine is $+7$, which is its maximum possible oxidation state. Since it cannot be oxidized any further, it cannot undergo disproportionation.
In $\text{ClO}^-$, $\text{ClO}_2^-$, and $\text{ClO}_3^-$, the oxidation states ($+1$, $+3$, and $+5$) are intermediate.
Therefore, these species can be both oxidized to higher states (like $+5$ or $+7$) and reduced to lower states (like $-1$), meaning they readily undergo disproportionation reactions.
itemize
Step 4: Final Answer:
Therefore, only species (i), (ii), and (iii) can undergo disproportionation (Option A).
Was this answer helpful?
0
0
Top NEST Chemistry Questions
- Consider the following reaction: \[ xMnO_4^- + yI^- + zH^+ \rightarrow pMn^{2+} + qIO_3^- + rH_2O \] The correct ratio \(x:y\) in the balanced equation is:
- Negatively charged monodentate strong field ligand \(X^-\) and weak field ligand \(Y^-\) form the complexes \[ [MnX_6]^{4-} \quad \text{and} \quad [MnY_6]^{4-} \] respectively. Under certain reaction conditions, let the crystal field splitting energies for \[ [MnX_6]^{4-} \quad \text{and} \quad [MnY_6]^{4-} \] be \(\Delta_{o1}\) and \(\Delta_{o2}\), respectively. Which of the following statements is/are correct?
Consider the following reaction:
In this context, the correct statement is:
- The reaction that produces butanoic acid as a product is:
- The compound that produces aromatic species upon deprotonation of the indicated hydrogen is:
View More Questions
Top NEST Redox Reactions Questions
- Consider the following reaction: \[ xMnO_4^- + yI^- + zH^+ \rightarrow pMn^{2+} + qIO_3^- + rH_2O \] The correct ratio \(x:y\) in the balanced equation is:
- Two galvanic cells, Cell I and Cell II, operate at the same temperature and involve two different redox systems. It is observed that at a particular value of the reaction quotient $\text{Q}_{\text{I}} = \text{Q}_{\text{II}} = \text{Q}_0$, the ratio of EMF of the two cells ($\text{E}_{\text{I}}/\text{E}_{\text{II}}$) as well as the ratio of change of Gibbs free energy for the two cells ($\Delta\text{G}_{\text{I}}/\Delta\text{G}_{\text{II}}$) are both 1/3. Assuming standard cell reactions written in the spontaneous direction and reversible operation, the pair of cells can satisfy these conditions is:
- $\text{PQ}_2$ is a sparingly soluble salt with solubility product $\text{K}_{sp} = 4 \times 10^{-12}$ in aqueous medium at some given temperature. It is observed that upon addition of a highly soluble salt RQ at the same temperature, the solubility of $\text{PQ}_2$ drops by a factor of 100. The concentration (in millimoles per litre) of added RQ in the solution is closest to:
- Consider the following chemical reactions performed at identical temperature and volume:
Experiment I: $\text{X(g)} \rightleftharpoons 2\text{Y(g)}$
Experiment II: $\text{X(g)} \rightleftharpoons \text{Z(g)}$
The partial pressure of the reactant and product are denoted, respectively, by $p_{\text{reactant}}$ and $p_{\text{product}}$ during the course of the reactions. Assuming ideal gas behaviour, the correct plot of $p_{\text{reactant}}$ versus $p_{\text{product}}$ for Experiments I and II is:
- Consider two reactions.
I. A zero order reaction: $\text{X} \rightarrow P_1$.
II. A first order reaction: $\text{Y} \rightarrow P_2$.
Both reactions begin simultaneously with the same initial concentration `C', that is $\text{X}_0 = \text{Y}_0 = \text{C}$. It is observed that the concentrations of the reactants fall to one-third of their initial values at the same time, provided the rate constants satisfy a certain ratio. If $k_0$ and $k_1$ denote, respectively, the zero and first order rate constants then the ratio $k_0/k_1$ necessary for this condition to be satisfied is:
Top NEST Questions
- In the 2024 Olympics, two runners, Noah Lyles and Kishane Thompson, finished the 100.00 m race in 9.784 s and 9.789 s respectively. Assuming that they move with constant speed throughout the race, the distance between the two runners when Noah Lyles touched the finishing line would be closest to:
- The Balmer series of hydrogenic spectral lines refers to an electron transitioning from \(n\geq 3\) to \(n=2\). A hydrogenic atom with atomic number \(Z=24\) undergoes a Balmer transition of the largest possible wavelength. The emitted photon has energy \(E_0\). Another hydrogenic atom with atomic number \(Z=25\) undergoes a similar Balmer transition with emitted photon energy \(E_1\). Then \(|E_1-E_0|\), in eV, is closest to:
- Three positive and two negative charges each of magnitude \(q\) are placed at five of the six vertices of a regular hexagon of side \(R\). The work done to bring a negative charge \(-q\) from infinity to the centre of the hexagon is:
- Consider a metal cylinder of heat capacity \(C\), thermal conductivity \(K\), mass \(m\) and radius \(r\). The combination that will have the dimension of time is:
- Let \(\theta \neq \dfrac{(2n+1)\pi}{2}\), for \(n\in\mathbb{Z}\). Consider the system of equations \[ x\cos\theta+y\sec\theta=0 \] \[ x\sin\theta+y\tan\theta=0 \] This system does not have a unique solution if and only if \(\theta\) belongs to:
View More Questions