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The oxidation states of the most electronegative element in the products of the reaction of \(\mathrm{BaO}_{2}\) with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) are (a) 0 and \(-1\) (b) \(-1\) and \(-2\) (c) \(-2\) and 0 (d) \(-2\) and \(+1\)

Short Answer

Expert verified
The correct option is (b) \(-1\) and \(-2\).

Step by step solution

01

Identify the Reactants

The given reaction involves barium peroxide (\(\mathrm{BaO}_{2}\)) reacting with dilute sulphuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)). These are the starting substances for our reaction.
02

Determine Products of Reaction

The products of the reaction between \(\mathrm{BaO}_2\) and \(\mathrm{H}_2\mathrm{SO}_4\) are barium sulfate (\(\mathrm{BaSO}_4\)), hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_2\)), and water. The balanced chemical equation is: \[ \mathrm{BaO}_2 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4 + \mathrm{H}_2\mathrm{O}_2 \]},
03

Assign Oxidation States

The most electronegative element in the products is oxygen. Assign the oxidation states for oxygen in \(\mathrm{BaSO}_4\), \(\mathrm{H}_2\mathrm{O}_2\), and \(\mathrm{H}_2\mathrm{O}\): - In \(\mathrm{BaSO}_4\), the oxidation state of oxygen is \(-2\).- In \(\mathrm{H}_2\mathrm{O}_2\), the oxidation state of oxygen is \(-1\).- In \(\mathrm{H}_2\mathrm{O}\), the oxidation state of oxygen is also \(-2\).
04

Identify the Most Common Oxidation States

From the products generated, identify the most common oxidation states of oxygen. These states are \(-2\) (in water and barium sulfate) and \(-1\) (in hydrogen peroxide).
05

Check Different Given Options

Compare the identified oxidation states \(-2\) and \(-1\) with the provided options:- Option (a) 0 and \(-1\)- Option (b) \(-1\) and \(-2\)- Option (c) \(-2\) and 0- Option (d) \(-2\) and \(+1\)The correct match is option (b) \(-1\) and \(-2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electronegative Elements
Electronegative elements are essential in understanding chemical reactions and oxidation states. Electronegativity is the tendency of an atom to attract electrons in a bond, and it varies across the periodic table. Oxygen, a highly electronegative element, is particularly important in the context of oxidation states.

In the BaOâ‚‚ and dilute Hâ‚‚SOâ‚„ reaction, oxygen is the most electronegative element in the products: BaSOâ‚„, Hâ‚‚Oâ‚‚, and Hâ‚‚O. Because of its high electronegativity, oxygen usually attracts more electrons, giving it a negative oxidation state. Understanding electronegativity helps determine the redistribution of electrons during chemical reactions, indicating which elements get reduced or oxidized.
  • Electronegativity influences the type and character of chemical bonds.
  • Highly electronegative elements often gain electrons, leading to a reduction.
  • The reactivity and type of chemical transformations an element undergoes can also be predicted using electronegativity.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products, often involving electron transfer. The reaction between BaOâ‚‚ and Hâ‚‚SOâ‚„ is an example where new substances are formed: BaSOâ‚„, Hâ‚‚Oâ‚‚, and Hâ‚‚O.

This particular reaction is interesting because it involves the breaking and forming of bonds and rearrangement of atoms.
Understanding the chemical reaction process helps in assigning oxidation states and predicting the behavior of the elements involved.
  • Reactants like BaOâ‚‚ and Hâ‚‚SOâ‚„ react to form new compounds by exchanging components.
  • Studying the products' formation offers insights into how atoms rearrange during reactions.
  • Chemical equations must be balanced to reflect the conservation of mass and charge.
Ensuring equations are balanced enables accurate predictions about the substances' behavior.
Oxidation and Reduction
Oxidation and reduction are two essential processes in chemical reactions involving electron transfer.

Oxidation refers to the loss of electrons, leading to an increase in oxidation state, while reduction involves gaining electrons, decreasing the oxidation state. In our exercise, identifying which elements are oxidized or reduced helps us understand the reaction better. Oxygen, being the most electronegative element, often undergoes reduction by gaining electrons.
  • Oxidation and reduction always occur simultaneously, termed as redox reactions.
  • Identifying oxidizing and reducing agents in reactions is crucial to understanding their complete dynamics.
  • The oxidation states give insights into how electrons are transferred between reactants.
The reaction between BaOâ‚‚ and Hâ‚‚SOâ‚„ showcases redox changes, with oxygen displaying noteworthy variations in oxidation states.
Assigning Oxidation Numbers
Assigning oxidation numbers is an important skill in understanding chemical reactions and redox processes.

Oxidation numbers help track the transfer of electrons and identify which elements are oxidized or reduced during a reaction.
For example, in our reaction's products, assigning oxygen an oxidation state of (-2) in BaSOâ‚„ and Hâ‚‚O indicates stability and complete electron gain, while (-1) in Hâ‚‚Oâ‚‚ suggests an incomplete gain.
  • Assignment rules generally apply: free elements have oxidation states of zero and oxygen often carries (-2) except in peroxides.
  • The sum of oxidation numbers in a neutral compound must equal zero.
  • Using consistent criteria for assigning values supports accurate analysis and predictions.
Practicing the assignment of oxidation numbers allows for a deeper understanding of chemical behavior and transformations in reactions.

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Most popular questions from this chapter

\(. \mathrm{Ag}\left|\mathrm{Ag}^{+}(\mathrm{IM}) \| \mathrm{Ag}^{+}(2 \mathrm{M})\right| \mathrm{Ag}\) 1 L solution 1 L solution \(0.5 \mathrm{~F}\) of electricity in the LHS (anode) the \(1 \mathrm{~F}\) electricity in the RHS (cathode) is first passed making them independent electro cells at \(298 \mathrm{~K}\). The emf of the cell after electrolysis will (a) increase (b) decrease (c) not change (d) time is also required

The equivalent conductances of two strong electrolytes at infinite dilution in \(\mathrm{H}_{2} \mathrm{O}\) (where ions move freely through a solution) at \(25^{\circ} \mathrm{C}\) are given below: [2007] \(\Lambda^{\circ}\left(\mathrm{CH}_{3} \mathrm{COONa}\right)=91.0 \mathrm{~S} \mathrm{~cm}^{2} /\) equiv. \(\Lambda^{\circ}(\mathrm{HCl})=426.2 \mathrm{~S} \mathrm{~cm}^{2} /\) equiv. What additional information/quantity one needs to calculate \(\Lambda^{\circ}\) of an aqueous solution of acetic acid? (a) \(\Lambda^{\circ}\) of \(\mathrm{CH}_{3} \mathrm{COOK}\) (b) The limiting equivalent conductance of \(\mathrm{H}^{+}\left(\lambda^{\circ}\right)\) (c) \(\Lambda^{\circ}\) of chloroacetic acid \(\left(\mathrm{ClCH}_{2} \mathrm{COOH}\right)\) (d) \(\Lambda^{\circ}\) of \(\mathrm{NaCl}\)

The reduction potential of hydrogen half-cell will be negative if: (a) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \mathrm{~atm}\) and \(\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}\) (b) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=1 \mathrm{~atm}\) and \(\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}\) (c) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=1 \mathrm{~atm}\) and \(\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) (d) \(\mathrm{p}\left(\mathrm{H}_{2}\right)=2\) atm and \(\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\)

Three faraday of electricity is passed through aqueous solutions of \(\mathrm{AgNO}_{3}, \mathrm{NiSO}_{4}\) and \(\mathrm{CrCl}_{3}\) kept in three vessels using inert electrodes. The ratio in moles in which the metals \(\mathrm{Ag}, \mathrm{Ni}\) and \(\mathrm{Cr}\) will be deposited is (a) \(1: 2: 3\) (b) \(2: 3: 6\) (c) \(6: 3: 2\) (d) \(3: 2: 6\)

If the aqueous solutions of the following salts are electrolysed for 1 hour with 10 ampere current, which solution will deposit the maximum mass of the metal t the cathode? The atomic weights are \(\mathrm{Fe}=56, \mathrm{Zn}=\) \(65, \mathrm{Ag}=108, \mathrm{Hf}=178\) and \(\mathrm{W}=184\) (a) \(\mathrm{ZnSO}_{4}\) (b) \(\mathrm{FeCl}_{3}\) (c) \(\mathrm{HfCl}_{4}\) (d) \(\mathrm{AgNO}_{3}\)

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