Test: Redox Reactions (October 7) - NEET MCQ

Explanation of Electron Releasing Tendency of Zinc, Copper, and Silver:

The electron releasing tendency of an element is determined largely by its ionization energy, that is, the energy required to remove an electron from an atom. The lower the ionization energy, the higher the electron releasing tendency of the element.

• Zinc (Zn): Zinc has the lowest ionization energy among the three metals. This is because it has a fully filled d orbital (3d¹⁰ 4s² configuration), and the removal of an electron would lead to a more stable electronic configuration (3d¹⁰ 4s¹).
• Copper (Cu): Copper has a higher ionization energy than zinc but lower than silver. Although copper has a partially filled d orbital (3d¹⁰ 4s¹ configuration), the removal of an electron would still lead to a more stable electronic configuration (3d⁹ 4s¹). However, the energy required to do so is higher than that for zinc, which results in a lower electron releasing tendency.
• Silver (Ag): Silver has the highest ionization energy among the three metals. It has a fully filled d orbital (4d10 5s1 configuration), and the removal of an electron would disrupt this stable configuration, thus requiring a lot of energy. This results in the lowest electron releasing tendency among the three metals.

Conclusion:
Based on the ionization energies, the electron releasing tendency of the metals, zinc, copper and silver, is in the order Zn > Cu > Ag, which correspond to option C.

Oxidation Number of Hydrogen in LiH, NaH and CaH₂
The oxidation number, or oxidation state, indicates the degree of oxidation (loss of electrons) of an atom in a chemical compound. In terms of simple ions, the oxidation number can be equal to the charge on the ion.

• In LiH, NaH, and CaH₂, the hydrogen is attached to a metal. In such cases, hydrogen generally exhibits an oxidation number of -1. This is because metals are more electropositive than hydrogen, so they donate electrons to the hydrogen atom.
• For LiH and NaH, each hydrogen atom is linked to one lithium or sodium atom. So, the oxidation number of hydrogen in these compounds is -1.
• In CaH₂, each hydrogen atom is linked to one calcium atom. However, as there are two hydrogen atoms, the total oxidation number for hydrogen is -2. But the oxidation number of each hydrogen atom remains -1.

In conclusion, the oxidation number of hydrogen in LiH, NaH, and CaH₂ is -1. Therefore, the answer is D: -1.0

Concept of Reducing Agent:
A reducing agent is a substance that loses or "donates" an electron to another substance in a redox chemical reaction. Therefore, a good reducing agent is the one that gets oxidized easily, or in other words, the one that can easily lose electrons.

Characteristics of a Good Reducing Agent:

• Electron Loss: A better reducing agent is the one that loses more electrons. This is because by losing electrons, the reducing agent gets oxidized and in turn reduces the other substance. This is the basic principle of a redox reaction.
• Reactivity: The reactivity of the metal also determines its capacity as a reducing agent. Metals that are high in the reactivity series are good reducing agents. This is because they can easily lose electrons and get oxidized.
• Stability: Metals that are less stable are better reducing agents because they can easily lose electrons to attain a stable state.

Hence, Option A is the correct answer - a better reducing agent is the one that loses more electrons.

• Oxidation Numbers of Halogens: The halogens (Fluorine, Chlorine, Bromine, Iodine, and Astatine) are a group in the periodic table. They are known for their high electronegativity and hence, when they combine with almost all other elements, they tend to have an oxidation number of -1 as they gain one electron to achieve a stable electronic configuration.
• Exception in the case of Oxygen: However, oxygen is more electronegative than all halogens except fluorine. So, when halogens (chlorine, bromine, and iodine) combine with oxygen, they tend to lose electrons to oxygen and hence, they show positive oxidation states.
• Conclusion: Therefore, when chlorine, bromine, and iodine are combined with oxygen, they have an oxidation number of +1 or any positive number (depending on the number of oxygen atoms they are combined with). Hence, the correct answer is C: +1 or any positive number.

Preparation of Hydrogen from H₂O with Ca (Calcium)

• Hydrogen can be prepared from water (H₂O) through a chemical reaction with calcium (Ca).
• In this reaction, calcium acts as a reducing agent. This is because calcium has the ability to donate electrons during the reaction, which reduces the other reactant, in this case, water.
• The overall chemical reaction that occurs is: Ca + 2H₂O → Ca(OH)₂ + H₂
• In this reaction, calcium (Ca) reacts with water (H₂O) to form calcium hydroxide (Ca(OH)₂) and hydrogen gas (H₂).
• The hydrogen gas that is produced can be collected and used for various purposes.
• Thus, out of the given options, it is calcium that can act as a reducing agent to prepare hydrogen from water.

Why Other Options are Incorrect

• Ag (Silver) and Au (Gold) are noble metals and do not readily participate in chemical reactions. They cannot act as reducing agents to produce hydrogen from water.
• Al (Aluminium) can act as a reducing agent, but it does not react with water under normal conditions to produce hydrogen.

Conclusion

• Therefore, the correct answer is option B: 'Ca, which acts as reducing agent'.

as oxygen is more electronegative than Cl,Br and I. So they have positive oxidation state.

Explanation of Highest Value of Oxidation Number Changes from 1 to 7

The highest value of oxidation number changes from 1 to 7 across the third period in the periodic table. This is due to the following reasons:

• The Atoms of Transition Elements:Transition metals are those elements located in the d-block of the periodic table. They have varying oxidation states, but they do not usually reach an oxidation state of 7. Their oxidation states primarily range between +2 and +3, although some can reach states of +4 or +5.
• The First Three Groups:The first three groups of the periodic table include alkali metals, alkaline earth metals, and boron group elements. These groups generally have oxidation states of +1, +2, and +3 respectively. They do not reach an oxidation state of 7.
• In Alkaline Earth Metals:Alkaline earth metals are the elements in the second group of the periodic table. These elements generally have an oxidation state of +2 due to the presence of two valence electrons which are readily lost in chemical reactions.
• Across the Third Period in the Periodic Table:The elements in the third period of the periodic table show a wider range of oxidation states, which can vary from +1 to +7. This is due to the presence of both s and p orbitals in their valence shell, which allows the elements to lose or gain more electrons in chemical reactions. Therefore, across the third period in the periodic table, the highest value of oxidation number changes from 1 to 7.

The correct option is A Cu>Ag<Au
As we move from the first to the second transition series, ionization energy decreases due to the increase in the number of shells. However, as we move from the second to the third transition series, due to the poor shielding of f electrons, ionization energy increases. Thus the ionisation energies of the coinage metals follow the order Cu>Ag<Au

Oxidation Number of Oxygen

• Oxygen is a highly electronegative element with a strong tendency to attract shared electrons towards itself. It almost always has an oxidation number of -2 in its compounds, except in peroxides like H₂O₂ or Na₂O₂ where it has an oxidation number of -1.
• The oxidation number of oxygen in its elemental form (O₂) is 0.
• The reason why the oxidation number of oxygen is usually -2 is due to its electron configuration. Oxygen has 6 valence electrons and needs 2 more electrons to achieve the stable electron configuration of the nearest noble gas, neon. Therefore, in most of its compounds, oxygen gains 2 electrons, giving it an oxidation number of -2.
• So, the correct answer to the question is D: -2 because the oxidation state of oxygen in most of its compounds is -2. This is due to the fact that oxygen is highly electronegative and tends to gain 2 electrons to achieve a stable electron configuration.

Zn is oxidized to Zn⁺² while H+ is reduced to H₂. So H⁺ is acting as oxidant and Zn is acting as reductant.