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The correct statement(s) about Cr2+ and Mn3+ is(are)
[Atomic numbers of Cr = 24 and Mn = 25]
[Atomic numbers of Cr = 24 and Mn = 25]
- a)Cr2+ is a reducing agent
- b)Mn3+ is an oxidizing agent
- c)Both Cr2+ and Mn3+ exhibit d 4 electronic configuration
- d)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configuration
Correct answer is option 'A,B,C'. Can you explain this answer?
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The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of...
(1) Cr2+ is a reducing agent because Cr3+ is more stable.
(2) Mn3+ is an oxidizing agent because Mn2+ is more stable.
(3) Cr2+ and Mn3+ exhibit d4 electronic configuration.
(2) Mn3+ is an oxidizing agent because Mn2+ is more stable.
(3) Cr2+ and Mn3+ exhibit d4 electronic configuration.
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The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of...
Cr2 is a reducing agent:
- A reducing agent is a substance that donates electrons and undergoes oxidation in a chemical reaction.
- In the case of Cr2, the chromium atoms in the compound have a lower oxidation state compared to their elemental form (Cr^0).
- This means that Cr2 can donate electrons to another substance, causing it to be reduced.
- Therefore, Cr2 can act as a reducing agent.
Mn3 is an oxidizing agent:
- An oxidizing agent is a substance that accepts electrons and undergoes reduction in a chemical reaction.
- In the case of Mn3, the manganese atoms in the compound have a higher oxidation state compared to their elemental form (Mn^0).
- This means that Mn3 can accept electrons from another substance, causing it to be oxidized.
- Therefore, Mn3 can act as an oxidizing agent.
Both Cr2 and Mn3 exhibit d4 electronic configuration:
- The electronic configuration of Cr is [Ar] 3d^54s^1, and the electronic configuration of Mn is [Ar] 3d^54s^2.
- Both Cr2+ and Mn3+ ions lose their outermost s electrons, resulting in a d^4 electronic configuration.
- This is because the 4s orbital is higher in energy than the 3d orbitals, so it is preferentially lost when forming ions.
- Therefore, both Cr2 and Mn3 exhibit a d^4 electronic configuration.
When Cr2 is used as a reducing agent, the chromium ion attains d5 electronic configuration:
- When Cr2 acts as a reducing agent, it donates electrons to another substance.
- In the process, one of the chromium atoms in Cr2 is oxidized from a +2 oxidation state to a +3 oxidation state.
- This oxidation results in the formation of a Cr3+ ion, which has a d^5 electronic configuration.
- Therefore, when Cr2 is used as a reducing agent, the chromium ion attains a d^5 electronic configuration.
In summary, the correct statements about Cr2 and Mn3 are:
a) Cr2 is a reducing agent.
b) Mn3 is an oxidizing agent.
c) Both Cr2 and Mn3 exhibit a d^4 electronic configuration.
- A reducing agent is a substance that donates electrons and undergoes oxidation in a chemical reaction.
- In the case of Cr2, the chromium atoms in the compound have a lower oxidation state compared to their elemental form (Cr^0).
- This means that Cr2 can donate electrons to another substance, causing it to be reduced.
- Therefore, Cr2 can act as a reducing agent.
Mn3 is an oxidizing agent:
- An oxidizing agent is a substance that accepts electrons and undergoes reduction in a chemical reaction.
- In the case of Mn3, the manganese atoms in the compound have a higher oxidation state compared to their elemental form (Mn^0).
- This means that Mn3 can accept electrons from another substance, causing it to be oxidized.
- Therefore, Mn3 can act as an oxidizing agent.
Both Cr2 and Mn3 exhibit d4 electronic configuration:
- The electronic configuration of Cr is [Ar] 3d^54s^1, and the electronic configuration of Mn is [Ar] 3d^54s^2.
- Both Cr2+ and Mn3+ ions lose their outermost s electrons, resulting in a d^4 electronic configuration.
- This is because the 4s orbital is higher in energy than the 3d orbitals, so it is preferentially lost when forming ions.
- Therefore, both Cr2 and Mn3 exhibit a d^4 electronic configuration.
When Cr2 is used as a reducing agent, the chromium ion attains d5 electronic configuration:
- When Cr2 acts as a reducing agent, it donates electrons to another substance.
- In the process, one of the chromium atoms in Cr2 is oxidized from a +2 oxidation state to a +3 oxidation state.
- This oxidation results in the formation of a Cr3+ ion, which has a d^5 electronic configuration.
- Therefore, when Cr2 is used as a reducing agent, the chromium ion attains a d^5 electronic configuration.
In summary, the correct statements about Cr2 and Mn3 are:
a) Cr2 is a reducing agent.
b) Mn3 is an oxidizing agent.
c) Both Cr2 and Mn3 exhibit a d^4 electronic configuration.
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The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer?
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The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer?.
The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer?.
Solutions for The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer?, a detailed solution for The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? has been provided alongside types of The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The correct statement(s) about Cr2+ and Mn3+ is(are)[Atomic numbers of Cr = 24 and Mn = 25]a)Cr2+ is a reducing agentb)Mn3+ is an oxidizing agentc)Both Cr2+ and Mn3+ exhibit d 4 electronic configurationd)When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configurationCorrect answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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