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Question

In aqueous solution, $${ Cr }^{ 2+ }$$ is a stronger reducing agent than $${ Fe }^{ 2+ }$$. This is because:


A
Cr2+ ion is more stable than Fe2+.
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B
Cr3+ ion with d3 configuration has favourable crystal field stabilisation energy.
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C
Cr3+ has half-filled configuration and hence more stable
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D
Fe3+ in aqueous solution is more stable than Cr3+.
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E
Fe2+ ion with d6 configuration has favourable crystal field stabilisation energy.
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Solution

The correct option is B $${ Cr }^{ 3+ }$$ ion with $${ d }^{ 3 }$$ configuration has favourable crystal field stabilisation energy.
$$\because { Cr }^{ 3+ }$$ ion has $${ d }^{ 3 }$$ configuration $$\left( { t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 } \right) $$, which makes $${ Cr }^{ 3+ }$$ stable by providing crystal field stabilisation energy.
Thus, $${ Cr }^{ 2+ }$$ ion can easily changes to $${ Cr }^{ 3+ }$$ (i.e. get oxidised) and acts as a strong reducing agent.
$${ Cr }^{ 2+ }\rightarrow { Cr }^{ 3+ }+{ e }^{ - }$$

Chemistry

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