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Which one of the following is most easily reduced ?
- a)V(CO)6
- b)Cr(CO)6
- c)Fe(CO)5
- d)Ni(CO)4
Correct answer is option ''. Can you explain this answer?
Verified Answer
Which one of the following is most easily reduced ?a)V(CO)6b)Cr(CO)6c)...
V (CO)6 is Paramagnetic (has unpaired electron) and is very reactive V (CO)6 will easily gain an electron by reduction to give the anion 
Most Upvoted Answer
Which one of the following is most easily reduced ?a)V(CO)6b)Cr(CO)6c)...
Explanation:
Complexes and their reducibility:
- The reducibility of a complex depends on the metal and ligands present in it.
- Ligands like CO (carbon monoxide) are strong π-acceptors and thus stabilize the metal in a lower oxidation state.
Analysis of given complexes:
- V(CO)6: Vanadium in this complex is in a relatively higher oxidation state and is less easily reducible compared to other transition metals.
- Cr(CO)6: Chromium in this complex is in a lower oxidation state compared to V(CO)6, making it more easily reducible.
- Fe(CO)5: Iron in this complex is also in a lower oxidation state compared to V(CO)6, making it more easily reducible than V(CO)6.
- Ni(CO)4: Nickel in this complex is in the zero oxidation state, making it the most easily reducible among the given options.
Therefore, the most easily reducible complex among the given options is Ni(CO)4.
Complexes and their reducibility:
- The reducibility of a complex depends on the metal and ligands present in it.
- Ligands like CO (carbon monoxide) are strong π-acceptors and thus stabilize the metal in a lower oxidation state.
Analysis of given complexes:
- V(CO)6: Vanadium in this complex is in a relatively higher oxidation state and is less easily reducible compared to other transition metals.
- Cr(CO)6: Chromium in this complex is in a lower oxidation state compared to V(CO)6, making it more easily reducible.
- Fe(CO)5: Iron in this complex is also in a lower oxidation state compared to V(CO)6, making it more easily reducible than V(CO)6.
- Ni(CO)4: Nickel in this complex is in the zero oxidation state, making it the most easily reducible among the given options.
Therefore, the most easily reducible complex among the given options is Ni(CO)4.
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