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Question: Find which of the following is an oxidizing agent ? A) \[Mn{(CO)_5}\] B) \[Fe{(CO)_5}\] C) \[M...

Find which of the following is an oxidizing agent ?
A) Mn(CO)5Mn{(CO)_5}
B) Fe(CO)5Fe{(CO)_5}
C) Mn2(CO)10M{n_2}{(CO)_{10}}
D) Fe2(CO)9F{e_2}{(CO)_9}

Explanation

Solution

In order to solve this question, we have to know about the oxidation number of MnMn and FeFe atoms present in compounds. Then we should calculate the EAN i.e. Effective atomic number. An oxidizing element has the ability to oxidize the other substances.
Formula used:
Here we used the formula of calculating Effective atomic number:
EAN=Z±X+2nEAN = Z \pm X + 2n
Where,
ZZ= atomic mass of atom
XX= oxidation state
nn= coordination number

Complete answer:
At first we must know about oxidizing agents; oxidizing agents help to oxidize other substances. It also increases the oxidation state of other substances, while the oxidation state of this substance will be decreased. It undergoes reduction so as to produce the most stable configuration.
Mn(CO)5Mn{(CO)_5} accepts electron from
⇒Mn(CO)5+e‾\Rightarrow Mn{(CO)_5} + e\overline {}
⇒[Mn(CO)5]‾\Rightarrow \left[ {Mn{{(CO)}_5}} \right]\overline {}
Now we see, in Mn(CO)5Mn{(CO)_5} oxidation number of MnMn is
⇒x+5(0)=−1\Rightarrow x + 5\left( 0 \right) = - 1
⇒x=−1\Rightarrow x = - 1
Therefore, EANEAN of Mn(CO)5Mn{(CO)_5} is –
⇒25+1+2(5)\Rightarrow 25 + 1 + 2\left( 5 \right)
⇒36\Rightarrow 36
[it is the stable configuration of KrKr]

So, the right answer is option A)Mn(CO)5Mn{(CO)_5}.

Note:
Some examples of oxidizing agents are Iodine, Chlorine, Bromine, Oxygen etc. It is used in combustion of fuel, bleaching of fabric, and processing of rubber. An oxidizing element accepts the electrons from the substances.