
The element which never acts as reducing agent in a chemical reaction is
A. $O$
B. $Li$
C. $F$
D. $C$
Answer
163.2k+ views
Hint: A compound or element which can donate its electron is called a reducing agent. A strong oxidising agent means it helps to oxidise other substances and itself get reduced. The element which has strongly bound electrons never acts as a reducing agent.
Complete Step by Step Answer:
The electronic configuration of $F$$is-$$1s^22s^22p^5$.
$F$ has $5$ electrons in a 2p shell. So, to get stability it needs one more electron and accepts one electron. Thus it acts as an oxidising agent not as a reducing agent.
The electronic configuration of $O$$is-$$1s^22s^22p^4$.
$O$ has $4$ electron in 2p shell. So, to get stability it releases one electron and acts as a reducing agent.
The electronic configuration of Li$is-$$1s^22s^1$.
Li has $1$ electrons in 2s shell. So, to get stability it needs one more electron and accepts one electron. Thus it acts as an oxidising agent.
The electronic configuration of $C$$is-$$1s^22s^22p^2$.
$C$ has $2$ electrons in a 2p shell. So, to get stability it needs one more electron and accepts one electron. Thus it acts as an oxidising agent.
The element which never acts as a reducing agent in a chemical reaction is$F$.
Thus the correct option is C.
Note: Fluorine is the most reactive among the halogens due to strong electronegativity. Lithium is the most electropositive and highly reactive element due to presence of one electron in valence shell.
Complete Step by Step Answer:
The electronic configuration of $F$$is-$$1s^22s^22p^5$.
$F$ has $5$ electrons in a 2p shell. So, to get stability it needs one more electron and accepts one electron. Thus it acts as an oxidising agent not as a reducing agent.
The electronic configuration of $O$$is-$$1s^22s^22p^4$.
$O$ has $4$ electron in 2p shell. So, to get stability it releases one electron and acts as a reducing agent.
The electronic configuration of Li$is-$$1s^22s^1$.
Li has $1$ electrons in 2s shell. So, to get stability it needs one more electron and accepts one electron. Thus it acts as an oxidising agent.
The electronic configuration of $C$$is-$$1s^22s^22p^2$.
$C$ has $2$ electrons in a 2p shell. So, to get stability it needs one more electron and accepts one electron. Thus it acts as an oxidising agent.
The element which never acts as a reducing agent in a chemical reaction is$F$.
Thus the correct option is C.
Note: Fluorine is the most reactive among the halogens due to strong electronegativity. Lithium is the most electropositive and highly reactive element due to presence of one electron in valence shell.
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