
Which of the following halides is not oxidized by ? $ Mn{O_2} $
A. $ {F^ - } $
B. $ C{l^ - } $
C. $ B{r^ - } $
D. $ {I^ - } $
Answer
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Hint :When the removal of electrons takes place in an atom, then an increase in the oxidation state of that atom is observed. The process is known as oxidation and the atom is known as a reducing agent. If an atom gains an electron, then a decrease in the oxidation state of the atom is observed and the process is known as reduction and the atom is referred to as an oxidizing agent.
Complete Step By Step Answer:
Fluoride ions are difficult to oxidize into fluorine because the size of fluoride ions is so small that the electrons present in the orbitals experience strong repulsive forces from each other. Due to these repulsions, the effect of the electrons outweighs their closeness with the nucleus and it becomes very easy to remove an electron. Hence, it takes very less energy to remove an electron from a fluoride ion.
Hence, the fluoride ion is the strongest oxidizing agent and a weak reducing agent i.e., fluoride ion itself gets reduced by gaining electrons and oxidizing other molecules. Therefore, the halide which is not oxidized by $ Mn{O_2} $ is fluoride ion i.e., $ {F^ - } $ .
Thus, option (A) is the correct answer.
Note :
It is important to note that on moving top to bottom in a group, the atomic size of the halide ions increases due to which the charge density spreads uniformly over the atoms and a decrease in the inter electronic repulsions is observed. Due to this, on moving down the group, the oxidation of halide ions becomes easier. Therefore, except $ {F^ - } $ , all other halide ions can get oxidized by $ Mn{O_2} $ .
Complete Step By Step Answer:
Fluoride ions are difficult to oxidize into fluorine because the size of fluoride ions is so small that the electrons present in the orbitals experience strong repulsive forces from each other. Due to these repulsions, the effect of the electrons outweighs their closeness with the nucleus and it becomes very easy to remove an electron. Hence, it takes very less energy to remove an electron from a fluoride ion.
Hence, the fluoride ion is the strongest oxidizing agent and a weak reducing agent i.e., fluoride ion itself gets reduced by gaining electrons and oxidizing other molecules. Therefore, the halide which is not oxidized by $ Mn{O_2} $ is fluoride ion i.e., $ {F^ - } $ .
Thus, option (A) is the correct answer.
Note :
It is important to note that on moving top to bottom in a group, the atomic size of the halide ions increases due to which the charge density spreads uniformly over the atoms and a decrease in the inter electronic repulsions is observed. Due to this, on moving down the group, the oxidation of halide ions becomes easier. Therefore, except $ {F^ - } $ , all other halide ions can get oxidized by $ Mn{O_2} $ .
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