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Question

Which of the following elements shows +7 oxidation state?


A

Manganese

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B

Boron

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C

Carbon

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D

Oxygen

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Solution

The correct option is A

Manganese


The explanation for the correct option:

Option (A) Manganese

Oxidation state or oxidation number

  • It is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

Electronic configuration

  • The electron configuration of an element describes how electrons are distributed in its atomic orbitals.

The oxidation state of Manganese

  • The electronic configuration of Manganese is: [Ar]3d54s2
  • The electrons present in the d and s subshell participate in the bond formation.
  • Manganese can lose these electrons to form chemical bonds.
  • Thus, Manganese shows variable oxidation states of +2, +3, +4, +5, +6, and +7.
  • Hence, this option is correct.

Explanation of incorrect options:

Option B: Boron

  • The electronic configuration of Boron is: [He]2s22p1
  • The electrons present in the p and s subshell participate in the chemical reaction as after losing these electrons the boron atoms form a stable electronic configuration.
  • Hence, boron shows an oxidation state of +3.
  • Therefore, this option is incorrect.

Option C: Carbon

  • The electronic configuration of carbon is: [He]2s22p2
  • The electrons present in the p and s subshell participate in the chemical reaction as after losing these electrons the boron atoms form a stable electronic configuration.
  • Hence, carbon shows an oxidation state of +4.
  • Therefore, this option is incorrect.

Option D: Oxygen

  • The electronic configuration of carbon is: [He]2s22p4
  • There are two unpaired electrons in the p subshell, the oxygen atom readily loses these two electrons to form a relatively stable electronic configuration.
  • Hence, oxygen shows an oxidation state of +2.
  • Therefore, this option is incorrect.

Hence, option A, Manganese, is the correct answer.


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