The most common oxidation state of oxygen is $- 2$ . This is best explained as due to.

2

electrons in the outermost shell

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4

electrons in the outermost shell

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6

electrons in the outermost shell

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8

electrons in the outermost shell

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Solution

The correct option is C $6$ electrons in the outermost shell
Electronic configuration of O: ${1 s}^{2} {2 s}^{2} 2 p^{4}$ . It has 6 electrons in outermost shell, so it need only two electrons to complete its octet. So its most common oxidation state is -2.

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Can u pls explain this rule 2 :The outermost shell cannot have more than 8 electrons. The penultimate shell cannot have more than 18 electrons. A new shell is formed as soon as the outermost shell attains 8 electrons

The most common oxidation state of an element is -2. The number of electrons present in its outermost shell is:

Q. ___________ have

4

8

electrons in the outermost shell.

Name:

1. Three elements having a single electron in their outermost shell.

2. Three elements with filled outermost shell.

3. Three elements having 7 electrons in their outermost shell.

Q. Oxygen has _________ number of electrons in the outermost shell.

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The most common oxidation state of oxygen is −2. This is best explained as due to.

The correct option is C 6 electrons in the outermost shellElectronic configuration of O: 1s22s22p4. It has 6 electrons in outermost shell, so it need only two electrons to complete its octet. So its most common oxidation state is -2.

The most common oxidation state of oxygen is $- 2$ . This is best explained as due to.

The correct option is C $6$ electrons in the outermost shell
Electronic configuration of O: ${1 s}^{2} {2 s}^{2} 2 p^{4}$ . It has 6 electrons in outermost shell, so it need only two electrons to complete its octet. So its most common oxidation state is -2.