| Directions: (Q. 11 to 15) |
| Case III: Read the passage given below and answer the following questions from 11 to 15. |
| All the elements of group 16 have \[n{{s}^{2}}n{{p}^{4}}\] configuration in their outermost shell. Therefore, the atoms of these elements try to gain or share two electrons to achieve noble gas configuration. Sulphur and other elements of group 16 are less electronegative than oxygen, so, they cannot accept electrons easily. By sharing of two electrons with other elements, these elements acquire \[n{{s}^{2}}n{{p}^{6}}\] configuration and exhibit +2 oxidation state. Except oxygen, group 16 elements have vacant d-orbitals in their valence shell to which electrons can be promoted from p- and s-orbitals of the same shell. As a result, they can show +4 and +6 oxidation states also. |
A) \[O{{F}_{2}}\]
B) \[{{H}_{2}}O\]
C) \[C{{l}_{2}}O\]
D) \[{{H}_{2}}{{O}_{2}}\]
Correct Answer: A
Solution :
As fluorine is more electronegative than oxygen, so, oxygen exhibits +2 oxidation state in\[O{{F}_{2}}\].
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