Which element has the highest electronegativity?
A) ${\text{C}}$
B) ${\text{Mg}}$
C) ${\text{O}}$
D) ${\text{S}}$
Answer
644.7k+ views
Hint: Electronegativity of an element is dependent on its ability to attract the electrons present in a covalent bond. The electronegativity of an element can be identified by the position of the element in the periodic table.
Complete step by step answer:
1) First of all let's understand the concept of electronegativity where it is the phenomenon that attracts the shared electrons that are present in a covalent bond. The higher is the electronegative value of an element the higher will be the pull of electrons towards itself.
2) Now let's analyze each element given in the option one by one. The carbon atom is placed in period two and all other elements given in the option are placed in the periods below than period two which means the carbon atom has more electronegativity than oxygen and sulfur but less than the oxygen atom. Hence, this option is an incorrect choice.
3) The element magnesium is placed in period three and there are other elements given in the option which are present in the period above than period three that means the magnesium atom doesn’t have much electronegativity. Hence, this option is an incorrect choice.
4) The element oxygen is present in period two of the periodic table and is present after the element carbon. As oxygen atom is present after the element carbon and in period two, it has the highest electronegativity among all elements given in options. Hence, this option is the correct choice.
5) The sulfur atom is present in period three and just below the oxygen atom. Hence, it has a lower electronegativity than oxygen. This option is an incorrect choice.
6) Therefore, the ${\text{O}}$ element has the highest electronegativity.
Hence, option C is the correct answer.
Note:
Electronegativity increases from going left to right in a period and decreases going from top to bottom in a group that is present in a periodic table. The order of electronegativity of the given element will be ${\text{O}} > {\text{S}} > {\text{C}} > {\text{Mg}}$ and their values according Pauling scale of electronegativity are ${\text{3}} \cdot {\text{44, 2}}{\text{.58, 2}}{\text{.55, 1}}{\text{.31}}$ respectively.
Complete step by step answer:
1) First of all let's understand the concept of electronegativity where it is the phenomenon that attracts the shared electrons that are present in a covalent bond. The higher is the electronegative value of an element the higher will be the pull of electrons towards itself.
2) Now let's analyze each element given in the option one by one. The carbon atom is placed in period two and all other elements given in the option are placed in the periods below than period two which means the carbon atom has more electronegativity than oxygen and sulfur but less than the oxygen atom. Hence, this option is an incorrect choice.
3) The element magnesium is placed in period three and there are other elements given in the option which are present in the period above than period three that means the magnesium atom doesn’t have much electronegativity. Hence, this option is an incorrect choice.
4) The element oxygen is present in period two of the periodic table and is present after the element carbon. As oxygen atom is present after the element carbon and in period two, it has the highest electronegativity among all elements given in options. Hence, this option is the correct choice.
5) The sulfur atom is present in period three and just below the oxygen atom. Hence, it has a lower electronegativity than oxygen. This option is an incorrect choice.
6) Therefore, the ${\text{O}}$ element has the highest electronegativity.
Hence, option C is the correct answer.
Note:
Electronegativity increases from going left to right in a period and decreases going from top to bottom in a group that is present in a periodic table. The order of electronegativity of the given element will be ${\text{O}} > {\text{S}} > {\text{C}} > {\text{Mg}}$ and their values according Pauling scale of electronegativity are ${\text{3}} \cdot {\text{44, 2}}{\text{.58, 2}}{\text{.55, 1}}{\text{.31}}$ respectively.
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