Which is the correct order of electronegativities
(A) $ F > N < O > C $
(B) $ F > N > O > C $
(C) $ F < N < O < C $
(D) $ F > O > N > C $
Answer
554.4k+ views
Hint: First we have to know that Electronegativity refers to the ability of an atom to attract a bonding pair of electrons (shared electrons) in a covalent bond. The ability of an atom/ element to attract the shared electrons is strong when the value of the electronegativity is high. Also, electronegativity increases when atomic size decreases.
Complete Step By Step Answer:
The attractive force for a bonding electron pair increases when an increase in effective nuclear charge. In other words, as the number of protons (atomic number) in the nucleus increases, the electronegativity or attraction will also increase. Therefore, electronegativity increases from left to right in a row in the periodic table.
The atomic names of $ F $ , $ O $ , $ N $ , $ C $ are fluorine, oxygen, nitrogen, carbon respectively. Also, the atomic numbers of $ F $ , $ O $ , $ N $ , $ C $ are $ 9,8,7,6 $ respectively. Since, electronegativity increases across a period with an increase in atomic number. Hence the order of electronegativities of $ F $ , $ O $ , $ N $ , $ C $ is $ F > O > N > C $ . Thus, fluorine is most electronegative and carbon is least electronegative.
Hence the correct option is (D) $ F > O > N > C $ .
Note:
In the periodic table, the electronegativity increases from bottom to top in groups, and increases from left to right across periods. Also note that the fluorine has the highest electronegative element, while cesium is the lowest electronegative element in the periodic table.
Complete Step By Step Answer:
The attractive force for a bonding electron pair increases when an increase in effective nuclear charge. In other words, as the number of protons (atomic number) in the nucleus increases, the electronegativity or attraction will also increase. Therefore, electronegativity increases from left to right in a row in the periodic table.
The atomic names of $ F $ , $ O $ , $ N $ , $ C $ are fluorine, oxygen, nitrogen, carbon respectively. Also, the atomic numbers of $ F $ , $ O $ , $ N $ , $ C $ are $ 9,8,7,6 $ respectively. Since, electronegativity increases across a period with an increase in atomic number. Hence the order of electronegativities of $ F $ , $ O $ , $ N $ , $ C $ is $ F > O > N > C $ . Thus, fluorine is most electronegative and carbon is least electronegative.
Hence the correct option is (D) $ F > O > N > C $ .
Note:
In the periodic table, the electronegativity increases from bottom to top in groups, and increases from left to right across periods. Also note that the fluorine has the highest electronegative element, while cesium is the lowest electronegative element in the periodic table.
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