
Which of the following elements has highest electronegativity?
$
(a){\text{ Al}} \\
{\text{(b) S}} \\
{\text{(c) Si}} \\
{\text{(d) P}} \\
$
Answer
599.4k+ views
Hint – In this question use the concept that Aluminum, Sulphur, Silicon and Phosphorus belong to the third period of the periodic table. As we move from left to right in this period electronegativity increases. This approach will help finding the right answer as it will simply be the rightmost element of period 3 from the options given.
Complete answer:-
As we know the given elements are present on the third period in the periodic table.
Now as we know that on moving across the period from left to right, the electronegativity increases, because the elements on the right side of the periodic table have more tendency to attract the electron in a bond (as the outermost shell of these elements have atoms which is increasing in number when we move from left to right) so the right most element has a higher tendency to complete its outermost shell with 8 valence electrons.
So in the given elements the right most element in the third period of periodic table is Sulphur (S) as it is in group 16 and the atomic number of Sulphur is 16 and the outermost shell of Sulphur has 6 electrons which is highest as compare to other electrons in the given elements as Al has 3, Si has 4 and P has 5 valence electrons in the outermost shell so Sulphur has highest tendency among the given elements to complete its outermost shell so among the given elements Sulphur (S) is most electronegative.
Hence option (B) is the correct answer.
Note – The elements of period three involve Sodium, Magnesium, Aluminum, Silicon, Phosphorous, Sulphur, Chlorine and Argon. In period 3 sodium with 11 protons is the least electronegative and chlorine with 17 protons is the most electronegative element. Another main reason behind the increase in electronegativity from left to right is that the nuclear charge is increasing faster than the electron shielding so the attraction that the atoms have for valence electrons increases.
Complete answer:-
As we know the given elements are present on the third period in the periodic table.
Now as we know that on moving across the period from left to right, the electronegativity increases, because the elements on the right side of the periodic table have more tendency to attract the electron in a bond (as the outermost shell of these elements have atoms which is increasing in number when we move from left to right) so the right most element has a higher tendency to complete its outermost shell with 8 valence electrons.
So in the given elements the right most element in the third period of periodic table is Sulphur (S) as it is in group 16 and the atomic number of Sulphur is 16 and the outermost shell of Sulphur has 6 electrons which is highest as compare to other electrons in the given elements as Al has 3, Si has 4 and P has 5 valence electrons in the outermost shell so Sulphur has highest tendency among the given elements to complete its outermost shell so among the given elements Sulphur (S) is most electronegative.
Hence option (B) is the correct answer.
Note – The elements of period three involve Sodium, Magnesium, Aluminum, Silicon, Phosphorous, Sulphur, Chlorine and Argon. In period 3 sodium with 11 protons is the least electronegative and chlorine with 17 protons is the most electronegative element. Another main reason behind the increase in electronegativity from left to right is that the nuclear charge is increasing faster than the electron shielding so the attraction that the atoms have for valence electrons increases.
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