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Hint : The most electronegative element is a halogen with the smallest size. It is the lightest halogen. It is pale yellow in colour and highly toxic. It belongs to the 2nd period. It is highly reactive in nature.
Complete answer :
We have studied that electronegativity is the tendency of an atom to attract a shared pair of electrons. It depends majorly on the nuclear charge and atomic size.
On analysing the options given to us; we see that Fluorine is the most electronegative element because it requires only one electron to complete its octet and by completing its octet it will be extra stable. The fluorine has high charge density because of low size which helps it in attracting the electron.
The Cesium is the least electronegative atom because it can easily donate one electron and by donating that electron, it will achieve noble gas configuration and will be extra stable.
The Helium has 2 electrons in its 1s orbital which is fully filled. So, it will not be much electronegative.
All the points that are in favour of Fluorine are the same for Iodine but Iodine has larger size as compared to Fluorine; so, charge density will be low and Fluorine will only be most electronegative.
Thus, option a.) is the correct answer.
Note :
The electronic configuration for Fluorine is : $1{s^2}2{s^2}2{p^5}$
Thus, by acquiring one electron it will achieve noble gas configuration and it will be more stable.
Fluorine has an electronegativity of 3.98 on the Pauling electronegativity scale.
Complete answer :
We have studied that electronegativity is the tendency of an atom to attract a shared pair of electrons. It depends majorly on the nuclear charge and atomic size.
On analysing the options given to us; we see that Fluorine is the most electronegative element because it requires only one electron to complete its octet and by completing its octet it will be extra stable. The fluorine has high charge density because of low size which helps it in attracting the electron.
The Cesium is the least electronegative atom because it can easily donate one electron and by donating that electron, it will achieve noble gas configuration and will be extra stable.
The Helium has 2 electrons in its 1s orbital which is fully filled. So, it will not be much electronegative.
All the points that are in favour of Fluorine are the same for Iodine but Iodine has larger size as compared to Fluorine; so, charge density will be low and Fluorine will only be most electronegative.
Thus, option a.) is the correct answer.
Note :
The electronic configuration for Fluorine is : $1{s^2}2{s^2}2{p^5}$
Thus, by acquiring one electron it will achieve noble gas configuration and it will be more stable.
Fluorine has an electronegativity of 3.98 on the Pauling electronegativity scale.
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