
Assertion: Fluorine is the most electronegative element.
Reason: Fluorine can form only ionic bonds.
(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(B) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(C) Assertion is true but Reason is false.
(D) Both assertion and Reason are false statements.
Answer
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Hint: To check the validity of assertion and reason we need to check the electronegative character of fluorine and the types of bonds formed by fluorine with other elements present in the periodic table. Fluorine is a diatomic molecule.
Complete Step-by-step Solution
Fluorine is the first member of the Halogen family. Its atomic number is $ 7 $ . It lies in the second period and group $ 17 $ of the periodic table. Its electronic configuration is $ 1{s^2}2{s^2}2{p^5} $ .
The outermost shell is a p-subshell which contains $ 5 $ electrons. To achieve the Noble gas configuration of Neon it needs one more electron.
Fluorine has $ 5 $ electrons in its \[2p\] shell. The ideal configuration of the \[p\] orbital has 6 electrons, since Fluorine is so close to attaining the noble gas configuration, the electrons are held very strongly to the nucleus. Thus it makes the fluorine most electronegative because of its small radius as the positive protons hold a very strong attraction to the electrons.
Being an element of the group $ 17 $ fluorine tends to form polar covalent bonds with non-metal. Eg: $ H - F $ . Having a $ - 1 $ oxidation state fluorine can also form an ionic bond.
Hence we can say that both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Therefore, Option (B) is the correct answer.
Note
Electronegativity is defined as the tendency of an atom in a molecule to draw the shared pair of electrons towards itself. Electronegativity of an element increases from moving down the group and along the period. Fluorine has a high electron affinity and high ionization energy. Fluorine exists as a gas at room temperature.
Complete Step-by-step Solution
Fluorine is the first member of the Halogen family. Its atomic number is $ 7 $ . It lies in the second period and group $ 17 $ of the periodic table. Its electronic configuration is $ 1{s^2}2{s^2}2{p^5} $ .
The outermost shell is a p-subshell which contains $ 5 $ electrons. To achieve the Noble gas configuration of Neon it needs one more electron.
Fluorine has $ 5 $ electrons in its \[2p\] shell. The ideal configuration of the \[p\] orbital has 6 electrons, since Fluorine is so close to attaining the noble gas configuration, the electrons are held very strongly to the nucleus. Thus it makes the fluorine most electronegative because of its small radius as the positive protons hold a very strong attraction to the electrons.
Being an element of the group $ 17 $ fluorine tends to form polar covalent bonds with non-metal. Eg: $ H - F $ . Having a $ - 1 $ oxidation state fluorine can also form an ionic bond.
Hence we can say that both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Therefore, Option (B) is the correct answer.
Note
Electronegativity is defined as the tendency of an atom in a molecule to draw the shared pair of electrons towards itself. Electronegativity of an element increases from moving down the group and along the period. Fluorine has a high electron affinity and high ionization energy. Fluorine exists as a gas at room temperature.
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