
Statement I: The fluoride ion has a larger radius than the fluoride atom.
Statement II: The fluoride ion has 8 electrons and 9 protons,
(A) Both statement I and statement II are correct and statement II is the correct explanation of statement I.
(B) Both statement I and statement II are correct, but statement II is not the correct explanation of statement I.
(C) Statement I is correct, but statement II is incorrect.
(D) Statement I is incorrect, but statement II is incorrect.
Answer
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Hint: To answer this question we should be aware of the atomic number of fluorine. Depending on the change in atomic number the atomic radius varies, especially with respect to valence electrons.
Complete step by step answer:
The atomic number of fluorine is 9 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$. The fluoride ion means ${{F}^{-}}$. Any element having positive charge means it has lost an electron and if an element has negative charge on it then, it means the element has gained an electron. The atomic number of fluorine ion is 10 and its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$. Since there is an increase in the number of electrons in the valence shell from fluoride to fluorine ions. There will be a slight increase in atomic radius of fluorine in comparison to fluorine atoms. Hence statement I is correct.
The fluorine atom has 9 electrons. Any atom loses or gains electrons only in order to attain the stable configuration that is to complete its octet. The fluorine needs one electron to attain the stable configuration. It will prefers to gain an electron instead of losing five electrons. So, formation of ${{F}^{-}}$ is very easy. The atomic number of ${{F}^{-}}$ is 10 that is it has 10 electrons.
Z = p = e …………………………..(1)
Where, Z = atomic number
p = number of protons
e = number of electrons
So, number of electrons = number of protons
10= number of protons.
Hence, Fluorine ions have 10 electrons and 10 protons.
Therefore, statement II is incorrect.
Thus, from the above discussion we can conclude that option C is the correct answer.
Note: Two important points to remember are:
-Any atom having positive charge means it has lost an electron and if an element has negative charge on it then, it means the element has gained an electron.
-Any atom loses or gains electrons only in order to attain the stable configuration that is to complete its octet.
Complete step by step answer:
The atomic number of fluorine is 9 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$. The fluoride ion means ${{F}^{-}}$. Any element having positive charge means it has lost an electron and if an element has negative charge on it then, it means the element has gained an electron. The atomic number of fluorine ion is 10 and its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$. Since there is an increase in the number of electrons in the valence shell from fluoride to fluorine ions. There will be a slight increase in atomic radius of fluorine in comparison to fluorine atoms. Hence statement I is correct.
The fluorine atom has 9 electrons. Any atom loses or gains electrons only in order to attain the stable configuration that is to complete its octet. The fluorine needs one electron to attain the stable configuration. It will prefers to gain an electron instead of losing five electrons. So, formation of ${{F}^{-}}$ is very easy. The atomic number of ${{F}^{-}}$ is 10 that is it has 10 electrons.
Z = p = e …………………………..(1)
Where, Z = atomic number
p = number of protons
e = number of electrons
So, number of electrons = number of protons
10= number of protons.
Hence, Fluorine ions have 10 electrons and 10 protons.
Therefore, statement II is incorrect.
Thus, from the above discussion we can conclude that option C is the correct answer.
Note: Two important points to remember are:
-Any atom having positive charge means it has lost an electron and if an element has negative charge on it then, it means the element has gained an electron.
-Any atom loses or gains electrons only in order to attain the stable configuration that is to complete its octet.
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