
Highest electron affinity is shown by:
(A) ${{\text{O}}^{\text{ - }}}$
(B) ${{\text{F}}^{\text{ - }}}$
(C) ${\text{C}}{{\text{l}}_{\text{2}}}$
(D) ${{\text{F}}_2}$
Answer
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Hint: Electron affinity is the energy released when an electron is added to the neutral gaseous atom. The species which does not possess complete electronic configuration in its valence shell can give more amount of energy thus have the highest electron affinity in the given options.
Complete Step by step answer:
The highest electron affinity in the periodic table is exhibited by two elements which are electronegative in nature. Those two elements are oxygen and fluorine besides all the electronegative elements halogens have highest electro negatives.
The given question consists of all the elements which generally exhibit highest electronegativities. We know that halogens (F,Cl,Br,I) have highest electronegativities and oxygen(O) also has highest electronegativity. But the options contain ions besides neutral atoms. Let us check all the options with their valence shell electronic configurations to know whether energy released (electronic affinity in this case) is highest or not in all the given species.
In ${{\text{O}}^{\text{ - }}}$(Oxygen ion) , the number of atoms are 9, the negative charge adds one electron to its valence shell. In the oxygen atom the atomic number is 8 and its electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{4}}}$. So the oxygen contains 6 electrons in the valence shell. Which means the octet is not complete and oxygen atom readily accepts electrons thus releases a higher amount of energy upon addition of an electron. ${{\text{O}}^{\text{ - }}}$contains 8 + 1 = 9electrons. Now it needs only one electron to fill the octet hence the electron affinity of ${{\text{O}}^{\text{ - }}}$would be high.
In ${{\text{F}}^{\text{ - }}}$ the total number of electrons in valence shell are (Z = 9) and the negative charge contributes one more electron thus 10 electrons which filled its valence shell with octet. Therefore,
${{\text{F}}^{\text{ - }}}$cannot easily accept electrons thus releasing less amount of energy. Same is the case with $C{l_2}$ and ${{\text{F}}_{\text{2}}}$ where they are already stable in their molecular state with octet filled in the valence shells so they also or may not release energy to gain the electrons as they require energy externally to break their existing bonds.
From the above explanations the species which has higher electron affinity is option (A).
Note: The molecules or ions which have their outermost shells completely filled by the electrons are stable. They do not readily gain electrons or lose electrons. If the given species needs one or two electrons to get state inert configuration they accept electrons to release higher energies thus they have higher electron affinities.
Complete Step by step answer:
The highest electron affinity in the periodic table is exhibited by two elements which are electronegative in nature. Those two elements are oxygen and fluorine besides all the electronegative elements halogens have highest electro negatives.
The given question consists of all the elements which generally exhibit highest electronegativities. We know that halogens (F,Cl,Br,I) have highest electronegativities and oxygen(O) also has highest electronegativity. But the options contain ions besides neutral atoms. Let us check all the options with their valence shell electronic configurations to know whether energy released (electronic affinity in this case) is highest or not in all the given species.
In ${{\text{O}}^{\text{ - }}}$(Oxygen ion) , the number of atoms are 9, the negative charge adds one electron to its valence shell. In the oxygen atom the atomic number is 8 and its electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{4}}}$. So the oxygen contains 6 electrons in the valence shell. Which means the octet is not complete and oxygen atom readily accepts electrons thus releases a higher amount of energy upon addition of an electron. ${{\text{O}}^{\text{ - }}}$contains 8 + 1 = 9electrons. Now it needs only one electron to fill the octet hence the electron affinity of ${{\text{O}}^{\text{ - }}}$would be high.
In ${{\text{F}}^{\text{ - }}}$ the total number of electrons in valence shell are (Z = 9) and the negative charge contributes one more electron thus 10 electrons which filled its valence shell with octet. Therefore,
${{\text{F}}^{\text{ - }}}$cannot easily accept electrons thus releasing less amount of energy. Same is the case with $C{l_2}$ and ${{\text{F}}_{\text{2}}}$ where they are already stable in their molecular state with octet filled in the valence shells so they also or may not release energy to gain the electrons as they require energy externally to break their existing bonds.
From the above explanations the species which has higher electron affinity is option (A).
Note: The molecules or ions which have their outermost shells completely filled by the electrons are stable. They do not readily gain electrons or lose electrons. If the given species needs one or two electrons to get state inert configuration they accept electrons to release higher energies thus they have higher electron affinities.
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