
Halogen with highest electron affinity is:
A. \[F\]
B. \[Cl\]
C. \[Br\]
D. \[I\]
Answer
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Hint: In periodic table as we move down the group , atomic size of elements increases so that electron affinity of elements decreases gradually from top to bottom. Size increases in the order as \[F < Cl < Br < I\] .
Complete step by step answer:
First , we take the case of \[F\] . It is considered as a small atom due to its small atomic size which lies on top. Electrons are carried in the outermost \[2p\] orbital which is having a small space (sub shell electronic configuration of\[F\] is \[1{s^2}\]\[2{s^2}\]\[2{p^5}\] ) . There is a chance of attachment of foreign electron in its outermost orbital. Due to small space availability in the outermost orbital , it cannot accommodate a foreign electron. So the coming electron experiences a lower electron affinity from the remaining electrons already available in the \[2p\] orbital.
Next comes the case of \[Cl\] atoms . Electrons are carried in the outermost \[3p\] orbital (sub shell electronic configuration of \[Cl\] is \[1{s^2}\]\[2{s^2}\] \[2{p^6}\] ) . Since it is the outermost orbital , it contains more space as compared to the outermost \[2p\] orbital of \[F\] . Here due to more space , electrons in \[3p\] orbital are ready to accommodate a foreign electron. Therefore , \[Cl\] has a higher electron affinity than \[F\] . The reason behind this is the orbital structure which made it to have highest electron affinity among halogen
Electron affinity decreases down the group, and hence Bromine and Iodine will have lesser electron affinity than chlorine.
By convention, fluorine should have had the highest electron affinity but it does not due to its small size and is an exception.
Hence , electron affinity of \[Cl\] is higher in halogen .
So, the correct answer is Option B.
Note: As size increases down the group , electron affinity decreases. So as \[F\] sits at top , we expect it to have the highest electron affinity . But as coming to the orbital structure case , the principle of electronic affinity gradually changes due to space availability in the outermost shell.
Complete step by step answer:
First , we take the case of \[F\] . It is considered as a small atom due to its small atomic size which lies on top. Electrons are carried in the outermost \[2p\] orbital which is having a small space (sub shell electronic configuration of\[F\] is \[1{s^2}\]\[2{s^2}\]\[2{p^5}\] ) . There is a chance of attachment of foreign electron in its outermost orbital. Due to small space availability in the outermost orbital , it cannot accommodate a foreign electron. So the coming electron experiences a lower electron affinity from the remaining electrons already available in the \[2p\] orbital.
Next comes the case of \[Cl\] atoms . Electrons are carried in the outermost \[3p\] orbital (sub shell electronic configuration of \[Cl\] is \[1{s^2}\]\[2{s^2}\] \[2{p^6}\] ) . Since it is the outermost orbital , it contains more space as compared to the outermost \[2p\] orbital of \[F\] . Here due to more space , electrons in \[3p\] orbital are ready to accommodate a foreign electron. Therefore , \[Cl\] has a higher electron affinity than \[F\] . The reason behind this is the orbital structure which made it to have highest electron affinity among halogen
Electron affinity decreases down the group, and hence Bromine and Iodine will have lesser electron affinity than chlorine.
By convention, fluorine should have had the highest electron affinity but it does not due to its small size and is an exception.
Hence , electron affinity of \[Cl\] is higher in halogen .
So, the correct answer is Option B.
Note: As size increases down the group , electron affinity decreases. So as \[F\] sits at top , we expect it to have the highest electron affinity . But as coming to the orbital structure case , the principle of electronic affinity gradually changes due to space availability in the outermost shell.
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