
The order of first electron affinity of $O$, $S$ and $Se$ is
A.$O > S > Se$
B.$S > Se > O$
C.$Se > O > S$
D.$S > O > Se$
Answer
570.9k+ views
Hint:
First we should be aware of the term electron affinity, what it refers to and the basic phenomenon. Then we have to mention the basic change in the electron affinity on increase and decrease in the size of the atom. Though in this case there is an exception where the over small size of $O$ makes it very easily ionizable, thus changing the trend for the lower two elements.
Complete step by step answer:
The electron affinity is the phenomenon of the change in the potential energy of the atom when an electron is added to an isolated neutral gaseous atom to form a negative ion.
The given group of the elements is the $16$ group of the modern periodic table as the electron affinity is fairly higher for that group when compared to the other ones.
As moving down the group electron affinity of elements decreases at every step of the table due to the increase in size of the atom. This is the general trend to observe the electron affinity for the atoms.
When it comes to the $O$ , it is such a small atom when compared to the other ones. It has a very small amount of space available within its 2p orbital. Due to the presence of this lesser space, any new electron trying to approach and attach to oxygen experiences lower electron affinity from the electrons already living in the element's 2p orbital. Therefore it’s electron affinity is close to $ - 141kJmo{l^{ - 1}}$ .
Whereas in the case of $S$ , sulfur's outermost orbital is a 3p orbital where there is ample amount of space. Due to the availability of this space the electrons in this orbital easily share this space with an extra electron.
Therefore the answer would be option B, $S > Se > O$ .
Note:
The change in the electron affinity trend due to $O$ is due to its size. Because the size itself creates a strong binding power of the nucleus over the electrons, thus making the nature of ionization very easy. This is because it holds the electron more firmly.
First we should be aware of the term electron affinity, what it refers to and the basic phenomenon. Then we have to mention the basic change in the electron affinity on increase and decrease in the size of the atom. Though in this case there is an exception where the over small size of $O$ makes it very easily ionizable, thus changing the trend for the lower two elements.
Complete step by step answer:
The electron affinity is the phenomenon of the change in the potential energy of the atom when an electron is added to an isolated neutral gaseous atom to form a negative ion.
The given group of the elements is the $16$ group of the modern periodic table as the electron affinity is fairly higher for that group when compared to the other ones.
As moving down the group electron affinity of elements decreases at every step of the table due to the increase in size of the atom. This is the general trend to observe the electron affinity for the atoms.
When it comes to the $O$ , it is such a small atom when compared to the other ones. It has a very small amount of space available within its 2p orbital. Due to the presence of this lesser space, any new electron trying to approach and attach to oxygen experiences lower electron affinity from the electrons already living in the element's 2p orbital. Therefore it’s electron affinity is close to $ - 141kJmo{l^{ - 1}}$ .
Whereas in the case of $S$ , sulfur's outermost orbital is a 3p orbital where there is ample amount of space. Due to the availability of this space the electrons in this orbital easily share this space with an extra electron.
Therefore the answer would be option B, $S > Se > O$ .
Note:
The change in the electron affinity trend due to $O$ is due to its size. Because the size itself creates a strong binding power of the nucleus over the electrons, thus making the nature of ionization very easy. This is because it holds the electron more firmly.
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