
Which of the following group of element eliminates electron easily
(A) $N,P,As$
(B) $O,S,Se$
(C) $Li,Na,K$
(D) $Cl.Br,I$
Answer
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Hint: An element loses or gains electrons to attain the noble gas configuration. Noble gas configuration refers to group \[18\] configuration. Group $1$ elements have one electron in its s orbital which they try to lose and become positively charged. Group $14$ onwards losing four electrons is difficult thus, they opt to share electrons. Group $17$ element easily accepts electrons rather than losing \[7\]electrons from its valence shell.
Complete Step by Step Solution:
Option A: $N,P,As$belong to group $15$ and have general valence shell electronic configuration as\[n{s^2}{\text{ }}n{p^3}\].
Thus, to attain noble gas configuration there are two ways. One is to lose the five electrons and carry a $ + 5$ charge. Another is to accept $3$ electrons and carry $ - 3$ charge. Both the situations are not supported due to high positive and negative charge. Hence, group $15$ usually forms covalent bonds instead of ionic.
Option B: $O,S,Se$ belong to group $16$. Their general electronic configuration of valence shells is \[n{s^2}{\text{ }}n{p^4}\]. They easily accept two electrons to form electronic configuration \[n{s^2}{\text{ }}n{p^6}\].
Option C: $Li,Na,K$ belong to group $1$. The general electronic configuration is \[n{s^1}\]. The easy way out is to remove an electron and carry $ + 1$ charge to attain noble gas configuration.
Option D: $Cl.Br,I$ belong to group $17$ having valence shell electronic configuration as \[n{s^2}{\text{ }}n{p^5}\]. Thus, to gain noble gas configuration it accepts one electron and bears $ - 1$ charge.
Hence option C is only correct.
Note: Various groups are known by different names. For example, group $1$is known as alkali as they form water soluble bases, group $2$ is known as alkaline earth metal as they are found in earth crust. Group $16$ is chalcogen because of their ore forming tendency. Group $17$ is known as halogen because of the salt producing tendency. Group $18$ are known as noble gases. There is also a pseudo gas configuration which involves complete filled d orbitals.
Complete Step by Step Solution:
Option A: $N,P,As$belong to group $15$ and have general valence shell electronic configuration as\[n{s^2}{\text{ }}n{p^3}\].
Thus, to attain noble gas configuration there are two ways. One is to lose the five electrons and carry a $ + 5$ charge. Another is to accept $3$ electrons and carry $ - 3$ charge. Both the situations are not supported due to high positive and negative charge. Hence, group $15$ usually forms covalent bonds instead of ionic.
Option B: $O,S,Se$ belong to group $16$. Their general electronic configuration of valence shells is \[n{s^2}{\text{ }}n{p^4}\]. They easily accept two electrons to form electronic configuration \[n{s^2}{\text{ }}n{p^6}\].
Option C: $Li,Na,K$ belong to group $1$. The general electronic configuration is \[n{s^1}\]. The easy way out is to remove an electron and carry $ + 1$ charge to attain noble gas configuration.
Option D: $Cl.Br,I$ belong to group $17$ having valence shell electronic configuration as \[n{s^2}{\text{ }}n{p^5}\]. Thus, to gain noble gas configuration it accepts one electron and bears $ - 1$ charge.
Hence option C is only correct.
Note: Various groups are known by different names. For example, group $1$is known as alkali as they form water soluble bases, group $2$ is known as alkaline earth metal as they are found in earth crust. Group $16$ is chalcogen because of their ore forming tendency. Group $17$ is known as halogen because of the salt producing tendency. Group $18$ are known as noble gases. There is also a pseudo gas configuration which involves complete filled d orbitals.
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