Question

1) An atom of an element has 2
electrons in its outermost shell.
What will be the type of ion
formed and the value of the
charge on the ion produced?​

Answer 1

Answer:

Element X can lose only one electron which is the outer most electron and so it has valency 1. If it is removed then the number of electrons is one less than the number of protons so there is one proton extra than electron. So a one positive charge is created.

Answer 2

Given that ,a atom has 2 electrons in its outermost shell . We need to find the type of ion formed and the value of charged produced. So , we know that , for a atom to be stable , it must have 8 electrons in the outermost shell . Here , the atom has 2 electrons , so ,

• It can gain 6 electrons to complete the octet , or else it should loose 2 electrons to complete the octet / duplet configuration . It will be rather easy for the atom to loose 2 electrons than , to gain 6. Hence the atom will loose 2 electrons .

Since the atom will loose two electrons to gain stability ,it will become positive and will be cation .

$\dashrightarrow\sf Atom\qquad \xrightarrow{\sf loses \ 2 \ e^{-} }\qquad \red{A^{2+}}$

Hence the charge on the atom will be 2+ . Hence , the final answer is ,

$\boxed{\begin{array}{c|c}\sf \red{ Type \ of \ ion }&\sf Cation \\\\\sf\red{ Charge}&\sf 2+ \end{array}}$

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$\underline{\underline{\blacksquare\pink{\textsf{ The Electronic Configuration table for first 20 Elements :- }}}}$

$\boxed {\begin{array}{c|c|c|c|cccc}\cline {1-8}\sf Element & \sf Symbol & \sf Valency & \sf{Atomic\:Number} & & \sf{Electron} & \sf Configuration & \\ \cline {5-8} &&&&\bf K&\bf L &\bf M&\bf N \\ \cline {1-8}\rm Hydrogen &\sf H & 1 & 1 & 1 &&& \\ \cline {1-8}\rm Helium &\sf He & 0 & 2 & 2 &&& \\ \cline {1-8}\rm Lithium &\sf Li & 1& 3 &2 &1 && \\ \cline {1-8}\rm Beryllium &\sf Be & 2 & 4 & 2 & 2 && \\ \cline {1-8}\rm Boron &\sf B & 3 & 5 & 2 & 3 && \\ \cline {1-8}\rm Carbon & \sf C & 4 & 6 & 2 & 4 && \\ \cline {1-8}\rm Nitrogen &\sf N & 3 & 7 & 2 & 5 && \\ \cline {1-8}\rm Oxygen &\sf O & 2 & 8 & 2 & 6 && \\ \cline {1-8}\rm Fluorine &\sf F & 1 & 9 & 2 & 7 && \\ \cline {1-8}\rm Neon & \sf Ne & 0 & 10 & 2 & 8 && \\ \cline {1-8}\rm Sodium &\sf Na & 1 & 11 & 2 & 8 & 1 &\\ \cline {1-8}\rm Magnesium &\sf Mg & 2 & 12 & 2 & 8 & 2 & \\ \cline {1-8}\rm Aluminium &\sf Al & 3 & 13 & 2 & 8 & 3 & \\ \cline{1-8}\rm Silicon &\sf Si & 4 & 14 & 2 & 8 & 4 & \\ \cline {1-8}\rm Phosphorus &\sf P & 3 & 15 & 2 & 8 & 7 & \\ \cline {1-8}\rm Sulphur &\sf S & 2 & 16 & 2 & 8 & 8 & \\ \cline{1-8}\rm Chlorine &\sf Cl & 1 & 17 & 2 & 8 & 8 &1 \\ \cline {1-8}\rm Argon &\sf Ar & 0 & 18 & 2 & 8 & 8 & 2 \\ \cline {1-8}\rm Potassium &\sf K & 1 & 19 & 2 & 8 & 8 &3 \\ \cline {1-8}\rm Calcium &\sf Ca & 2 & 20 & 2 & 8 & 8 & 4 \\ \cline{1-8} \end {array}}$

$\rule{200}2$