
Explain the element tin and mention its electronic configuration?
Answer
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Hint: For writing the electronic configuration of any atom we have to know about the atomic number of that atom as atomic number (${\text{Z}}$) of any element is equal to the number of protons or number of electrons present in that atom.
Complete answer:
As we know that in the periodic table tin (${\text{Sn}}$) is situated in the ‘IV A’ group of the p – block and it is denoted as ${}_{{\text{50}}}^{{\text{118}}{\text{.7}}}{\text{Sn}}$ in the table.
-Upper quantity which is mention in ${}_{{\text{50}}}^{{\text{118}}{\text{.7}}}{\text{Sn}}$ describes the Atomic mass (${\text{A}}$) and lower quantity describes the Atomic number (${\text{Z}}$) of the gallium element.
-As we know that relation between Atomic number (${\text{Z}}$) and number of electrons present inside an atom are as follow:
Atomic Number (${\text{Z}}$) = No. of electrons = No. of protons
So that in tin (${\text{Sn}}$) $50$ electrons are present as the atomic number of gallium is also $50$.
-Now electronic configuration of Gallium is written as ${\text{1}}{{\text{s}}^{\text{2}}}{\text{,2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{,3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{,4}}{{\text{s}}^{\text{2}}}{\text{4}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{d}}^{{\text{10}}}}{\text{,5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{2}}}$($2,8,18,18,4$) or $\left[ {{\text{Kr}}} \right]{\text{4}}{{\text{d}}^{{\text{10}}}}{\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{2}}}$ and from this it is clear that in the outermost shell four electrons are present.
Note:
Here some of you may think that in the electronic configuration krypton atoms are used, so the reason is that krypton is a noble gas whose atomic number is $36$ and till its outermost shell it bears stable configuration. That’s why for shorthand notation we will use a noble gas symbol in place of $36$ electrons.
Complete answer:
As we know that in the periodic table tin (${\text{Sn}}$) is situated in the ‘IV A’ group of the p – block and it is denoted as ${}_{{\text{50}}}^{{\text{118}}{\text{.7}}}{\text{Sn}}$ in the table.
-Upper quantity which is mention in ${}_{{\text{50}}}^{{\text{118}}{\text{.7}}}{\text{Sn}}$ describes the Atomic mass (${\text{A}}$) and lower quantity describes the Atomic number (${\text{Z}}$) of the gallium element.
-As we know that relation between Atomic number (${\text{Z}}$) and number of electrons present inside an atom are as follow:
Atomic Number (${\text{Z}}$) = No. of electrons = No. of protons
So that in tin (${\text{Sn}}$) $50$ electrons are present as the atomic number of gallium is also $50$.
-Now electronic configuration of Gallium is written as ${\text{1}}{{\text{s}}^{\text{2}}}{\text{,2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{,3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{,4}}{{\text{s}}^{\text{2}}}{\text{4}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{d}}^{{\text{10}}}}{\text{,5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{2}}}$($2,8,18,18,4$) or $\left[ {{\text{Kr}}} \right]{\text{4}}{{\text{d}}^{{\text{10}}}}{\text{5}}{{\text{s}}^{\text{2}}}{\text{5}}{{\text{p}}^{\text{2}}}$ and from this it is clear that in the outermost shell four electrons are present.
Note:
Here some of you may think that in the electronic configuration krypton atoms are used, so the reason is that krypton is a noble gas whose atomic number is $36$ and till its outermost shell it bears stable configuration. That’s why for shorthand notation we will use a noble gas symbol in place of $36$ electrons.
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