An element has $ 2 $ electrons in its N shell.
$ (i) $ What is its atomic number $ ? $
$ (ii) $ State its position in the periodic table.
$ (iii) $ Is it a metal or nonmetal $ ? $
$ (iv) $ State the name assigned to this group.
Answer
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Hint: An element contains the shells in order of K, L, M, N. Each shell contains a fixed number of electrons. We will calculate the sum of all these electrons and predict the atomic number of given elements. The position of elements in the group can be predicted by knowing the number of shells and number of valence electrons. If the element has tendency to lose electrons then it will be a metal, otherwise it will be a non-metal.
Complete answer:
The element contains K, L, M, N shells in the order from nucleus to outermost shell. K shell is nearest to the nucleus and the N shell is farthest from the nucleus. Also we know that the K shell can contain a maximum of two electrons. Similarly, the L shell contains eight electrons. M shell can contain 18 electrons. Then according to the question, the N shell contains only two electrons. Therefore we can calculate the total number of electrons present in the element.
Total number of electrons $ = {\text{ 2}}\left( K \right){\text{ + 8}}\left( L \right){\text{ + 18}}\left( M \right){\text{ + 2}}\left( N \right) $
Total number of electrons $ = {\text{ 2 + 8 + 18 + 2}} $
Total number of electrons $ = {\text{ 30}} $
$ (i) $ Thus, we also know that the atomic number of an element is equal to the number of electrons present in the element. Therefore we can say that the atomic number of the given element $ = {\text{ 30}} $ .
$ (ii) $ We can write its electronic configuration as: $ 1{s^2}{\text{, 2}}{{\text{s}}^2}{\text{ , 2}}{{\text{p}}^6}{\text{ , 3}}{{\text{s}}^2}{\text{ , 3}}{{\text{p}}^6}{\text{ , 3}}{{\text{d}}^{10}}{\text{ , 4}}{{\text{s}}^2} $
Thus we can observe that it contains a maximum of four shells. Thus it is a $ {4^{th}} $ period element. Also it contains electrons in $ d - $ subshell. So it is also a $ d - $ block element. Hence we can say that the group of the element is $ {12^{th}} $ . Thus it is present at $ {4^{th}} $ period and the $ {12^{th}} $ group of the periodic table.
$ (iii) $ It is a metal because it can donate a pair of electrons to make $ Z{n^{ + 2}} $ ion. Thus it is treated as metal.
$ (iv) $ The name assigned to the Zinc is $ II{\text{ B}} $ .
Note:
The number of electrons in a shell is given by the rule of $ 2{n^2} $ . Here n is the shell number. This will give us the maximum number of electrons present in a particular shell. For writing the electronic configuration we use $ Aufbau $ Principle. The elements which form cation are known as metals.
Complete answer:
The element contains K, L, M, N shells in the order from nucleus to outermost shell. K shell is nearest to the nucleus and the N shell is farthest from the nucleus. Also we know that the K shell can contain a maximum of two electrons. Similarly, the L shell contains eight electrons. M shell can contain 18 electrons. Then according to the question, the N shell contains only two electrons. Therefore we can calculate the total number of electrons present in the element.
Total number of electrons $ = {\text{ 2}}\left( K \right){\text{ + 8}}\left( L \right){\text{ + 18}}\left( M \right){\text{ + 2}}\left( N \right) $
Total number of electrons $ = {\text{ 2 + 8 + 18 + 2}} $
Total number of electrons $ = {\text{ 30}} $
$ (i) $ Thus, we also know that the atomic number of an element is equal to the number of electrons present in the element. Therefore we can say that the atomic number of the given element $ = {\text{ 30}} $ .
$ (ii) $ We can write its electronic configuration as: $ 1{s^2}{\text{, 2}}{{\text{s}}^2}{\text{ , 2}}{{\text{p}}^6}{\text{ , 3}}{{\text{s}}^2}{\text{ , 3}}{{\text{p}}^6}{\text{ , 3}}{{\text{d}}^{10}}{\text{ , 4}}{{\text{s}}^2} $
Thus we can observe that it contains a maximum of four shells. Thus it is a $ {4^{th}} $ period element. Also it contains electrons in $ d - $ subshell. So it is also a $ d - $ block element. Hence we can say that the group of the element is $ {12^{th}} $ . Thus it is present at $ {4^{th}} $ period and the $ {12^{th}} $ group of the periodic table.
$ (iii) $ It is a metal because it can donate a pair of electrons to make $ Z{n^{ + 2}} $ ion. Thus it is treated as metal.
$ (iv) $ The name assigned to the Zinc is $ II{\text{ B}} $ .
Note:
The number of electrons in a shell is given by the rule of $ 2{n^2} $ . Here n is the shell number. This will give us the maximum number of electrons present in a particular shell. For writing the electronic configuration we use $ Aufbau $ Principle. The elements which form cation are known as metals.
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