
The element whose group number and period number is equal is:
a) $ Sodium $
b) $ Beryllium $
c) $ Boron $
d) $ Potassium $
Answer
530.1k+ views
Hint: The Group number can be calculated by knowing the electronic configuration of elements as the valence electron number gives the group number. Period number can be calculated by knowing about the highest occupied orbital in the electronic configuration.
Complete step by step solution:
In the periodic table the vertical columns are called as the groups and the horizontal lines are called the periods.
There are eighteen groups and seven periods in the modern periodic table.
$ Sodium $ Metal has the electronic configuration - $ 1{s^2}2{s^2}2{p^6}3{s^1} $ . Hence the group number of this element is one and the period number is three. As the element belongs to the s-block family.
$ Beryllium $ Has the electronic configuration - $ 1{s^2}2{s^2} $ . Hence the group number of this element is two and the period number is two. It has an equal number of groups and periods in the periodic table. As the element belongs to the s-block family.
$ Boron $ Has the electronic configuration - $ 1{s^2}2{s^2}2{p^1} $ . Hence the period number is two and the group number is thirteen. As the element belongs to the p-block family.
$ Potassium $ Has the electronic configuration - $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1} $ . Hence the period number is two and the group number is four. As the element belongs to the s-block family.
Hence as we know the elements of group two are known as the alkaline earth metals, so $ Beryllium $ is the first member of its group. It has a small size and high ionization energy. The group number and period number are equal, i.e. two.
The electronegativity of this group is low but in most cases $ Beryllium $ forms the ionic compound. For example, the electronegativity difference in $ Be{F_2} $ compounds is very high.
Thus the correct option is (b).
Note:
The group $ 2 $ atoms which are alkaline earth metals are large but smaller than group $ 1 $ atoms which are alkali metals. This is because as the number of electrons increases in the outer orbital, the effective nuclear charge increases which attracts the outer shell closer and the size of the atom decreases.
Complete step by step solution:
In the periodic table the vertical columns are called as the groups and the horizontal lines are called the periods.
There are eighteen groups and seven periods in the modern periodic table.
$ Sodium $ Metal has the electronic configuration - $ 1{s^2}2{s^2}2{p^6}3{s^1} $ . Hence the group number of this element is one and the period number is three. As the element belongs to the s-block family.
$ Beryllium $ Has the electronic configuration - $ 1{s^2}2{s^2} $ . Hence the group number of this element is two and the period number is two. It has an equal number of groups and periods in the periodic table. As the element belongs to the s-block family.
$ Boron $ Has the electronic configuration - $ 1{s^2}2{s^2}2{p^1} $ . Hence the period number is two and the group number is thirteen. As the element belongs to the p-block family.
$ Potassium $ Has the electronic configuration - $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1} $ . Hence the period number is two and the group number is four. As the element belongs to the s-block family.
Hence as we know the elements of group two are known as the alkaline earth metals, so $ Beryllium $ is the first member of its group. It has a small size and high ionization energy. The group number and period number are equal, i.e. two.
The electronegativity of this group is low but in most cases $ Beryllium $ forms the ionic compound. For example, the electronegativity difference in $ Be{F_2} $ compounds is very high.
Thus the correct option is (b).
Note:
The group $ 2 $ atoms which are alkaline earth metals are large but smaller than group $ 1 $ atoms which are alkali metals. This is because as the number of electrons increases in the outer orbital, the effective nuclear charge increases which attracts the outer shell closer and the size of the atom decreases.
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