Question

Write down the outermost electronic configuration of alkali metals. How will you justify their placement in group $1$ of the periodic table?

Answer 1

Analysing the electronic configuration of different alkali metals and justifying their correct placement in periodic table:

•Alkali metals belong to group number one which is placed in s- block. The outermost electronic configuration of alkali metals is- $n{s}^1$ .They has one electron in the outermost orbital.

•The elements are placed in the same group on the basis of the outermost electronic configuration. The elements in group number $1$ should have the same configuration.

•The alkali metals are: $Li,Na,K,Rb,Cs$ and $Fr.$

•Alkali metals belong to group number one which is placed in s- block. The outermost electronic configuration of alkali metals is- ns1 .They has one electron in the outermost orbital.

•The elements are placed in the same group on the basis of the outermost electronic configuration. The elements in group number $1$ should have the same configuration.

•The alkali metals are: $Li,Na,K,Rb,Cs$ and $Fr.$

The electronic configurations of these elements can be given as:

$Li\left(3\right)-1s^{2}2s^1$

$Na\left(11\right)-1s^{2}2s^{2}2p^{6}3s^1$

$K\left(19\right)\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^1$

$Rb\left(37\right)-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^1$

$Cs\left(55\right)-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$

$4d^{10}5p^{6}6s^1$

$Fr\left(87\right)-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$

$5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^{6}7s^1$

• As, in all above electronic configuration, they have similar electronic configuration with one electron in outermost shell. Hence, because of this they are placed in group 1.

Conclusion:

As we have seen that an electronic configuration where all alkali metals have 1 electron in their outermost shell that's why they are placed in group 1 in the periodic table.