Question

Based on the group valency of elements state the formula for the following giving justification for each: -
(a)- Oxides of 1st group elements.
(b)- Halides of the elements of group 13
(c)- Compounds formed when an element of group 2 combines with an element of group 16

Answer 1

Hint: The formula can be formulated by the crisscross method of the valency of the elements. The valency of group 1 is +1. The valency of group 13 is +3. The valency of group 2 is +2. The valency of group 16 is -2.

Complete step by step answer:
The chemical properties and the physical properties of a group are similar. They have common outer electronic configuration due to which they have common valence or oxidation state.
So, with the valency of the compounds, we can formulate the formula of elements of a group with the element of other groups. This can be done by crisscross.
(i)- Oxides of 1st group elements.
The outer electronic configuration of 1st group elements is $n{{s}^{1}}$. So, their valency is 1. The oxygen has a valency of -2. So it can be crisscrossed.

So, it has a formula${{M}_{2}}O$. It is done to complete the octet of the elements.
An example $L{{i}_{2}}O$ is a compound of the oxide of group 1.


(ii)- Halides of the elements of group 13
The outer electronic configuration of group 13 is $n{{s}^{2}}n{{p}^{1}}$ hence, their valency is +3. The halogens have a valency of -1. So, it can be crisscrossed.

Hence, the formula is $M{{X}_{3}}$. It is done to complete the octet of the elements.
An example $B{{F}_{3}}$ is a halide of group 13.

(iii)- Compounds formed when an element of group 2 combines with an element of group 16
The outer electronic configuration of group 2 is $n{{s}^{2}}$ so it has a valency +2 and the outer electronic configuration of group 16 is $n{{s}^{2}}n{{p}^{4}}$ so it has valency -2.
Hence the electrons donated by group 2 will be accepted by group 16, therefore, the formula will be $MO$.
$CaO$ is an example of group 2 with group 16.

Note: The sodium of group forms peroxide with oxygen ($N{{a}_{2}}{{O}_{2}}$ ) and the other metal like potassium, rubidium, and francium forms peroxide with oxygen ($M{{O}_{2}}$ ) because the reactivity of oxygen increases down the group.