Question

A non-metal ‘M’ forms ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$ but does not form ${\text{M}}{{\text{I}}_{\text{6}}}$. The incorrect statement regarding non-metal ‘M’ is-
A.M can form multiple bond
B.M is of second period element
C.Atomicity of non-metal is 4
D.The range of oxidation number for M is +5 to -3

Answer 1

In the compounds ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$, determine the oxidation number of ‘M’. Also, determine the structures of ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$ so that we can determine the bonds formed by the non-metal.

Complete step by step answer:
We are given compounds formed by non-metal ‘M’ which are ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$.
In compounds ${\text{MC}}{{\text{l}}_{\text{3}}}$, we can see that non-metal ‘M’ forms three bonds with three chlorine atoms, in compound ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ non-metal ‘M’ forms more than one bond with the five oxygen atoms and in compound ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$ non-metal ‘M’ forms more than one bond with the three magnesium atoms.
We can determine the valency of non-metal ‘M’ by drawing the structures of the compounds ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$.
The structures of ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$ are as follows:

From the structures we can see that the non-metal ‘M’ forms multiple bonds. Thus, the statement ‘M can form multiple bonds’ is correct. Thus, option (A) is not correct.
The compounds formed by the non-metal ‘M’ are ${\text{MC}}{{\text{l}}_{\text{3}}}$, ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ and ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$.
The oxidation state of the non-metal ‘M’ in ${\text{MC}}{{\text{l}}_{\text{3}}}$ is +3. The oxidation state of the non-metal ‘M’ in ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is +5. And the oxidation state of the non-metal ‘M’ in ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{M}}_{\text{2}}}$ is -3
Thus, the statement ‘the range of oxidation numbers for M is +5 to -3’ is correct. Thus, option (D) is correct.
If the non-metal ‘M’ is a second period element then the atomicity of the non-metal ‘M’ is 2.
Thus, the statement ‘M is of second period element’ is correct. Thus, option (B) is not correct.
Thus, the statement ‘atomicity of non-metal is 4’ is incorrect.

Thus, the correct option is option (C).

Note:
The number of atoms that make up or compose its molecule is known as atomicity. The non-metal ‘M’ is a second period element and thus, the atomicity of the non-metal ‘M’ is 2. Thus, the molecule of non-metal is ${{\text{M}}_2}$.