Question

An element X(2,8,2) combines separately with $N{{O}_{3}}$ and ${{\left( S{{O}_{4}} \right)}_{2}}$, ${{\left( P{{O}_{4}} \right)}_{3}}$ radicals. Write the formulae of the three compounds so formed. To which group of the periodic table does the element ‘X’ belong? Will it form a covalent or ionic compound? Why?

Answer 1

Hint: From the electronic configuration, we can see that the element X has 2 electrons in the valence shell. So it will react with the given ions to achieve a noble gas electronic configuration state to become stable and unreactive.

Complete answer:
-First of all let us discuss the element X given in the question. Its electronic configuration is given. According to it, there are 2 electrons in the outermost shell of the atom which is called valence shell. Thus, it belongs to group 2 of the periodic table.
The configuration can be understood as $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$
-Counting the total number of electrons from the given electronic configuration, we see that the answer is 2+8+2=12. As there are a total 12 electrons in the element X, its atomic number is also 12. This means that the element X is none other than Magnesium.
-As the valence electrons are 2, it becomes easier for the atom to lose its electrons to achieve the noble gas configuration and this is why such elements are called metals. Metals are those elements which lose their electrons to become stable. Group 2 elements are called alkali earth metals.
-As the element becomes stable by losing its electrons, the type of bond formed will be an ionic bond as it involves the generation of ions- cation and anion. Covalent bonds are formed when the atoms complete their octet by sharing electrons, not by losing/gaining electrons from other atoms.
-Now coming to the reaction of Magnesium with different ions given.
We know that the valency of Mg is 2. If the valencies are the same, then the ions react in equal proportions to form a molecule or a compound. If not, then they follow criss-cross rules. In that, the valencies appear in the subscript of the other ion.
-To complete the reaction, we must first find out how many electrons the given ions lose.
$N{{O}_{3}}$loses 1 electron and becomes $N{{O}_{3}}^{1-}$
${{\left( S{{O}_{4}} \right)}_{2}}$ loses 2 electron and becomes $S{{O}_{4}}^{2-}$
${{\left( P{{O}_{4}} \right)}_{3}}$ loses 3 electron and becomes $P{{O}_{4}}^{3-}$
-Now by criss-cross method, we can find the reaction products.
     \[\begin{align}

& M{{g}^{2+}}\searrow N{{O}_{3}}^{1-} \\
& \text{ 1} \quad \quad \quad \quad {2} \\
\end{align}\]
$\begin{align}
& M{{g}^{2+}}\searrow S{{O}_{4}}^{2-} \\
& \text{ 2} \quad \quad \quad \quad {2} \\
\end{align}$

$\begin{align}
& M{{g}^{2+}}\searrow P{{O}_{4}}^{3-} \\
& \text{ 3}\quad \quad \quad \quad {2} \\
\end{align}$
Thus the reactions can be written as
     \[Mg+S{{O}_{4}}\to MgS{{O}_{4}}\]
     \[Mg+2N{{O}_{3}}\to Mg{{\left( N{{O}_{3}} \right)}_{2}}\]
     \[3Mg+2P{{O}_{4}}\to M{{g}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}\]
Thus, we study about the element X, its type of bond and the reactions with different ions.

Note:
The numbers shown in the reactant side of the reactions are just for balancing the equations. They tell us the ratios in which the reactants need to be combined for the reaction to take place. The numbers on the product side give us the actual product formed and it cannot be changed.