Question

For the following reactions which of the following options is correct ? ${\text{Mg}}\,{\text{ + }}\,{{\text{N}}_{\text{2}}}\, \to \,{\text{A}}\,{\text{compound}}$
${\text{A}}\,{\text{ + }}\,6\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,{\text{Mg(OH}}{{\text{)}}_{\text{2}}}\,{\text{ + }}\,{\text{B}}\,\,{\text{gas}}$
A and B are respectively
A.${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}{\text{,}}\,{\text{N}}{{\text{O}}_{\text{2}}}$
B. ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}{\text{,}}\,{\text{NO}}$
C. ${\text{MgO,}}\,{\text{N}}{{\text{H}}_3}$
D. ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}{\text{,}}\,{\text{N}}{{\text{H}}_3}$

Answer 1

To determine A and B we will write the equation of the reaction of magnesium with nitrogen. Then the equation for the reaction of the product with water. Then we will balance the equation.

Complete Step by step answer: Magnesium reacts with nitrogen to give magnesium nitride. So, the product will be ${\text{MgN}}$. Magnesium shows valancey $+ 2$and nitrogen atom shows valancey $- 3$so, the charge is not balanced. So, to balanced charge, we will add three as the subscript for Mg and two as the subscript for N.
So, the product will be ${\text{M}}{{\text{g}}_3}{{\text{N}}_2}$.
The equation for the reaction of magnesium with nitrogen is as follows:
${\text{Mg}}\,{\text{ + }}\,{{\text{N}}_{\text{2}}}\, \to \,{\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}$
At the right side of the equation, one magnesium atom is present, and left side three so, add three as coefficients in front of Mg.
${\text{3}}\,{\text{Mg}}\,{\text{ + }}\,{{\text{N}}_{\text{2}}}\, \to \,{\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}$
So, product A is magnesium nitride ${\text{M}}{{\text{g}}_3}{{\text{N}}_2}$.
Product A means magnesium nitride is reacting with water to give magnesium hydroxide and B which is a gas.
Water dissociate as ${{\text{H}}^{\text{ + }}}$ and ${\text{O}}{{\text{H}}^ - }$, out of these ions, ${\text{O}}{{\text{H}}^ - }$ is reacting with magnesium to give magnesium hydroxide so, left hydronium${{\text{H}}^{\text{ + }}}$ion will reacts with nitrogen. The reaction of nitrogen with hydrogen in $1:3$ gives the product ammonia${\text{N}}{{\text{H}}_{\text{3}}}$.
The equation for the reaction of magnesium nitride with water is as follows:
${\text{M}}{{\text{g}}_3}{{\text{N}}_2}{\text{ + }}\,6\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,{\text{Mg(OH}}{{\text{)}}_{\text{2}}} + \,{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$
On the right side of the equation, three magnesium atoms are present, and left side one so, add three as a coefficient in front of${\text{Mg(OH}}{{\text{)}}_{\text{2}}}$.
${\text{M}}{{\text{g}}_3}{{\text{N}}_2}{\text{ + }}\,6\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,3\,{\text{Mg(OH}}{{\text{)}}_{\text{2}}} + \,{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$
At the right side of the equation, twelve hydrogen atoms are present and left side six so, add two as a coefficient in front of${\text{N}}{{\text{H}}_{\text{3}}}$.
${\text{M}}{{\text{g}}_3}{{\text{N}}_2}{\text{ + }}\,6\,{{\text{H}}_{\text{2}}}{\text{O}}\, \to \,3\,{\text{Mg(OH}}{{\text{)}}_{\text{2}}} + \,2\,{\text{N}}{{\text{H}}_{\text{3}}}{\text{(g)}}$
So, product B is ammonia${\text{N}}{{\text{H}}_{\text{3}}}$.
So, A and B are respectively are${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}{\text{,}}\,{\text{N}}{{\text{H}}_3}$.

Therefore, option (D) ${\text{M}}{{\text{g}}_{\text{3}}}{{\text{N}}_{\text{2}}}{\text{,}}\,{\text{N}}{{\text{H}}_3}$ is correct.

Note: By the reaction, neutral compounds are formed, so charge balancing is necessary. Magnesium is an alkaline earth metal. All alkaline earth metals on reaction with nitrogen form nitrides. Hydrolysis of nitrides gives hydroxides. During balancing a reaction, first we balance the atom which is present in a minimum number. Hydrogen and oxygen get balanced at the last step.