What is the simplest formula of the ionic compound formed when magnesium reacts with an element X whose ground state electronic configuration is $1s^2 2s^2 2p^3$?

1. Identify Element X: The electronic configuration $1s^2 2s^2 2p^3$ corresponds to an atomic number of 7 ($2+2+3$), which is Nitrogen (N). Element X belongs to Group 15 of the periodic table .
2. Determine Valency of X: Elements in Group 15 have 5 valence electrons ($ns^2 np^3$). To attain a stable noble gas configuration (octet), element X tends to gain 3 electrons, exhibiting a valency of 3 and forming an anion $X^{3-}$ (Nitride ion) .
3. Determine Valency of Magnesium: Magnesium (Mg) is an alkaline earth metal (Group 2) with valence shell configuration $3s^2$. It loses 2 electrons to form a stable cation $Mg^{2+}$ .
4. Formulate Compound: To form a neutral ionic compound, the total positive charge must equal the total negative charge. By the criss-cross method, the valency of Mg (2) becomes the subscript of X, and the valency of X (3) becomes the subscript of Mg.

• Cation: $Mg^{2+}$
• Anion: $X^{3-}$ (or $N^{3-}$)
• Formula: $Mg_3X_2$ (Checking charges: $3 \times (+2) + 2 \times (-3) = 0$).

This is consistent with the formation of magnesium nitride ($Mg_3N_2$) as shown in the redox reaction section of the NCERT text: $3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$ .