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NEET CHEMISTRYEasy

What is the simplest formula of the ionic compound formed when magnesium reacts with an element X whose ground state electronic configuration is 1s22s22p31s^2 2s^2 2p^3?

A

Mg2X3Mg_2X_3

B

MgX2MgX_2

C

Mg2XMg_2X

D

Mg3X2Mg_3X_2

Step-by-Step Solution

  1. Identify Element X: The electronic configuration 1s22s22p31s^2 2s^2 2p^3 corresponds to an atomic number of 7 (2+2+32+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 (ns2np3ns^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 X3X^{3-} (Nitride ion) .
  3. Determine Valency of Magnesium: Magnesium (Mg) is an alkaline earth metal (Group 2) with valence shell configuration 3s23s^2. It loses 2 electrons to form a stable cation Mg2+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: Mg2+Mg^{2+}
  • Anion: X3X^{3-} (or N3N^{3-})
  • Formula: Mg3X2Mg_3X_2 (Checking charges: 3×(+2)+2×(3)=03 \times (+2) + 2 \times (-3) = 0).

This is consistent with the formation of magnesium nitride (Mg3N2Mg_3N_2) as shown in the redox reaction section of the NCERT text: 3Mg(s)+N2(g)Mg3N2(s)3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s) .

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