Solveeit Logo
Login

Question

Question: Correct formula for a trivalent metal nitride is ________. A. \({M_3}{N_2}\) B. \({M_3}{N_3}\) ...

Correct formula for a trivalent metal nitride is ________.
A. M3N2{M_3}{N_2}
B. M3N3{M_3}{N_3}
C. MNMN
D. M2N3{M_2}{N_3}

Explanation

Solution

A trivalent metal cation means the metal in its ground state loses three electrons to oxidize to a +3 oxidation state. The metal which loses three electrons must have undergone, simultaneously, three ionization enthalpy heating processes. The nitride ion is a trivalent anion with an overall charge of -3 on the nitrogen atom to complete its octet.

Complete step by step answer:

The metal nitride compound formation takes place through two steps. They are:
(i) Metal cation formation: A metal forms a cation when it accepts ionization energy of certain value in its neutral gaseous isolated atomic form to lose an electron to get converted into a positively charged ion. A trivalent metal cation is formed when it undergoes simultaneous ionization thrice. The reaction is as follows:
M→I.E1M++e−→I.E2M2++e−→I.E3M3++e−M\xrightarrow{{I.{E_1}}}{M^ + } + {e^ - }\xrightarrow{{I.{E_2}}}{M^{2 + }} + {e^ - }\xrightarrow{{I.{E_3}}}{M^{3 + }} + {e^ - }
Thus, the overall reaction of formation of a trivalent metal cation is:
M→(IE)totalM3++3e−M\xrightarrow{{{{(IE)}_{total}}}}{M^{3 + }} + 3{e^ - }
The examples of trivalent cations include Al3+,Ga3+,etc.A{l^{3 + }},G{a^{3 + }},etc.
(ii) Nitride anion formation: The electron affinity is the amount of energy released when a neutral gaseous isolated atom gains an electron to get converted into a negative charged anion. Thus, when a nitrogen atom is present in a neutral isolated form and gains an electron, it gets converted into a nitride anion. The reaction follows as:
N+e−→(E.A)1N−+e−→(E.A)2N2−+e−→(E.A)3N−3N + {e^ - }\xrightarrow{{{{(E.A)}_1}}}{N^ - } + {e^ - }\xrightarrow{{{{(E.A)}_2}}}{N^{2 - }} + {e^ - }\xrightarrow{{{{(E.A)}_3}}}{N^{ - 3}}
Thus, the overall reaction of formation of a nitride anion is:
N+3e−→(EA)totalN3−N + 3{e^ - }\xrightarrow{{{{(EA)}_{total}}}}{N^{3 - }}
Now, when the trivalent metal cation and the nitride anion combine with each other, they form the following compound:
M3++N3−→MN{M^{3 + }} + {N^{3 - }} \to MN
Thus, the correct option is C. MNMN

Note:
The ionization enthalpy and the electron gain enthalpy or electron affinity are just two energies which are opposite to each other. The greater the value of ionization enthalpy, the lower is the value of electron gain enthalpy for the same atom.