Consider an ionic compound, MX3 , composed of generic metal M and generic gaseous halogen X. The enthalpy of formation of MX3 is Δ∘f=−789 kJ/mol.
The enthalpy of sublimation of M is Δsub=179 kJ/mol.
The first, second, and third ionization energies of M are IE1=585 kJ/mol, IE2=1643 kJ/mol, and IE3=2719 kJ/mol .

The electron affinity of X is ΔEA=−333 kJ/mol . (Refer to the hint).

The bond energy of X2 is BE=245 kJ/mol.
Determine the lattice energy of MX3.

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answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

the answer your looking for is c).