What is the ∆G for the following reaction under nonstandard conditions where the following data is given? 2NH4Cl(s) + CaO(s) CaCl2(s)+ H2O + 2NH3(g)

Given:
CaO: ∆Hf = -635.09 kJ, ∆Sf =39.75 J/K
CaCl2: ∆Hf = -795.8 kJ, ∆Sf =104.6 J/K.
NH3: ∆Hf = -46.11 kJ, ∆Sf = 192.45,
H2O: ∆Hf = -285.830 kJ, ∆Sf = 69.91 ,
NH4Cl: ∆Hf = -314.43 kJ, ∆Sf = 94.6
A.
2.27 J
B.
-4.54 J
C.
4.54 J
D.
9.08 J
E.
-9.08 J

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.