How many electrons can be held in the second orbital
2 answers:
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Answer:
Explanation:
One way to calculate the lattice energy is to use Hess's Law.
The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:
Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?
We must generate this reaction rom the equations given.
(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹
(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹
(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹
(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹
(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹
Now, we put these equations together to get the lattice energy.
<u>E/kJ </u>
(5) Li⁺(g) +e⁻ ⟶ Li(g) 520
(6) Li(g) ⟶ Li(s) -161
(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409
(8) Cl(g) ⟶ ½Cl₂(g) -121.5
(9) Cl⁻(g) ⟶ Cl(g) + e⁻ <u>+349</u>
Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862
The lattice energy of LiCl is .