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Answer to Question #2632 in Inorganic Chemistry for Mandy

Question #2632
Write down the Born-Haber cycle for SrBr2 formation.
Expert's answer
1. Sr(s) → Sr(g) + VBr2(g) → 2Br(g) + B
2. Sr(g) → Sr(g)+ + e- + IEM1
Sr+(g) → Sr(g)2++ e- + IEM2
Br + e → Br- - EAx
3. Sr2+ + 2Br-→ SrBr2 + UL
4. Br2(g) + Sr(s) → SrBr2 + ∆Hf
Therefore,
∆Hf = V + B + IEM1 + IEM2 - EAx + UL
V is the enthalpy of sublimation for metal atoms
B is the bond energy of halogen atom
IEM is the ionization energy of the metal atom
EAX is the electron affinity of halogen atom
UL is the lattice energy (defined as exothermic here)

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