Calculate second electron affinity of oxygen for the process, O ⁻( g )+e⁻( g ) O ²⁻( g ) by using the following data (i) Heat of sublimation of Mg (s)=+147.7 ~kJ ~mol ⁻¹ (ii) Ionisation energy of Mg (g) to form Mg ²⁺(g)=+2189.0 ~kJ ~mol ⁻¹ (iii) Bond dissociation energy for O ₂=+498.4 ~kJ ~mol ⁻¹ (iv) First electron affinity of O ( g )=-141.0 ~kJ ~mol ⁻¹ (v) Heat formation of MgO =-601.7 ~kJ ~mol -1 (vi) Lattice ener

Step-by-step solution

Stepwise formation of MgO involves array l|l|l Step & |c| Reaction & H in kJ mol ⁻¹ (i) & Mg ( s ) Mg ( g ) & 147.7 (ii) & Mg (g) Mg ²⁺+2 e ⁻ & 2189.0 (iii) & 1 2 O ₂(g) O ( g ) & 498.4 / 2 (iv) & O (g)+1 e⁻ O ⁻(g) & -141.0 (v) & O ⁻(g) t e⁻ O ²⁻(g) & q (vi) & Mg ²⁺(g)+ O ²⁻(g) MgO (s) & -3791.0 array Mg ( s )+ 1 2 O ₂(g) MgO ( s ) H =-1346.1+q Mg ( s )+ 1 2 O ₂(g) MgO ( s ) H₂=-601.7 By Born Herber cycle (based on Hes s law) H₁= H₂-1346.1+q=-601.7q=744.4 Jmol ⁻¹