Calculate the energy required to convert all atoms in (4.8 ~g ) of ( Mg ) to ( Mg ²⁺ ) in the vapour state. ( IE ₁ ) and ( IE ₂ ) of ( Mg ) are (740 ~kJ / mol ) and (1450 ~kJ / mol ) respectively.

Step-by-step solution

For 1 mole of ( ( Mg Mg ²⁺ ) ) ( IE = E E ₁+ IE ₂=(740+1450)=2190 ~kJ / mol ). Number of mole in (4.8 ~g ) of ( Mg =4.8 / 24=0.2 ~mol ) For 1 mole energy required (=2190 ~kJ / mol ) For 0.2 mole energy required (=2190 0.2=438 ~kJ ) Thus, for (4.8 ~g ) of ( Mg ) to ( Mg ²⁺ ) conversion, energy required is (438 ~kJ ). Hence, the correct option is (d).