The 1^{st} ionization enthalpy for Mg is +737 | vedclass

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Question

(C) The electronic configuration of $Mg$ is $[Ne]3s^2$. Removing the first electron requires $737 	ext{ kJ/mol}$ to form the $Mg^+$ ion with configur

Vedclass · Accepted Answer

+$1450$ kJ/mol

Vedclass · Answer

(C) The electronic configuration of $Mg$ is $[Ne]3s^2$. Removing the first electron requires $737 	ext{ kJ/mol}$ to form the $Mg^+$ ion with configuration $[Ne]3s^1$. The second ionization enthalpy involves removing the second electron from the $Mg^+$ ion. Since the electron is being removed from a positively charged species, the electrostatic attraction is stronger, requiring significantly more energy than the first ionization. Therefore, the $2^{nd}$ ionization enthalpy must be greater than the $1^{st}$ ionization enthalpy $(737 	ext{ kJ/mol})$ and must be positive. Among the given options, $+1450 	ext{ kJ/mol}$ is the only value that is greater than $737 	ext{ kJ/mol}$ and positive, making it the most probable experimental value.

The $1^{st}$ ionization enthalpy for $Mg$ is $+737 \text{ kJ/mol}$. The most probable estimated value of the $2^{nd}$ ionization enthalpy of $Mg$ is . . . . . . .

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