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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 . . . . . . .

Similar Questions

The incorrect statement among the following is:

The correct order of the first ionization enthalpies of the following elements is:

Ionization energy of gaseous $Na$ atoms is $495.5 \ kJ \ mol^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is $(h = 6.626 \times 10^{-34} \ J \ s, N_A = 6.022 \times 10^{23} \ mol^{-1})$.

Which of the following statements is incorrect?

Which of the following has the least ionization potential?

Vedclass Test Series

Exam Paper Generator

Online Exam Module