Question

$I{P}_1$ and $I{P}_2$ of Mg are 178 and 348 K cal ${mol}^{-1}$. The energy required for the reaction Mg $\to$ ${Mg}^{2+}$ + 2$e^{-}$ is

• A. 435 K cal
• B. 536 K cal
• C. 635 K cal
• D. 178 K cal

Answer 1

The correct option is B

536 K cal

$I{P}_1$ = 178 K cal ${mol}^{-1}$

$I{P}_2$ = 348 K cal ${mol}^{-1}$

Now, take a look at the reaction

Mg $\to$ ${Mg}^{2+}$ + 2$e^{-}$

So, to take out two electrons from Mg, we have to provide energy.

Therefore, total energy required will be

= $I{P}_1$ + $I{P}_2$

= 178 + 348

= + 526 K. cal