Question

Which of the following transitions involves maximum amount of energy?
A. $$M^{-}(g)\stackrel{}{\to }M(g)$$
B. $$M(g)\stackrel{}{\to }{M}^{+}(g)$$
C. $$M^{+}(g)\stackrel{}{\to }{M}^{2+}(g)$$
D. $$M^{2+}(g)\stackrel{}{\to }{M}^{3+}(g)$$

Answer 1

Hint: Here, every reaction will take some energy to be performed, we need to check which reaction takes the most amount of energy to perform. Thus, relative energy of the reactions will be important to check in this question.

Complete answer:
The correct answer to this question is option D, the reaction $$M^{2+}(g)\stackrel{}{\to }{M}^{3+}(g)$$ involves maximum amount of energy to perform. Let us see how by seeing every equation and comparing the energies required for them to be performed.
The reaction given in option A, that is $$M^{-}(g)\stackrel{}{\to }M(g)$$, shows that the anion of an element called M is getting converted back to its neutral self. So, here we are taking one electron from the natural state of that element.
Thus, in this reaction not much energy will have to be supplied for it to be performed as we are taking the atom of the element back to its original and natural state.
Now, the reaction in option B, that is $$M(g)\stackrel{}{\to }{M}^{+}(g)$$, shows that we are taking one electron from the outer orbit of the atom of that element from its natural state. Thus, it is implied that the energy that we have to supply for this reaction to be performed will definitely be more than that in the first reaction.
Now, the reaction in option C, that is $$M^{+}(g)\stackrel{}{\to }{M}^{2+}(g)$$, shows that we are taking one electron out of the cation of the atom of that element. Thus, it shows that we are further taking out one electron from the outermost orbit of the atom of that element.
Thus, more energy will have to be applied for this reaction to occur as the attraction force of nucleus on $$M^{+}$$ will be more than that on $$M$$. So, the energy that has to be applied for this reaction to perform will be more than that of option B.
Now, the reaction in option D, that is $$M^{2+}(g)\stackrel{}{\to }{M}^{3+}(g)$$, shows that we are further taking one electron from the outermost orbit of the cation$$M^{2+}$$. Also, the attraction energy of the nucleus on the outermost orbit of $$M^{2+}$$ will be more than that of $$M^{+}$$.
Therefore, it is clear that the energy that has to be applied in the reaction of option D will be more than that in option C. So, the energy that has to be applied for the reaction of option D to occur will be the largest amongst the three.
So the correct answer is option D.

Note:
Another approach to this question might be through the concept of ionisation energy. Let us suppose that the ionisation of the reaction in option B is $$I.E{.}_1$$, similarly that of the reaction in option C and D is $$I.E{.}_2$$ and $$I.E{.}_3$$ respectively. And it is known that the successive ionisation energy of any element increases. Thus, the ionisation energy of the third transition will be more than second and that of the second will be more than the first. Therefore, $$I.E{.}_3$$ > $$I.E{.}_2$$ > $$I.E{.}_1$$ .