Which of the following elements acts as the best reducing agent? (a) \(\mathrm{Na}\) (b) \(\mathrm{Cl}\) (c) \(\mathrm{Mg}\) (d) \(\mathrm{F}\)

Short Answer

Expert verified
Answer: (a) Na (Sodium)

Step by step solution

01

Determine the electron loss tendencies

We need to look at the periodic table and see where these elements are placed. This way, we can get an idea of their willingness to lose or gain electrons and act as reducing agents. The elements are as follows: (a) Na (Sodium) - Group 1, Alkali metal (b) Cl (Chlorine) - Group 17, Halogen (c) Mg (Magnesium) - Group 2, Alkaline Earth metal (d) F (Fluorine) - Group 17, Halogen
02

Identify the characteristics of the groups

In general, elements in Group 1 and Group 2 have a greater tendency to lose electrons compared to other groups due to their low ionization energies, making them good reducing agents. On the other hand, elements from Group 17 have high electron affinity and gain electrons easily, making them poor reducing agents.
03

Compare elements within the same group

Since Na and Mg belong to Group 1 and Group 2 respectively, we now need to compare their electron loss tendencies. In general, elements in Group 1 have a higher tendency to lose electrons (greater reducing ability) compared to Group 2 elements.
04

Determine the best reducing agent

Based on the comparison of elements and their electron loss tendencies from their respective groups, we can infer that \(\mathrm{Na}\) (Sodium) from Group 1 has the highest tendency to lose electrons and hence, acts as the best reducing agent among the given elements. So, the correct answer is (a) \(\mathrm{Na}\).

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