91Ó°ÊÓ

In chemistry, reducing strength refers to the ability of a substance to donate electrons during a chemical reaction. The stronger a reducing agent, the more easily it loses electrons to other substances.

In our exercise, metal D forms hydrogen gas bubbles when placed in hot water. This means that metal D is a strong enough reducing agent to reduce water by giving electrons, thus releasing hydrogen gas.

Next, we saw that metal D could not react with a salt solution of metal E, indicating that metal D is a weaker reducing agent than metal E.

Finally, since metal D gets coated and discolored in a salt solution of metal F, metal F is a stronger reducing agent than metal D. By placing E in a solution of F's salt, no reaction happens, showing that metal F is still stronger.

Displacement reactions happen when a more reactive metal displaces a less reactive metal from its compound. These reactions tell us a lot about the reactivity and reducing strength of metals.

When metal D is placed in hot water, it displaces the hydrogen, indicating its reactivity.
In the scenario with metal E's salt, no reaction means metal D cannot displace E, showing E is more reactive.

The discoloration of metal D in F's salt solution demonstrates a displacement reaction where metal F is clearly more reactive and reduces D ions by taking their place.

The metal reactivity series helps us understand how different metals will behave in reactions. This series arranges metals in order of decreasing reactivity.

In the provided exercise:

• Metal D reacts with hot water and releases hydrogen, showing it’s quite reactive.
• Metal D does not cause a reaction with metal E's salt, which means E is less reactive than D.
• Metal F is highly reactive as evidenced by displacing D’s ions resulting in discoloration.
• When metal E is placed in the salt of F, no reaction occurs.

Thus, the overall reactivity order from least to most would be D < E < F.