Four metals A, B, C and D were taken in five different solutions of silver nitrate, zinc sulphate, copper sulphate and iron sulphate and the observations have been listed below:
Reactivity order of metals A, B, C and D is:
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Reactivity of a metal \(\propto\) Number of metals it can displace from their salt solutions.
- Metal B displaces 3 metals.
- Metal A displaces 2 metals.
- Metal C displaces 1 metal.
- Metal D displaces 0 metals.
Hence, the order is directly B $\gt $ A $\gt $ C $\gt $ D.
Step 1: Focus on who displaces iron, the toughest test here. Only Metal-B is able to displace iron out of iron sulphate solution. Displacing iron needs more reactivity than displacing silver, zinc or copper, so Metal-B must sit at the very top of the reactivity order. Step 2: Look at who fails to displace anything at all. Metal-D cannot displace silver, zinc, copper or iron from any of their solutions. A metal that cannot push out even silver, the easiest of these to displace, has to be the weakest of the four, placing Metal-D at the very bottom. Step 3: Rank the remaining two, A and C, by how many metals they can displace. Metal-A manages to displace both silver and copper, two successful displacements. Metal-C only manages to displace silver, just one successful displacement. Since displacing more ions signals higher reactivity, Metal-A must rank above Metal-C. Step 4: Put it all together. From strongest to weakest, the order is Metal-B, then Metal-A, then Metal-C, then Metal-D. \[ \boxed{B > A > C > D} \]
