Question

Sunil studied the characteristics of four metals W, X, Y, and Z, which were exposed to the atmosphere. He noted down the following observations:
$$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{W}& \text{No layer}\\ \text{X}& \text{Greenish layer}\\ \text{Y}& \text{Black layer}\\ \text{Z}& \text{Reddish-brown layer}\end{array}$$
Identify W, X, Y, and Z.

• A. W:- Copper ($\text{Cu}$)
  X:- Silver ($\text{Ag}$)
  Y:- Iron ($\text{Fe}$)
  Z:- Platinum ($\text{Pt}$)
• B. W:- Copper ($\text{Cu}$)
  X:- Iron ($\text{Fe}$)
  Y:- Silver ($\text{Ag}$)
  Z:- Platinum ($\text{Pt}$)
• C. W:- Platinum ($\text{Pt}$)
  X:- Copper ($\text{Cu}$)
  Y:- Silver ($\text{Ag}$)
  Z:- Iron ($\text{Fe}$)
• D. W:- Platinum ($\text{Pt}$)
  X:- Iron ($\text{Fe}$)
  Y:- Silver ($\text{Ag}$)
  Z:- Copper ($\text{Cu}$)

Answer 1

The correct option is C W:- Platinum ($\text{Pt}$)

X:- Copper ($\text{Cu}$)
Y:- Silver ($\text{Ag}$)
Z:- Iron ($\text{Fe}$)
W:- Platinum ($\text{Pt}$): It does react with the atmosphere and it does not form any layer over it.

X:- Copper ($\text{Cu}$): It forms a greenish coating of basic copper carbonate,
(${\text{CuCO}}_3.{\text{Cu(OH)}}_2$), when exposed to ${\text{CO}}_2$, ${\text{O}}_2$ and ${\text{H}}_2\text{O}$. It is called tarnishing of copper.

Y:- Silver ($\text{Ag}$): When silver is exposed to an atmosphere containing hydrogen sulphide (${\text{H}}_2\text{S}$) gas, it forms a black layer of silver sulphide (${\text{Ag}}_2\text{S}$). It is called tarnishing of silver.

Z:- Iron ($\text{Fe}$): It forms a reddish brown layer of rust, (${\text{Fe}}_2{\text{O}}_3.{\text{nH}}_2\text{O}$), when it comes in contact with moisture and oxygen. This is called rusting of iron.