Question

Which of the following elements show variable valencies?
(A) Tin
(B) Iron
(C) Copper
(D) All of the above

Answer 1

Hint:. To solve this question we should know the nature of each compound. It is important to be aware of when an element is said to have variable valencies. Understanding these two points helps us to come to a conclusion within no time.

Complete step by step answer:
Valency is the possible number of bonds an element can share with other elements or other molecules. Always an atom loses or gains an electron in order to attain the stable configuration this is noble gas configuration. The atomic number is 3 and it has only 1 electron in the valence shell; it readily loses an electron to attain stable configuration rather than gaining 7 electrons. Likewise some electrons prefer to gain over losing electrons. When an element loses an electron it possesses a positive charge on it. Some elements with higher atomic number exhibit variable valencies as they can readily lose or gain electrons to form stable configuration; these types of elements form variable valencies.

Option A is tin. Tin is a p-block element. The atomic number of tin is 50 and its electronic configuration is $[Kr]4{d^{10}}5{s^2}5{p^2}$. The tin shows +4 and +2 oxidation state. The elements of the boron family show these variable valencies (oxidation number +2 and +4). Hence option A is the correct answer.

Option B is Iron. Iron is a d-block element. The atomic number of iron is 26 and its electronic configuration is $[Ar]3{d^6}4{s^2}$. Iron shows both +3 oxidation state and +2 oxidation state. When it forms ferrous oxide its oxidation state will be in +2 states whereas when it forms ferric oxide its oxidation state will be in +3 oxidation state. Hence it is evident that iron exhibits variable valencies. Therefore option B is the correct answer.

Option C is copper. Copper is also a d-block element. The atomic number is 29 and its electronic configuration is $[Ar]3{d^{10}}4{s^1}$. Copper shows +1 oxidation state when cuprous oxide is formed and shows +2 oxidation states when cupric oxide is formed. Hence option C is the correct answer.
So, the correct answer is “Option D”.

Note: The members of the boron family show both +2 oxidation state and +4 oxidation states. d-block elements exhibit variable valencies due to which their components are coloured. Almost all the elements show variable valencies. S-block elements don’t show variable valencies property at all.