Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?
A.
$$3{d^3},4{s^2}$$
B.
$$3{d^5},4{s^1}$$
C.
$$3{d^5},4{s^2}$$
D.
$$3{d^2},4{s^2}$$
Answer :
$$3{d^5},4{s^2}$$
Solution :
The sum of number of electrons (unpaired) in $$d$$-orbitals and number of electrons in $$s$$-orbital gives the number of oxidation states $$(os)$$ exhibited by a $$d$$-block element. Therefore,
$$\eqalign{
& \left( {\text{A}} \right)3{d^3},4{s^2} \Rightarrow OS = 3 + 2 = 5 \cr
& \left( {\text{B}} \right)3{d^5},4{s^1} \Rightarrow OS = 5 + 1 = 6 \cr
& \left( {\text{C}} \right)3{d^5},4{s^2} \Rightarrow OS = 5 + 2 = 7 \cr
& \left( {\text{D}} \right)3{d^2},4{s^2} \Rightarrow OS = 2 + 2 = 4 \cr} $$
Hence, element with $$3{d^5},4{s^2}$$ configuration exhibits largest number of oxidation states.
Releted MCQ Question on Inorganic Chemistry >> Classification of Elements and Periodicity in Properties
Releted Question 1
The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is