Which one of the element with the following outer orbital configuration may exhibit the largest number of oxidation states?
-
3d5,4s2
-
3d5,4s1
-
3d5,4s2
-
3d2,4s2
C.
3d5,4s2
A number of oxidation states exhibited by d-block elements is the sum of the number of electrons (unpaired) in d-orbitals and number of electrons in s - orbital.
a) 3d3, 4s2 ⇒ Oxidation state = 3+2 = 5
b) 3d5, 4s1 ⇒ Oxidation state = 5+1 = 6
c) 3d5, 4s2 ⇒ Oxidation state = 5+2 = 7
d) 3d4, 4s2 ⇒ Oxidation state = 2+ 2 = 4
Hence, an element with 3d5, 4s2 configuration exhibits the largest number of oxidation states.