Arrange the following transition metals in increasing order of their highest oxidation state exhibited by them:
A. Co
B. Mn
C. Cu
D. Cr
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C, A, D, B
In the first transition series, the highest oxidation state of an element up to manganese equals the total number of 4s and 3d electrons, because all of them can take part in bonding. After manganese, the 3d electrons start pairing and are held more firmly, so the highest oxidation state falls again.
Working it out for each metal from its configuration:
(A) Co (Z = 27): [Ar]3d74s2. The paired d electrons are not all used; its oxidation states are +2, +3 and at most +4.
(B) Mn (Z = 25): [Ar]3d54s2. All 2 + 5 = 7 electrons can be used: highest state +7 (as in \( \ce{MnO4^{-}} \)).
(C) Cu (Z = 29): [Ar]3d104s1. The 3d subshell is full; it shows +1 and at most +2.
(D) Cr (Z = 24): [Ar]3d54s1. All 1 + 5 = 6 electrons can be used: highest state +6 (as in \( \ce{Cr2O7^{2-}} \)).
Increasing order of highest oxidation state:
Cu (+2) < Co (+4) < Cr (+6) < Mn (+7)
NCERT (2026-27 edition):
Oxidation States – …
