NEETd and f-Block ElementsHard
Question
For the four successive transition elements (Cr, Mn, Fe and Co), the stability of + 2 oxidation state will be there in which of the following order ?
(At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)
(At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)
Options
A.Fe > Mn > Co > Cr
B.Co > Mn > Fe > Cr
C.Cr > Mn > Co > Fe
D.Mn > Fe > Cr > Co
Solution
This can be understood on the basis of Eo values for M2+/M.
Eo / V Cr Mn Fe Co
M2+/M - 0.90 - 1.18 - 0.44 - 0.28
Eo values for Mn is more negative than expected from general trend due to extra stability of half-filled Mn2+ ion.
THus the correct order should be
Mn > Cr > Fe > Co
An examination of Eo values for redox couple M3+/M2+ shows that Cr2+ is strong reducing agent (EoM3+/M2+ = 0.41 V) and liberates H2 from dilute acids.
2Cr2+ (aq) + 2H+ (aq) → 2Cr3+ (aq) + H2 ↑ (g)
∴ The correct order is Mn > Fe > Cr > Co.
Eo / V Cr Mn Fe Co
M2+/M - 0.90 - 1.18 - 0.44 - 0.28
Eo values for Mn is more negative than expected from general trend due to extra stability of half-filled Mn2+ ion.
THus the correct order should be
Mn > Cr > Fe > Co
An examination of Eo values for redox couple M3+/M2+ shows that Cr2+ is strong reducing agent (EoM3+/M2+ = 0.41 V) and liberates H2 from dilute acids.
2Cr2+ (aq) + 2H+ (aq) → 2Cr3+ (aq) + H2 ↑ (g)
∴ The correct order is Mn > Fe > Cr > Co.
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