NEETChemistryCoordination Compounds
Match List I with List II. List I (Standard electrode potential) List II (Reason for the value) (A) E^ ( Mn ³⁺/ Mn ²⁺) (I) Positive value because high energy to transform metal to ion is not balanced by hydration enthalpy (B) E^ ( Cr ³⁺/ Cr ²⁺) (II) Comparatively lower positive value due to extra stability of d^5 configuration of +3 state (C) E^ ( Cu ²⁺/ Cu ) (III) Negative value due to extra stability of t_ 2g ^3 co
Options
- A(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
- B(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
- C(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- D(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Correct answer
C. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Step-by-step solution
E^ ( Mn ³⁺/ Mn ²⁺) has a high positive value because the reduction of Mn ³⁺ ( d^4 ) to Mn ²⁺ ( d^5 ) leads to a highly stable half-filled d -subshell. E^ ( Cr ³⁺/ Cr ²⁺) has a negative value because Cr ³⁺ ( t_ 2g ^3 ) is highly stable in aqueous solution, so Cr ²⁺ ( d^4 ) readily oxidises to Cr ³⁺ . E^ ( Cu ²⁺/ Cu ) is positive because the high enthalpy of atomisation and ionisation energy of copper is not balanced by its hydration enthalpy. E^ ( Fe ³⁺/ Fe ²⁺) has a comparatively lower positive value than Mn becaus