Question

Which one of the following substances has three unpaired d electrons? A) [V(H2O)6]4+ B) [Zn(NH3)4]2+ C) [Cu(NH3)4]2+D) [Ag(NH3)2]+ E) [Cr(CN)6]3- The answer is E, but how? please explain.

Answer 1

Answer:

The correct option is: E) [Cr(CN)₆]³⁻

Reason: In this complex, Cr is present in +3 oxidation state and the electron configuration of Cr³⁺ is [Ar] 3d³

Therefore, the number of unpaired electrons = 3      

       

Explanation:

A) [V(H₂O)₆]⁴⁺

Central metal: Vanadium (V)

Let the oxidation state of V be x

Oxidation state of H₂O = 0

x + 6 × (0) = +4

x = +4

Ground state electron configuration: V = [Ar] 3d³4s²

Excited state electron configuration: V⁴⁺ = [Ar] 3d¹

Therefore, the number of unpaired electrons = 1

B) [Zn(NH₃)₄]²⁺

Central metal: Zinc (Zn)

Let the oxidation state of Zn be x

Oxidation state of NH₃ = 0

x + 4 × (0) = +2

x = +2

Ground state electron configuration: Zn = [Ar] 3d¹⁰4s²

Excited state electron configuration: Zn²⁺ = [Ar] 3d¹⁰

Therefore, the number of unpaired electrons = 0

C) [Cu(NH₃)₄]²⁺

Central metal: Copper (Cu)

Let the oxidation state of Cu be x

Oxidation state of NH₃ = 0

x + 4 × (0) = +2

x = +2

Ground state electron configuration: Cu = [Ar] 3d¹⁰4s¹

Excited state electron configuration: Cu²⁺ = [Ar] 3d⁹

Therefore, the number of unpaired electrons = 1

D) [Ag(NH₃)₂]⁺

Central metal: Silver (Ag)

Let the oxidation state of Ag be x

Oxidation state of NH₃ = 0

x + 2 × (0) = +1

x = +1

Ground state electron configuration: Ag = [Ar] 4d¹⁰5s¹

Excited state electron configuration: Ag⁺ = [Ar] 4d¹⁰

Therefore, the number of unpaired electrons = 0

E) [Cr(CN)₆]³⁻

Central metal: Chromium (Cr)

Let the oxidation state of Cr be x

Oxidation state of CN⁻ = -1

x + 6 × (-1) = -3

x = +3

Ground state electron configuration: Cr = [Ar] 3d⁵ 4s¹

Excited state electron configuration: Cr³⁺ = [Ar] 3d³

Therefore, the number of unpaired electrons = 3