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
Step-by-step 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