Answer :
For a) [Ru(NH₃)₅Cl]SO₄
Ru configuration = d⁶s²
In this complex Ru oxidation number is +3
Ru³⁺ configuration = d⁵
number of [tex] d^{n} [/tex] electrons = 5
For b) Na₂[Os(CN)₆]
Os configuration = d⁶s²
In this complex Os oxidation number is +4
Os⁴⁺ configuration = d⁴
number of [tex] d^{n} [/tex] electrons = 4
Ru configuration = d⁶s²
In this complex Ru oxidation number is +3
Ru³⁺ configuration = d⁵
number of [tex] d^{n} [/tex] electrons = 5
For b) Na₂[Os(CN)₆]
Os configuration = d⁶s²
In this complex Os oxidation number is +4
Os⁴⁺ configuration = d⁴
number of [tex] d^{n} [/tex] electrons = 4
Answer:
(a) [Ru(NH₃)₅Cl]SO₄
The number of d electrons in the valence shell of Ru³⁺ is 5.
Also the total number of d electrons in Ru³⁺ is 15
(b) Na₂[Os(CN)₆]
The number of d electrons in the valence shell of Os⁴⁺ is 4.
Also the total number of d electrons in Os⁴⁺ is 24
Explanation:
In a coordination complex, a central metal is bonded to neutral or charged atoms or molecules, called ligands via coordinate covalent bonds.
(a) [Ru(NH₃)₅Cl]SO₄
The central metal atom is Ru
The electron configuration of Ru in ground state: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d⁷ 5s¹
Let the oxidation state of Ru in the given complex be x
∴ [x × 1] + [(0) × 5] + [(-1) × 1] + [(-2) × 1] = 0
⇒ x + 0 - 1 - 2 = 0
⇒ x = +3
Electron configuration of Ru³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d⁵
The number of d electrons in the valence shell of Ru³⁺ is 5
Also the total number of d electrons in Ru³⁺ is 15
(b) Na₂[Os(CN)₆]
The central metal atom is Os
The electron configuration of Os in ground state: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d⁶ 6s²
Let the oxidation state of Os in the given complex be x
∴ [x × 1] + [(0) × 5] + [(-1) × 1] + [(-2) × 1] = 0
⇒ x + 0 - 1 - 2 = 0
⇒ x = +3
Electron configuration of Os⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d⁴
The number of d electrons in the valence shell of Os⁴⁺ is 4
Also the total number of d electrons in Os⁴⁺ is 24