Chemistry - Coordination Chemistry Question with Solution | TestHub
Examine the following species and determine the total number of distinct species in which the metal centre have 16 valence electrons.
[Pt(C2H4)Cl3]- , [Ir(CO)Cl(PPh₃)₂] , [Pt(PPh₃)4]2+ , Fe(CO)₅ , [Pt(NH₃)6]4+,
[Ag(NH₃)₂]⁺, [PdCl₄]²⁻, [Co(CO)3NO] , Ti(η1-C₅H₅)₂(η5-C₅H₅)₂
Answer:
Solution:
To determine the number of valence electrons, we use the 18-electron rule (or 16-electron rule for d⁸ metals).
1. [Pt(C₂H₄)Cl₃]⁻: Pt(II) is d⁸. C₂H₄ (2e⁻), 3 Cl⁻ (6e⁻). Total = 8 + 2 + 6 = 16e⁻.
2. [Ir(CO)Cl(PPh₃)₂]: Ir(I) is d⁸. CO (2e⁻), Cl⁻ (2e⁻), 2 PPh₃ (4e⁻). Total = 8 + 2 + 2 + 4 = 16e⁻.
3. [Pt(PPh₃)4]2+: Pt(0) is d¹⁰. 4 PPh₃ (8e⁻). Total = 8 + 8 = 16e⁻.
4. Fe(CO)₅: Fe(0) is d⁸. 5 CO (10e⁻). Total = 8 + 10 = 18e⁻.
5. [Pt(NH₃)₆]⁴⁺: Pt(IV) is d⁶. 6 NH₃ (12e⁻). Total = 6 + 12 = 18e⁻.
6. [Ag(NH₃)₂]⁺: Ag(I) is d¹⁰. 2 NH₃ (4e⁻). Total = 10 + 4 = 14e⁻.
7. [PdCl₄]²⁻: Pd(II) is d⁸. 4 Cl⁻ (8e⁻). Total = 8 + 8 = 16e⁻.
8. [Co(CO)₃NO]: Co(0) is d⁹. 3 CO (6e⁻), NO (linear, 3e⁻). Total = 9 + 6 + 3 = 18e⁻.
9. Ti(η¹-C₅H₅)₂(η⁵-C₅H₅)₂: Ti(IV) is d⁰. 2 η¹-C₅H₅ (4e⁻), 2 η⁵-C₅H₅ (12e⁻). Total = 12 + 4 = 16e⁻.
Species with 16 valence electrons: [Pt(C₂H₄)Cl₃]⁻, [Ir(CO)Cl(PPh₃)₂], [Pt(PPh₃)4]2+, [PdCl₄]²⁻ and
Ti(η1-C₅H₅)₂(η5-C₅H₅)₂
There are 5 such species.