Answer to Question #99275 in Inorganic Chemistry for joeh

Question

Count the number of valence electron per metal of the reaction products. Are they obeying the 18 valence-electron rule? Give comments.
(a) (PR3)2 Ir (CO) Cl + dppe --------> Ir (dppe)2 Cl + 2PR3 + CO
(b) (PR3)2 Ir (CO) Cl + CO ----------> (PR3)2 Ir (CO)2 Cl
(c) V(CO)6 + Na-------------------------> [V(CO)6]- + Na+
(d) (η5 - C5H5)2 Ni + NO------------> η5 -C5H5 Ni (NO) + C5H5

Na/Hg
(e) (C5H5)2 Ni -------------------------> η5 - C5H5 Ni (η3 - C5H7)
EtOH

Answer 1

a) IrCl gives 10 electrons, and 2*(PR₃) give 4 electrons, CO gives 2 electrons, so (PR3)₂Ir(CO)Cl has 16 electrons.

After reaction: IrCl gives 10 electrons, 2*(dope) give 8 electrons, in sum: 18 electrons.

b) Before the reaction: (PR3)2Ir(CO)Cl has 16 electrons

After: we add 2 electrons from CO, and in sum we have 18 electrons.

c) V(0) gives 5 electrons, and 6*CO give 12 electrons, in sum: 17 electrons.

But after reaction we add 1 electrons from Na, and V^-1 gives 6 electrons, so we have 18 electrons in this ion.

d) Ni(0) has 8 electrons, and 2*(η5 - C5H5) gives 10 electrons. In sum: 20 electrons.

After reaction we add 3 electrons from NO and leave 5 from (η5 - C5H5), so in sum we have 18 electrons.

e) Before the reaction Ni(C5H5)₂ has 20 electrons, 10 from Ni and 10 from 2*C5H5.

After the reaction, we have 10 electrons from Ni and 5 electrons from η5 - C5H5, η3 - C5H7 gives 3 electrons, in sum: 18 electrons

Ca) Ir