Coordination Compounds · Question 6

Among V(CO)6, Cr(CO)5, Cu(CO)3, Mn(CO)5, Fe(CO)5, [Co(CO)3]3-, [Cr(CO)4]4-, and Ir(CO)3, the total number of species isoelectronic with Ni(CO)4 is . [Given, atomic number: V = 23, Cr = 24, Mn = 25, Fe = 26, Co = 27, Ni = 28, Cu = 29, Ir = 77]

Answer

1

Solution

Isoelectronic species have the same total number of electrons (each CO has 6 + 8 = 14 electrons).

Ni(CO)4: 28 + 4 × 14 = 84.

V(CO)6 = 23 + 84 = 107; Cr(CO)5 = 24 + 70 = 94; Cu(CO)3 = 29 + 42 = 71; Mn(CO)5 = 95; Fe(CO)5 = 96.

[Co(CO)3]3- = 27 + 42 + 3 = 72; [Cr(CO)4]4- = 24 + 56 + 4 = 84; Ir(CO)3 = 77 + 42 = 119.

Only [Cr(CO)4]4- has 84 electrons.

Answer: 1