What is the oxidation state, d electron count, and coordination number respectively of the cobalt ion in [cobr(nh3)4]po4? what is the oxidation state, d electron count, and coordination number respectively of the cobalt ion in [cobr(nh3)4]po4? +3, 6, and 5 +3, 6, and 6 +4, 5, and 6 +4, 5, and 5 0, 9, and 6

The correct option is: +4, 5, and 5

Explanation:

A coordination complex consists of a central metal atom or ion bonded to ligands by coordinate covalent bonds. The ligands and the central metal together form the coordination sphere.

[CoBr(NH₃)₄]PO₄

Central metal atom - Co

Coordination sphere consists of 1 Br⁻ and 4 NH₃ ligands

Therefore, the coordination number= 5

Let the oxidation state of Co = x

Oxidation state of Br⁻ ligand = (-1)

Oxidation state of NH₃ ligand = 0    

Oxidation state of PO₄³⁻ ligand = -3

∴ x + (-1) × 1 + 0 × 4 + (-3) × 1 = 0

⇒ x = +4

Oxidation state of Co = +4

Electron configuration of Co in ground state: [Ar] 3d⁷ 4s²

Electron configuration of Co in +4 oxidation state: [Ar] 3d⁵ 4s⁰

Therefore, the number of d electrons in cobalt ion = 5

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.