Answer :

Explanation:

The paramagnetic or diamagnetic behavior of a molecule depends on the number of unpaired electrons in its molecular orbitals.

For \(CuF_6^{3-}\), copper is in the +3 oxidation state, which means it has lost three electrons. Copper typically has a configuration of [Ar] 3d^10 4s^1, but in the +3 oxidation state, it loses all its 4s and two 3d electrons, leaving behind a configuration of [Ar] 3d^7. Since there are unpaired electrons in the d orbitals (specifically, one unpaired electron in the 3d orbital), \(CuF_6^{3-}\) is paramagnetic.

For \(AuF_6^{-}\), gold is also in the +3 oxidation state, which means it has lost three electrons. The electron configuration of gold in the +3 oxidation state is [Xe] 4f^14 5d^8. Since all the electrons are paired in the d orbitals, there are no unpaired electrons, making \(AuF_6^{-}\) diamagnetic.

In summary, \(CuF_6^{3-}\) is paramagnetic because it has unpaired electrons in its molecular orbitals, while \(AuF_6^{-}\) is diamagnetic because all its electrons are paired.

Answer:

The (CuF6)3- complex, where copper is in the +3 oxidation state, is diamagnetic. Diamagnetic compounds do not possess any unpaired electrons in their electronic configurations. In the case of (CuF6)3-, the copper ion (Cu3+) has an electronic configuration of [Ar]3d^8. When it forms a complex with six fluoride (F-) ions, the resulting complex has a fully filled 3d orbital, meaning that all the electrons are paired. This complete pairing of electrons results in a diamagnetic nature for (CuF6)3-.

On the other hand, the (AuF6)- complex, where gold is in the +3 oxidation state, is paramagnetic. Paramagnetic compounds contain unpaired electrons in their electronic configurations. In the case of (AuF6)-, the gold ion (Au3+) has an electronic configuration of [Xe]4f^14 5d^8 6s^2. When it forms a complex with six fluoride (F-) ions, there are two unpaired electrons in the 5d orbital of the gold ion. These unpaired electrons make (AuF6)- paramagnetic, as it exhibits weak attraction in the presence of a magnetic field.

Therefore, even though both (CuF6)3- and (AuF6)- have a +3 oxidation state, their paramagnetic or diamagnetic nature is determined by the presence or absence of unpaired electrons in their respective complexes.

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