Answer the following questions about the complex [RhCl6]3−, where Rh is rhodium [Kr]4d84s1.
a Deduce the oxidation state of rhodium in the complex ion.
b Deduce the electron configuration of the metal cation.
c Deduce the number of unpaired electrons present in the complex ion.
d State the shape of the complex and show the arrangement of electrons.
The Correct Answer and Explanation is :
Let’s break down each question about the complex ([RhCl_6]^{3-}):
a) Oxidation State of Rhodium in the Complex
The complex is ([RhCl_6]^{3-}). Chlorine (Cl) is a monodentate ligand and typically carries a charge of -1. Since there are six chloride ions, the total charge contributed by the ligands is (6 \times (-1) = -6). The overall charge of the complex is (-3).
To find the oxidation state of Rh, we set up the equation:
[
\text{Oxidation state of Rh} + \text{Charge of chloride ligands} = \text{Charge of complex}
]
[
x + (-6) = -3
]
Solving for (x):
[
x = +3
]
Thus, the oxidation state of Rh in the complex is (+3).
b) Electron Configuration of the Metal Cation (Rh(^{3+}))
The atomic electron configuration of neutral Rhodium (Rh) is:
[
[Kr] 4d^8 5s^1
]
For the Rh(^{3+}) cation, it has lost 3 electrons. Typically, the 5s orbital loses electrons first, followed by the 4d orbital. Removing three electrons would result in:
[
[Kr] 4d^6
]
Thus, the electron configuration of Rh(^{3+}) is ([Kr] 4d^6).
c) Number of Unpaired Electrons in the Complex Ion
Rhodium has a (+3) oxidation state, and its electron configuration in the 4d orbital is (4d^6). In a coordination complex, the ligands influence the splitting of the d-orbitals, creating two sets of energy levels.
For the ([RhCl_6]^{3-}) complex, chloride is a weak field ligand, meaning it does not cause a strong splitting of the d-orbitals. Therefore, in this case, we would expect the electrons to occupy the higher energy orbitals first, resulting in unpaired electrons.
For a (d^6) configuration in a weak field, we typically expect 2 unpaired electrons.
Thus, the number of unpaired electrons in the complex is 2.
d) Shape of the Complex and Electron Arrangement
The complex ([RhCl_6]^{3-}) consists of six chloride ions arranged around a central Rh(^{3+}) ion. With six ligands, the complex adopts an octahedral geometry, which is the most common shape for a 6-coordinate complex.
In an octahedral field, the (d)-orbitals split into two sets: (e_g) (higher energy) and (t_{2g}) (lower energy). Since chloride is a weak field ligand, it does not cause strong splitting, leading to unpaired electrons in the higher-energy (e_g) orbitals.
Below is an image showing the electron arrangement in an octahedral field for Rh(^{3+}), considering the (4d^6) configuration.
I will generate the image to show the electron arrangement in the complex.
The image above shows the octahedral complex ([RhCl_6]^{3-}) with Rh(^{3+}) in the center and six chloride ions surrounding it. The electron configuration of Rh(^{3+}) is (4d^6), and in this weak field complex, the electrons are arranged with two unpaired electrons in the higher-energy (e_g) orbitals and the rest paired in the lower-energy (t_{2g}) orbitals.
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