Problem 57 - Chemistry The Central Science In Si Units

Question

Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2}\) (b) \(\mathrm{K}_{2}\left[\mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]\) (c) \(\left[\mathrm{Os}(\mathrm{en})_{3}\right] \mathrm{Cl}_{3}\) (d) \([\mathrm{Cr}(\mathrm{EDTA})] \mathrm{SO}_{4},(\mathbf{e})\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

(a) 6 \(d\) electrons, (b) 8 \(d\) electrons, (c) 5 \(d\) electrons, (d) 4 \(d\) electrons, (e) 10 \(d\) electrons.

Step 1: Determine Oxidation States

First, we need to determine the oxidation state of the central metal in each complex. For (a) \([\text{Pt}(\text{NH}_3)_2 \text{Cl}_2]\text{Cl}_2\): The neutral ammonia ligands \((\text{NH}_3)\) and chloride ligands \((\text{Cl}^-)\) contribute 0 and -1 charges, respectively. If the complex overall has no charge, the calculation for Pt will be 0 (from \(\text{NH}_3\)) - 2 (from 2 \(\text{Cl}^-\)) = +2, due to the external 2 chlorides balancing the charge. Thus, Pt has a +2 oxidation state.For (b) \(\text{K}_2[\text{Cu}(\text{C}_2\text{O}_4)_2]\): Each oxalate ligand \((\text{C}_2\text{O}_4)^{2-}\) carries a -2 charge, totaling -4 for two ligands. K contributes +1 for each K, totaling +2, requiring Cu to be in the +3 oxidation state to balance the charge.For (c) \([\text{Os}(\text{en})_3]\text{Cl}_3\): Ethylenediamine \((\text{en})\) is neutral, so if the complex has no charge, Os must be +3 to balance the 3 negative charges from chlorides.For (d) \([\text{Cr}(\text{EDTA})]\text{SO}_4\): EDTA is a 4- ligand, SO sulfate contributes -2 charge. Since the whole entity is neutral, Cr must be in the +2 oxidation state to balance. Lastly, for (e) \([\text{Cd}(\text{H}_2\text{O})_6]\text{Cl}_2\): The water is neutral, and Cl contributes -2 in total when 2 chloride anions are present, so Cd must be +2.

Step 2: Determine Electron Configuration of the Metal

For each metal, determine the electron configuration of its neutral atom, then account for the oxidation state to find the number of valence \(d\) electrons.(a) Pt electronic configuration is \([Xe] 6s^2 4f^{14} 5d^8\). In Pt(II), you remove 2 electrons from 5d, leaving 6 \(d\) electrons.(b) Cu electronic configuration is \([Ar] 4s^1 3d^{10}\). In Cu(III), you remove 2 from 4s and 1 from 3d: total 8 \(d\) electrons.(c) Os electronic configuration is \([Xe] 6s^2 4f^{14} 5d^6\). In Os(III), remove 3 electrons from 5d, leaving 5 \(d\) electrons.(d) Cr electronic configuration is \([Ar] 4s^1 3d^5\). In Cr(II), remove 2 electrons from 3d, leaving 4 \(d\) electrons.(e) Cd electronic configuration is \([Kr] 5s^2 4d^{10}\). In Cd(II), you remove 2 from 5s, leaving 10 \(d\) electrons.

Step 3: Report Result for Each Complex

Now list the d electrons for each complex: (a) \(\text{Pt}^{2+}\) in \([\text{Pt}(\text{NH}_3)_2 \text{Cl}_2]\text{Cl}_2\) has 6 \(d\) electrons. (b) \(\text{Cu}^{3+}\) in \(\text{K}_2[\text{Cu}(\text{C}_2\text{O}_4)_2]\) has 8 \(d\) electrons. (c) \(\text{Os}^{3+}\) in \([\text{Os}(\text{en})_3]\text{Cl}_3\) has 5 \(d\) electrons. (d) \(\text{Cr}^{2+}\) in \([\text{Cr}(\text{EDTA})]\text{SO}_4\) has 4 \(d\) electrons. (e) \(\text{Cd}^{2+}\) in \([\text{Cd}(\text{H}_2\text{O})_6]\text{Cl}_2\) has 10 \(d\) electrons.

Key Concepts

Oxidation States in Coordination Complexes

Understanding oxidation states is crucial for determining the properties of coordination complexes. An oxidation state represents the hypothetical charge of an atom if all bonds were ionic, rather than covalent.
This helps us track electron transfer in metals at the center of complexes. Take, for example, the coordination complex \([ ext{Pt}( ext{NH}_3)_2 ext{Cl}_2] ext{Cl}_2\). This complex features platinum (Pt) with electrons shared with other atoms.
To determine the oxidation state, evaluate each ligand's contribution: neutral ammonia \(( ext{NH}_3)\) gives no charge, while each chloride ion \(( ext{Cl}^-)\) contributes a \(-1\) charge. With two chloride anions outside the bracket balancing the charge, Pt must be in the \(+2\) oxidation state.

• Ligands like \(( ext{NH}_3)\) and neutral water \(( ext{H}_2 ext{O})\) do not affect charge.
• On the other hand, other anionic ligands, such as \(SO_4^{2-}\) in complexes, require the central metal to adjust its oxidation state appropriately.

Recognizing each coordination complex's oxidation state allows for next steps in comprehending electron distributions.

Electron Configuration

Once oxidation states are known, electron configurations provide insight into an element’s electron arrangement. Understanding configurations is vital for knowing how many valence electrons, particularly \(d\) electrons, remain in complexes.
Electrons fill orbitals based on energy levels, following principles like Aufbau's rule. For transition metals, after establishing an initial electron configuration, adjust based on the oxidation state.
Take platinum in \( ext{Pt}^{2+}\) from \([ ext{Pt}( ext{NH}_3)_2 ext{Cl}_2] ext{Cl}_2\). Its neutral electron configuration is \([Xe] 6s^2 4f^{14} 5d^8\). After accounting for the \(+2\) oxidation state, remove two 5d electrons: \([Xe] 6s^2 4f^{14} 5d^6 \). Thus, 6 \(d\) electrons remain.

• Varying electron configurations impact each element's metallic characteristics, especially within transition metals.
• This configuration helps predict chemical behavior in reactions.

Configurations illuminate how electrons occupy or leave orbitals when forming complexes.

Transition Metals Chemistry

Transition metals showcase unique properties due to their \(d\) orbitals. These metals, found in the center of the periodic table, include familiar elements like iron, copper, and platinum. Transition metals frequently act as central elements in complexes due to their ability to accommodate multiple oxidation states.
These metals are characterized by their \(d\) orbitals. The skeleton of properties is affected by the electron filling in these orbitals, leading to diverse phenomena such as variable oxidation states, catalytic behavior, and colorful compounds.
Such phenomenon is primarily because:

• These \(d\) orbitals are partially filled, as seen in copper’s \(Cu^{3+}\) with 8 \(d\) electrons of \([ ext{Cu}( ext{C}_2 ext{O}_4)_2]^{2-}\).
• Electrons can be easily shifted to form stable complexes, allowing for flexibility in chemistry.

While transitioning electrons doesn’t majorly affect the core nucleus, it impacts reactivity significantly.
Transition metals truly shine in their role of forming coordination complexes.