Question

What is the electron configuration for the transition metal ion(s) in each of the following compounds? a. \(\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{4}\right]\) b. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\right] \mathrm{I}_{2}\) c. \(\mathrm{Na}_{2}\left[\mathrm{TaF}_{7}\right]\) d. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{Pt} \mathrm{I}_{4}\right]\) Pt forms \(+2\) and \(+4\) oxidation states in compounds.

Short answer

The electron configurations for the transition metal ions in the given compounds are: a. Fe: \(1s^22s^22p^63s^23p^63d^6\) b. Co: \(1s^22s^22p^63s^23p^63d^7\) c. This exercise contains an error in the composition of the substance. d. Pt(II): \(1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^8\) Pt(IV): \(1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^7\)

Step-by-step solution

Identify the transition metal

In this compound, the transition metal is iron (Fe).

Find the oxidation state

We can find the oxidation state of Fe by considering the charges of the other species within the compound: \[2 \cdot (\mathrm{NH}_{4})^{+} + [\mathrm{Fe}(H_2O)_{2}\mathrm{Cl}_{4}]^{x} = \mathrm{neutral}\] \[+2 + x = 0\] Solving for x, we find that the oxidation state of Fe is +2.

Determine the electron configuration

As a neutral atom, Fe has an electron configuration of: \[\mathrm{Fe}:\,\, 1s^22s^22p^63s^23p^63d^64s^2\] Now that we know the oxidation state of Fe is +2, we will remove two electrons from the configuration: \[\mathrm{Fe}^{2+}:\,\, 1s^22s^22p^63s^23p^63d^6\] So, the electron configuration for the iron ion in this compound is \(1s^22s^22p^63s^23p^63d^6\). #b. \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{NH}_{2}\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{2}\right] \mathrm{I}_{2}\) #

Identify the transition metal

In this compound, the transition metal is cobalt (Co).

Find the oxidation state

We can find the oxidation state of Co by considering the charges of the other species within the compound: \[[\mathrm{Co}(NH_3)_{2}(NH_{2}CH_{2}CH_{2}NH_{2})_{2}]^{x} + 2 \cdot (\mathrm{I}^{-}) = \mathrm{neutral}\] \[x - 2 = 0\] Solving for x, we find that the oxidation state of Co is +2.

Determine the electron configuration

As a neutral atom, Co has an electron configuration of: \[\mathrm{Co}:\,\, 1s^22s^22p^63s^23p^63d^74s^2\] Now that we know the oxidation state of Co is +2, we will remove two electrons from the configuration: \[\mathrm{Co}^{2+}:\,\, 1s^22s^22p^63s^23p^63d^7\] So, the electron configuration for the cobalt ion in this compound is \(1s^22s^22p^63s^23p^63d^7\). #c. \(\mathrm{Na}_{2}\left[\mathrm{TaF}_{7}\right]\)#

Identify the transition metal

In this compound, the transition metal is tantalum (Ta).

Find the oxidation state

We can find the oxidation state of Ta by considering the charges of the other species within the compound: \[2 \cdot (\mathrm{Na}^{+}) + [\mathrm{TaF}_7]^{x} = \mathrm{neutral}\] \[+2 + x = 0\] Solving for x, we find that the oxidation state of Ta is -2.

Determine the electron configuration

As a neutral atom, Ta has an electron configuration of: \[\mathrm{Ta}:\,\, 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^36s^2\] Now that we know the oxidation state of Ta is -2, we will add two electrons to the configuration: \[\mathrm{Ta}^{-2}:\,\, 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^36s^2\] However, elements of group 5 do not form negative oxidation states in compounds; thus, this exercise contains an error in the composition of the substance. #d. \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{I}_{2}\right]\left[\mathrm{Pt} \mathrm{I}_{4}\right]\) # In this compound, we have two different platinum ions, Pt(II) and Pt(IV).

Identify the transition metal.

In both parts of this compound, the transition metal is platinum (Pt).

Pt(II) Determine the electron configuration

As a neutral atom, Pt has an electron configuration of: \[\mathrm{Pt}:\,\, 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^96s^1\] Now that we know the oxidation state of the first Pt ion is +2, we will remove two electrons from the configuration: \[\mathrm{Pt}^{2+}:\,\, 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^8\] So, the electron configuration for the first platinum ion (Pt(II)) in this compound is \(1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^8\).

Pt(IV) Determine the electron configuration

In the neutral atom, Pt has an electron configuration of: \[\mathrm{Pt}:\,\, 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^96s^1\] Now that we know the oxidation state of the second Pt ion is +4, we will remove four electrons from the configuration: \[\mathrm{Pt}^{4+}:\,\, 1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^7\] So, the electron configuration for the second platinum ion (Pt(IV)) in this compound is \(1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^64f^{14}5d^7\).

Key concepts

Electron Configuration

Electron configuration is the arrangement of electrons in an atom’s electron shells. This configuration provides insight into the chemical properties and reactivity of elements, especially transition metals. Transition metals generally fill their 3d orbitals, with electrons often removed first from the 4s orbital when forming cations. For example, neutral iron (Fe) has an electron configuration of \[\mathrm{Fe}: 1s^22s^22p^63s^23p^63d^64s^2\]In the +2 oxidation state, electrons are removed primarily from the 4s orbital:\[\mathrm{Fe^{2+}}: 1s^22s^22p^63s^23p^63d^6\]Where two electrons are removed, leading to a partially filled 3d shell.

Oxidation States

Oxidation states, or oxidation numbers, refer to the number of electrons involved or absorbed during a chemical reaction. For transition metals, this value indicates the electrons lost, commonly affecting the electron configuration. In coordination compounds, the overall charge is neutral, which can be calculated by accounting for the charges of the ligands and the counterion. For instance, in a complex with cobalt:

• The oxidation state of cobalt in \([\mathrm{Co}(\mathrm{NH}_3)_2(\mathrm{NH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{NH}_2)_2]^{x} \; \mathrm{I}_2\) is +2, as the two \\(\mathrm{I}^{-}\) ions stabilize the overall charge to zero.

This demonstrates how charge balance helps deduce oxidation states in these compounds.

Platinum Complexes

Platinum complexes are coordination compounds that feature platinum atoms as their central metal. The unique aspect of platinum is its ability to exist in different oxidation states, primarily +2 and +4. For example, in the following complex:

• \(\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{I}_2\right] \left[\mathrm{Pt} \mathrm{I}_4\right]\), one platinum is in the +2 oxidation state and the other in the +4 state.

Such complexes are significant in the field of chemistry and medicine, including usage in cancer therapies such as cisplatin.

Coordination Compounds

Coordination compounds consist of a central metal atom or ion bonded to surrounding molecules or ions called ligands. These compounds are notable for their specific structures and diverse reactivity due to the type and number of ligands. Ligands include neutral molecules like \(\mathrm{H}_2\mathrm{O}\) or \(\mathrm{NH}_3\), and anions such as \(\mathrm{Cl}^{-}\). For example:

• In \(\left(\mathrm{NH}_4\right)_2\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_2 \mathrm{Cl}_4\right]\), water and chloride ions serve as ligands.

Coordination compounds illustrate how metal ions coordinate with donor atoms of ligands, affecting the spectral and magnetic properties of the compound.