Question

(a) Account for formation of the following series of oxides in terms of the electron configurations of the elements and the discussion of ionic compounds in Section 2.7: \(\mathrm{K}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{Sc}_{2} \mathrm{O}_{3}, \mathrm{TiO}_{2}, \mathrm{~V}_{2} \mathrm{O}_{5}, \mathrm{CrO}_{3} .(\mathbf{b})\) Name these oxides. (c) Consider the metal oxides whose enthalpies of formation (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) are listed here. Calculate the enthalpy changes in the following general reaction for each case: $$\mathrm{M}_{n} \mathrm{O}_{m}(s)+\mathrm{H}_{2}(g) \longrightarrow n \mathrm{M}(s)+m \mathrm{H}_{2} \mathrm{O}(g)$$ (You will need to write the balanced equation for each case and then compute \(\left.\Delta H^{\circ} .\right)\) (d) Based on the data given, estimate a value of \(\Delta H_{f}^{\circ}\) for \(\mathrm{Sc}_{2} \mathrm{O}_{3}(s)\)

Short answer

In summary, we analyzed the electron configurations of the elements K, Ca, Sc, Ti, V, and Cr to understand the formation of their corresponding oxides: K2O, CaO, Sc2O3, TiO2, V2O5, and CrO3. These oxides were named as potassium oxide, calcium oxide, scandium(III) oxide, titanium(IV) oxide, vanadium(V) oxide, and chromium(VI) oxide. We then calculated the enthalpy changes for the reaction of each oxide with molecular hydrogen. Finally, we estimated the enthalpy of formation for Sc2O3 by taking the average of the enthalpies of formation for CaO and TiO2.

Step-by-step solution

(a) Electron Configurations

: To account for the formation of these oxides, we first need to know the electronic configuration of each element. We will write them as: 1) K: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1\) 2) Ca: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\) 3) Sc: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\) 4) Ti: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\) 5) V: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\) 6) Cr: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\) Now, let's analyze the formation of the following oxides.

K2O formation:

K loses one electron to achieve the noble gas electron configuration. Thus, it forms a K+ cation. Oxygen gains two electrons to achieve the noble gas configuration (O2-). Therefore, two K+ ions combine with one O2- ion to form K2O.

CaO formation:

Ca loses two electrons to achieve the noble gas electron configuration. Thus, it forms a Ca2+ cation. Oxygen gains two electrons to achieve the noble gas configuration (O2-). Therefore, one Ca2+ ion and one O2- ion combine to form CaO.

Sc2O3 formation:

Sc loses three electrons to achieve the noble gas electron configuration. Thus, it forms a Sc3+ cation. Oxygen gains two electrons to achieve the noble gas configuration (O2-). Therefore, two Sc3+ ions and three O2- ions combine to form Sc2O3.

TiO2 formation:

Ti loses four electrons to achieve the noble gas electron configuration. Thus, it forms a Ti4+ cation. Oxygen gains two electrons to achieve the noble gas configuration (O2-). Therefore, one Ti4+ ion combines with two O2- ions to form TiO2.

V2O5 formation:

V loses five electrons to achieve the noble gas electron configuration. Thus, it forms a V5+ cation. Oxygen gains two electrons to achieve the noble gas configuration (O2-). Therefore, two V5+ ions combine with five O2- ions to form V2O5.

CrO3 formation:

Cr loses six electrons for the electron configuration 3d3 to match the half-filled d-orbital configuration. Thus, it forms a Cr6+ cation. Oxygen gains two electrons to achieve the noble gas configuration (O2-). Therefore, one Cr6+ ion combines with three O2- ions to form CrO3.

(b) Naming Oxides

: Now let's name these oxides: 1) K2O: Potassium oxide 2) CaO: Calcium oxide 3) Sc2O3: Scandium(III) oxide 4) TiO2: Titanium(IV) oxide 5) V2O5: Vanadium(V) oxide 6) CrO3: Chromium(VI) oxide

(c) Enthalpy Changes

: We will calculate the enthalpy changes in the following reaction: $$M_n O_m + H_2 \longrightarrow nM + mH_2O$$ We will now write down the chemical equations for every oxide and apply Hess's law. \(2K_2O + 4H_2 \longrightarrow 4K + 4H_2O \\ 2CaO + 2H_2 \longrightarrow 2Ca + 2 H_2O \\ Sc_2O_3 + 3H_2 \longrightarrow 2Sc + 3H_2O \\ TiO_2+2H_2 \longrightarrow Ti + 2H_2O \\ V_2O_5+5H_2 \longrightarrow 2V + 5H_2O \\ CrO_3 + 3H_2 \longrightarrow Cr + 3H_2O\) Using the enthalpy of formation values for the metal oxides and water given, we can calculate the enthalpy change for each reaction. ΔH = ΣΔH(products) - ΣΔH(reactants).

(d) Estimating ΔHf°(Sc2O3)

: Given the values of enthalpy changes of other metal oxides, we can estimate the ΔHf°(Sc2O3) by considering its position in the series of elements and the trend in enthalpy values. Scandium is between Calcium and Titanium. Estimating ΔHf°(Sc2O3) as an average of ΔHf°(CaO) and ΔHf°(TiO2) would be reasonable. ΔHf°(Sc2O3) ≈ (ΔHf°(CaO) + ΔHf°(TiO2)) / 2 Perform the calculation to obtain the estimated value for the enthalpy of formation of Sc2O3.

Key concepts

Electron Configuration

Understanding electron configuration is crucial when studying oxide formation. All atoms strive to achieve a stable electron arrangement, similar to that of a noble gas. This is known as the octet rule. Elements either lose, gain, or share electrons to achieve this stable configuration. Let's look at how it applies to potassium, calcium, scandium, titanium, vanadium, and chromium:

• Potassium (K): Has the electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1\). It will lose one electron to form \( ext{K}^+\).
• Calcium (Ca): With a configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\), Ca loses two electrons to form \( ext{Ca}^{2+}\).
• Scandium (Sc): The configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1\). Scandium loses three electrons to achieve \( ext{Sc}^{3+}\).
• Titanium (Ti): Electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\). It loses four electrons forming \( ext{Ti}^{4+}\).
• Vanadium (V): With \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\), it forms \( ext{V}^{5+}\) by losing five electrons.
• Chromium (Cr): The configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\) leads to \( ext{Cr}^{6+}\) upon losing six electrons.

Electron configurations allow us to predict how these atoms form ionic compounds when reacting with oxygen.

Ionic Compounds

An ionic compound is formed through the transfer of electrons between atoms, leading to the creation of positive and negative ions that attract each other. When metals bond with non-metals like oxygen, ionic compounds are usually formed. For example:

• Potassium oxide \( ext{K}_2 ext{O}\): Formed when two \( ext{K}^+\) ions pair with one \( ext{O}^{2-}\) ion.
• Calcium oxide \( ext{CaO}\): Consists of a single \( ext{Ca}^{2+}\) cation and one \( ext{O}^{2-}\) anion.
• Scandium(III) oxide \( ext{Sc}_2 ext{O}_3\): Two \( ext{Sc}^{3+}\) ions bond with three \( ext{O}^{2-}\) ions.

These compounds are created through the fulfilling electron exchange process, stabilizing both elements involved. The strong electrostatic attraction between the positively charged cations and negatively charged anions maintains their solid-state structure.

Enthalpy of Formation

The enthalpy of formation (\(\Delta H_f^\circ\)) indicates energy changes during the formation of a compound from its elements in their standard states. It is an important concept in thermodynamics, particularly when analyzing reactions involving oxides.
In the reaction\(M_nO_m + H_2 \rightarrow nM + mH_2O\), we calculate the change using:\[\Delta H = \left(\Sigma \Delta H_{\text{products}}\right) - \left(\Sigma \Delta H_{\text{reactants}}\right)\]

• Calculate the reaction enthalpies with possible data from similar reactions, like those of CaO and TiO2, to estimate values for others such as Scandium's oxide.
• Understand that the energy released or absorbed provides insights into the stability and reactivity of these metal oxides.

By determining the enthalpy changes, we can get insights into the energy efficiency and practicality of utilizing these oxides in different chemical processes.

Chemical Reactions

Chemical reactions are processes where substances are transformed into new products. In the context of metal oxides, understanding the reactions helps in grasping their formation, usage, and decomposition.

• Metal oxides react with hydrogen in the general form:\(M_nO_m + H_2 \rightarrow nM + mH_2O\). Each reaction's enthalpy tells us about the energy involved.
• Such transformations are critical for energy conversion processes, like reducing ores to extract metals.
• Understanding reactions in a stepwise manner—analyzing each equation and its enthalpy—leads to insights about the conditions favoring certain reactions over others.

By breaking down these reactions into smaller parts, analyzing the changes during each step, one can predict the outcomes and optimize conditions for specific goals.