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: The formation of the series of oxides, K2O, CaO, Sc2O3, TiO2, V2O5, and CrO3 can be explained in terms of ionic compounds and the electron configurations of the individual elements involved in the formation: 1. \(\mathrm{K_2O}\): Potassium (K) has the electron configuration [Ar] \(4s^1\), and oxygen (O) has the electron configuration [He] \(2s^2 \ 2p^4\). Both of them need to achieve a stable configuration, so K loses one electron to become \(\mathrm{K^+}\) and O gains two electrons to become \(\mathrm{O^{2-}}\). Two potassium atoms donate one electron each to one oxygen atom to form the ionic compound \(\mathrm{K_2O}\). 2. \(\mathrm{CaO}\): Calcium (Ca) has the electron configuration [Ar] \(4s^2\), and oxygen (O) has the same configuration as in (1). Calcium loses two electrons to become \(\mathrm{Ca^{2+}}\), and oxygen gains two electrons to become \(\mathrm{O^{2-}}\). One calcium atom donates two electrons to one oxygen atom to form the ionic compound \(\mathrm{CaO}\). 3. \(\mathrm{Sc_2O_3}\): Scandium (Sc) has the electron configuration [Ar] \(3d^1 \ 4s^2\), and oxygen has the same configuration as in (2). Sc loses its three electrons to become \(\mathrm{Sc^{3+}}\), and oxygen gains two electrons to become \(\mathrm{O^{2-}}\). Two scandium atoms donate three electrons each to three oxygen atoms to form the ionic compound \(\mathrm{Sc_2O_3}\). 4. \(\mathrm{TiO_2}\): Titanium (Ti) has the electron configuration [Ar] \(3d^2 \ 4s^2\), and oxygen has the same configuration as in (3). Ti loses its four electrons to become \(\mathrm{Ti^{4+}}\), and oxygen gains two electrons to become \(\mathrm{O^{2-}}\). One titanium atom donates four electrons to two oxygen atoms to form the ionic compound \(\mathrm{TiO_2}\). 5. \(\mathrm{V_2O_5}\): Vanadium (V) has the electron configuration [Ar] \(3d^3 \ 4s^2\), and oxygen has the same configuration as in (4). V loses its five electrons to become \(\mathrm{V^{5+}}\), and oxygen gains two electrons to become \(\mathrm{O^{2-}}\). Two vanadium atoms donate five electrons each to five oxygen atoms to form the ionic compound \(\mathrm{V_2O_5}\). 6. \(\mathrm{CrO_3}\): Chromium (Cr) has the electron configuration [Ar] \(3d^5 \ 4s^1\) and oxygen has the same configuration as in (5). Cr loses its six electrons to become \(\mathrm{Cr^{6+}}\), and oxygen gains two electrons to become \(\mathrm{O^{2-}}\). One chromium atom donates six electrons to three oxygen atoms to form the ionic compound \(\mathrm{CrO_3}\).

: The names of the oxides are as follows: 1. \(\mathrm{K_2O}\) - Potassium oxide 2. \(\mathrm{CaO}\) - Calcium oxide 3. \(\mathrm{Sc_2O_3}\) - Scandium(III) oxide 4. \(\mathrm{TiO_2}\) - Titanium(IV) oxide 5. \(\mathrm{V_2O_5}\) - Vanadium(V) oxide 6. \(\mathrm{CrO_3}\) - Chromium(VI) oxide

: For each of the compounds, we need to write the balanced equation and then compute the enthalpy change \(\Delta H^{\circ}\) for the general reaction: $$ \mathrm{M}_{n} \mathrm{O}_{m}(s)+\mathrm{H}_{2}(g) \longrightarrow n\mathrm{M}(s)+m \mathrm{H}_{2} \mathrm{O}(g) $$ Assuming we are given the enthalpies of formation for the oxides (\(\Delta H^{\circ}_\mathrm{f}\)), we can use the overall equation for the change in enthalpy: $$ \Delta H^{\circ} = [n \Delta H^{\circ}_\mathrm{f}(\mathrm{M}) + m \Delta H^{\circ}_\mathrm{f}(\mathrm{H_2O})] - [\Delta H^{\circ}_\mathrm{f}(\mathrm{M_nO_m}) + \Delta H^{\circ}_\mathrm{f}(\mathrm{H_2})] $$ For each oxide, we can plug in the known enthalpy values and calculate \(\Delta H^{\circ}\).

: To estimate the value of \(\Delta H_{f}^{\circ}\) for Sc2O3, we can find the average value of the enthalpies of formation of its neighbors in the periodic table, namely, the oxides of Calcium (CaO) and Titanium (TiO2). Using the equation from part (c), we can find the enthalpy of formation of Sc2O3: $$\Delta H^{\circ}_\mathrm{f}(\mathrm{Sc_2O_3}) = [2\Delta H^{\circ}_\mathrm{f}(\mathrm{Sc}) + 3\Delta H^{\circ}_\mathrm{f}(\mathrm{H_2O})]-\Delta H^{\circ}_\mathrm{f}(\mathrm{CaO}) -\Delta H^{\circ}_\mathrm{f}(\mathrm{TiO_2})$$ By plugging in the known enthalpy values for CaO, TiO2, Sc, and H2O, we can compute the estimated \(\Delta H^{\circ}_f\) for Sc2O3.