(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 \(\mathrm{Sc}_{2} \mathrm{O}_{3}(s)\)

Step by step solution

01

(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.

02

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.

03

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.

04

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.

05

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.

06

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.

07

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.

08

(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

09

(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).

10

(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.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!