The second electron affinity of oxygen [ E A 2 ( O ) ] is to be determined. Concept introduction: Lattice energy is the amount of energy needed for the conversion of one mole of an ionic solid to its constituent ions in the gaseous form. The higher the value of lattice energy for a compound, the higher is its stability. Born–Haber cycle is the method of calculating the lattice energy of a compound using different values of its constituent elements. The steps to calculate lattice energy using Born–Haber cycle are given below: The enthalpies of the formation of the metal as well as the non-metal are to be added. The ionization energy of the metal is added, and electron affinity is subtracted from the calculated value. The mode of the enthalpy of formation of the required compound is added to the above value.

Question

Want to see the full answer?

Answer 1

Question

Chapter 8.3, Problem 1PPB

Interpretation Introduction

Interpretation:

The second electron affinity of oxygen [EA2(O)] is to be determined.

Concept introduction:

Lattice energy is the amount of energy needed for the conversion of one mole of an ionic solid to its constituent ions in the gaseous form.

The higher the value of lattice energy for a compound, the higher is its stability.

Born–Haber cycle is the method of calculating the lattice energy of a compound using different values of its constituent elements.

The steps to calculate lattice energy using Born–Haber cycle are given below:

The enthalpies of the formation of the metal as well as the non-metal are to be added.

The ionization energy of the metal is added, and electron affinity is subtracted from the calculated value.

The mode of the enthalpy of formation of the required compound is added to the above value.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Bookmarks Profiles Tab Window Help Chemical Formula - Aktiv Che X + → C 11 a app.aktiv.com Google Chrome isn't your default browser Set as default Question 12 of 16 Q Fri Feb 2 Verify it's you New Chrome availabl- Write the balanced molecular chemical equation for the reaction in aqueous solution for mercury(I) nitrate and chromium(VI) sulfate. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction. 3 Hg(NO3)2(aq) + Cг2(SO4)3(aq) → 3 Hg₂SO (s) + 2 Cr(NO3), (aq) ean Ui mate co ence an climate bility inc ulnerabili women, main critic CLIMATE-INI ernational + 10 O 2 W FEB 1 + 4- 3- 2- 2 2 ( 3 4 NS 28 2 ty 56 + 2+ 3+ 4+ 7 8 9 0 5 (s) (1) Ch O 8 9 (g) (aq) Hg NR CI Cr x H₂O A 80 Q A DII A F2 F3 FA F5 F6 F7 F8 F9 #3 EA $ do 50 % 6 CO & 7 E R T Y U 8 ( 9 0 F10 34 F11 川 F12 Subr + delete 0 { P }

Deducing the reactants of a Diels-Alder reaction n the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ • If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. >

Predict the major products of the following organic reaction: + Some important notes: A ? • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure.