introduction-to-inorganic-chemistry

A1.2 Hints for exercises, Section 5.2

(i) Use the Coulomb equation for the energy between two charges, q1 and q2, a distance d apart:

E = q1q2/4πε0d

Here ε0 is the permittivity of a vacuum, a constant.

(ii)Multiply by the Avogadro constant, L.

(iii)Include the repulsion between like charges. Remember that one mole of Na+ and Cl− ions gives ½ mole of (Na+Cl−)2 molecules.

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A1.3 Answers to questions and exercises, Chapters 6 - 9

Section 6.3

(i)[SiC4/4]∞

(ii)[SiO4/2]∞

Section 7.1

Series in order of decreasing electropositive character (values in volts):

Cations: K (–2.92), Ca (–2.87), Na (–2.71), Mg (–2.34), Al (–1.67), Zn (–0.76), Fe (–0.44), Sn (–0.14), H (0.00), Cu (+0.34)

Anions: I (+0.53), Br (+1.07), Cl (+1.36), F (+2.85)

Section 7.3, first exercise

For the elements in the previous exercise, Pauling’s electronegativities give:

K (0.8), Na (0.9), Ca (1.0), Mg (1.2), Al (1.5), Zn (1.6), Fe/Sn (1.8), Cu (1.9), H (2.1), I (2.5), Br (2.8), Cl (3.0), F (4.0)

The order is the same as in the previous exercise, except for inversions between Na and Ca, and Cu and H.

Section 7.3, second exercise

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Section 7.4

(i)H, +1; C, –4

(ii)Fe, +2; Cl, –1

(iii)Na, +1; H, –1

(iv)Si, +4; O, –2

(v)C, +4; O, –2; Cl, –1

(vi)H, +1; O, –2; N, –1

(vii)H, +1; O, –1

(viii)Si, +4; C, –4 [by analogy with (i) and (iv)]

Section 7.5

(i)tin tetrachloride or tin(IV) chloride

(ii)oxygen difluoride or oxygen(II) fluoride

(iii)dimagnesium stannide

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Section 8.1

(i)

(ii) 2/3

Section 8.2

The data give, respectively: [Hg+]2/[Hg2+] = 680, 436, 246 M [Hg22+]/[Hg2+] = 114, 110, 124

The figures in the second set are more constant.

Section 8.4

(i)ECl4, E2Cl6, etc.

(ii)ECl4 only.

Section 8.5

Left to reader (add electrovalencies to covalencies).

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Section 9.3

(iii)

(iv)

(v)

Ions (iii), (iv), and (v) can also be drawn with non-integral bonds, as in Section 8.5:

(iii)

(iv)

(v)

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A1.4 Answers to questions and exercises, Chapters 10 - 12

Section 10.1

Left to reader (see Sections 2.1 - 2.3).

Section 10.4

(i)Generalization holds with some exceptions.

(ii)Generalization holds.

(iii)Element 114 expected to be like lead, only somewhat more electropositive. Valencies = 2 and 4, with 2 much more stable.

Section 11.1

Section 11.2

(i) Lowest charge = +1

(ii)

Lowest charge = +1

(iii)

Lowest charge = +1

In (ii) and (iii), note that the bonding electrons are drawn closer to the oxygen atoms to indicate that the bonds are polarized away from the nitrogen atoms.

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Introduction to Inorganic Chemistry Appendix 1 Hints and answers

Section 11.3

(i)

Me:N:::C:

(ii)

Section 11.4

OCN–, NO2+, CN22–, FCN, N2O, N3–

Section 12.3.1

Section 12.3.2

(i)Dipotassium magnesium chloride

(ii)Potassium magnesium chloride

(iii)Triiron tetraoxide or iron(II) diiron(III) oxide

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A1.5 Answers to questions and exercises, Chapters 13 - 18

Section 13.1

(i) Data explained with coordination number = 4 = square planar.

(I) = [Pt(NH3)4]Cl2

(II), (III) = cis and trans [PtCl2(NH3)2]0 (IV) = K2[PtCl4]

(II) and (III) can be distinguished by reaction with a chelating agent (see Section 13.2). Chelating agents can replace ligands in a cis position, but not trans.

(ii)Magnus’s green salt = [Pt(NH3)4][PtCl4]

This has the same empirical formula as (II) and (III).

Section 13.3

These compounds are usually named as follows:

(I): tetraammineplatinum(II) chloride

(II), (III): cis- and trans-diamminedichloroplatinum(II) (IV): potassium tetrachloroplatinate(II)

Magnus’s green salt: tetraammineplatinum(II) tetrachloroplatinate(II)

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Section 13.4

As explained in the text, the bond formula of a coordination entity can be written in two ways. The formula of (i) is drawn below in the modern way, of (ii) and (iii) in the older way.

Section 15.2

(i)

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(ii)

(a)displacement

(b)decomposition

(c)combination, addition

(d)hydrolysis

(e)double decomposition, precipitation (BaSO4 is insoluble)

(f)combination

Section 16.1

CaCO3 → CaO + CO2

CaO + H2O → CaCO3

M2CO3 + Ca(OH)2 → 2MOH + CaCO3(ppt)

Section 16.6

Section 17.2

Fe2+ → Fe3+ for X = Cl, Br

Fe2+ ← Fe3+ for X = I

Section 17.3

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