s-Block Chemistry

Question 1

Ionic model

Answer 1

Ions re assumed to be hard spheres with fixed sizes held together by electrostatic interactions, the radii of spheres are known as ionic radii.

Question 2

Internuclear distance

Answer 2

Can be measured by x-ray crystallography but it is hard to determine individual radii as hard to know when electron density stops and starts. Ionic radii change depending on the compound.

Question 3

Trends in ionic radii

Answer 3

Anions>cations
Increase down a group; Increase in PQN
Dicationmonoanion

Question 4

Lattice Enthalpy

Answer 4

Enthalpy change for the conversion of 1 mol of the ionic solid into the gaseous state

Question 5

ΔU in the ionic model:

Answer 5

-ΔlattH° in the ionic model comes from electrostatic interactions between ions, by considering these interactions in an ionic solid you can obtain a theoretical value

ΔU = -(z- z+ e^2)/(4 π εo r)

Question 6

Madelung constant

Answer 6

in a crystal there are repulsive and attractive interactions in all directions to account for this the Madelung constant is introduced as well as Avogadro’s number

ΔUlatt = +(A Na z- z+ e^2)/(4 π εo r)

Lattice energy is the -ve of this potential so the +ve sign is now included

Question 7

Born-Lande Equation

Answer 7

The prior expression assumes ions are point charges; additional short range forces need to be included from overlapping electron clouds, these increase as)decreases.

+(A Na z- z+ e^2)/(4 π εo r)(1-1/n)

The born exponent is an average of an n value obtained by comparing the compressibility of a solid based on its occupied orbitals.

Question 8

ΔU/ΔH

Answer 8

All prior calculations give internal energy changes, ΔU, however ΔH is heat change at constant pressure, they are related by: ΔH = ΔU + pV; the difference is so small hence can be ignored

Question 9

Calculating the Madelung constant

Answer 9

The first attractive energy equation is used, times by the # of nearest neighbours, r is calculate during extended Pythagoras theorem. Include sign +/- !!!

d = (2r^2)^1/2
d = (3r^2)^1/2

ΔU = ΔU1 + ΔU2 + ΔU3 …

The terms alternate in sign and are of high but decreasing magnitude.

Question 10

Deviations from the ionic model

Answer 10

+ve charges distort electron clouds of anions, large and small charged ions are more easily polarisable. highly charged small cations bonded to large charge diffuse anions have the greatest degree of covalent character.

The BL eq. underestimates ΔUlatt high covalent character compounds.

ΔUlatt(BH) ≈ ΔUlatt(BL) -> low covalency
ΔUlatt(BH) > ΔUlatt(BL) -> high covalency

Question 11

Kapustinskii Equation

Answer 11

A simplification of the Born-Lande equation;

ΔUlatt = (k ν z+ z-)/(r+ + r-)

Where k = 107900, ν = no. ions in the formula units

Thermochemical radii of polyatomic ions can be calculated in this way

Question 12

van Arkel-Ketelaar triangles

Answer 12

Predict the bonding type for a binary compound;

Metallic ^^ Ionic ^^ Covalent

X-axis - Average electronegativity
Y-axis - Difference in electronegativity

When both χ’s are high –> covalent, when high Δχ –> ionic

Question 13

Hydrogen

Answer 13

H2 is colourless and odourless, MP: -259°C, BP: -253°C (low/weak IM forces), low Mr, lowest density of all gases (0.082gdm-3). Unreactive at RTP (H-H 436JKmol-1) with the exception of O2, F2 and Cl2.

Question 14

Production of H2

Answer 14

Steam reforming: methane + water -> carbon monoxide + 3hydrogen (NiO/850°C)

Shift reaction: CO + H2O -> CO2 + H2 (Fe/450°C)

Electrolysis of H2O: 2H20 -> 2H2 + O2

Question 15

Use of hydrogen

Answer 15

Mainly used in the Haber process (53%), in the petrochemical industry, for extraction of metals from ores and in ethanol production (17%)

Question 16

Hydrides

Answer 16

Hydrogen forms binary compounds called hydrides with most elements; χH>χR (R is electropositive) the H atom is hydride and has an oxidation state of -1. If χH

Question 17

Hydridic Hydrides

Answer 17

Hydridic hydrides are formed by group 1 & 2 and are ionic structures, strongly basic

RH + H2O -> ROH + H2

Reactivity with water increases down the group hence they are kept under inert atmospheres.

Question 18

Protic hydrides

Answer 18

Covalent not ionic as ionisation (1312KJmol-1) &raquo_space; electron gain (-73KJmol-1) hence H+ difficult to produce. Only formed when dissolved in a solvent that can solvate H+; the compound here acts as an acid

Question 19

Non-polar hydrides

Answer 19

(χH≈χR) may have a small dipole: i.e. in CH4 H=δδ+ and in B2H6 H=δδ-

Question 20

Nomenclature of hydrides

Answer 20

Hydridic hydrides -> hydridides
Protic hydrides/Non-polar hydrides -> R-ane
Group 16&17 ->hydrogen R-ide
Trivial names - Ammonia, water

Question 21

Covalent Hydrides

Answer 21

Electron precise compounds - All valance electrons are involved in bonding (group 14)
Electron deficient compounds - 3-centre-2-electron bonds (group 13)
Electron rich compounds - Not all electrons on the central atom are involved in bonding (lone pairs) hence can act as lewis bases (groups 15-17)

Question 22

Acidity of hydrides

Answer 22

Increases across a period (Δχ increases) and increases down group (decrease in BDE, decreasing attraction between X and H3O+)

HX + H2O –> H3O+ X-

Question 23

BDE of hydrides

Answer 23

Increases from left to right due to an increase in ionic contribution as χX increases. Decrease down group as valance orbitals get larger and more diffuse as PQN increases therefore ns np interactions with H 1s are reduced.

Question 24

Isotope effects of hydrides

Answer 24

Protium/deuterium/tritium

H^2 Has twice the mass of H^1 hence its properties will vary considerably, D2O ice sinks as it is more dense,

BDE of D2 443>H2 436 therefore more energy is needed to break X-D that X-H; the lowest vibrational energy state for D2 is lower than for H2

Question 25

Group 1 general properties

Answer 25

Li, Na, K, Rb, Cs, Fr (radioactive longest lived isotope 22mins, minor component in uranium minerals Low MPs, BPs, ΔaH° -> weak metallic bonding Low densities -> large atomic radii, open body-centered cubic structures

Question 26

Preparation of group 1 metals

Answer 26

Li and Na electrolysis of molten chlorides in a downs cell, CaCl2 added to lower the MP 2Na+ + 2e- -> 2Na (reduction) 2Cl- -> Cl2 + 2e- K, Rb, Cs reduction of molten salts with Na at high temps KCl + Na -> K NaCl eq lies to the left therefore K (volatile) removed by fractional distillation

Question 27

Group 1 oxides

Answer 27

All burn in air to form oxides: Li (O2- oxide), Na (O2^2- peroxide), K (O2- superoxide) All are basic and fact with water to form hydroxides, H2O2 and oxygen as you descend the group Large anions are stabilised by large cations; G1 cations increase down the group therefore are better at stabilising large peroxide and superoxide anions Li2O2 decomposes on heating whereas Na2O2 is stable to it (Born hater cycle using ΔG = ΔH-TΔS)

Question 28

Group 1 suboxides

Answer 28

Rb and Cs burn in limited amounts of O2 to form coloured suboxides i.e. Rb9O2; delocalised electrons cause metallic behaviour

Question 29

Group 1 hydroxides

Answer 29

Metal reacts with water to form hydroxide and hydrogen gas. ΔH is -ve, reactivity increases down the group. Li,Na, K less dense than H2O so react on the surface, the metal melts and, for K the H2 gas ignites. Rb and Cs are more dense so sink and rect explosively. NaOH produced industrially by electrolysis of NaCl (chloroalkali process) 2NaCl + 2H2O -> 2NaOH + H2 + Cl2

Question 30

Group 1 halides

Answer 30

Colourless ionic solids, high MPs, rock salt structure

Question 31

Group 1 ethynides

Answer 31

All metals react to with ethyne in liquid ammonia to form the NH2^- mono cation or the C2^2- dianion. Group 1 ethynides decompose in H2O to form LiOH and C2H2

Question 32

Group 1 nitrides

Answer 32

Li only metal to form a stable binary nitride (stability decreases down the group) Li3N decomposes in water to form LiOH and NH3

Question 33

Oxoanions

Answer 33

Group 1 metals form salts i.e. NO3-, CO3^2-, SO4^2-. Group 1 nitrates (NO3-) decompose on heating to nitrites (NO2-) and oxygen. For lithium decomposition gives 2Li2O + 4NO + 3O2 Nitrates become more stable down the group due to the decreasing difference between the lattice enthalpies of nitrate and nitrite making decomposition less favourable.. Compounds with small anions become less stable down the group (decrease in attic enthalpies of reactants). Compounds with large anions increase in stability down the group (decrease in lattice enthalpies of decomposition products).

Question 34

Solubilities of group 1 metals

Answer 34

A compound is soluble if ΔsolG < 0 as entropy cannot be ignored ΔsolG = ΔlattG + 2ΔhydH Large anions -> solubility decreases Small anions -> solubility increases Small cation and small anion/large cation and anion -> INSOLUABLE ΔlattG +ve and dominates as everything either all high or all low respectively For large and small combination ΔhydG of the smaller ion dominates hence -ve and dominates.

Question 35

Group 1 coordination chemistry

Answer 35

M+ are lewis acids so can react with lewis base ligands. Group 1 have low charge density and ar relatively large hence are weak lewis acids. Coordinate weakly to H2O, decreases down the group. Crown ether coordination; relationship between cavity size, canonic radius and stability -> optimum spatial fit. [n1]crown-n2 where n1=atoms in ring and n2=no oxygen atoms Cryptands are also size selective and have a central cavity