G14

Question 1

how does oxidation state trend down G14

Answer 1

ranges from +4 to -4 in hydrides.

+4 OS becomes less important with respect to the 2+ OS going down group

Question 2

what are the two common allotropes of carbon

Answer 2

Diamond and graphite

Question 3

describe properties of diamond

Answer 3

sp3 C, 4 sigma bonds per C, strong in all directions, dense solid
high thermal conductivity
electrical insulator
C-C length 1.54 angstrum

Question 4

describe properties of graphite

Answer 4

sp2 trig planar

3sigma 1pi bond per C

2D delocalised pi orbital

alternate layers lie above one another,

stronger intralayer and weaker interlayer bonding,

lubricant

electrical conductor parallel to layer

forms intercalation compounds

Question 5

what is the issue with diamond

Answer 5

diamond is metastable as graphite is thermodynamically most stable form and so diamond conversion in graphite is favoured. this is very slow

Question 6

what are examples and properties of fullerenes

Answer 6

C60,70,76,78,80,84

sp2 carob
hexagonal/pentagonal rings like a football
all C same env 1 CNMR peak

bond lengths: 1.39 and 1.46 due to localised bonds

localised c=c undergo addition

Question 7

what structures do Si and Ge form

Answer 7

diamond like structure a multiple bonds are weaker between 2nd/3rd row elements - graphite like structure unfavourable

Question 8

what polymorphs do Tin form

Answer 8

metallic white-tin exists at 298 K where Sn is 6-coordinate

grey-tin when white tin cooled to 286 K contrains 4-coordinate Sn in a diamond like structure

Question 9

what are the uses of Carbon

Answer 9

diamond very hard used in cutting tools and abrasives

graphite inert, thermal stability, electrical conductivities

carbon black = vulcanized rubber

Question 10

what are uses of Si, Sn and Pb

Answer 10

Si used in steal and electronics as semiconductor

Silica (SiO2) glass and construction

Sn coat metals corrosion prevention and used in superconducting magnets

Pb in plumbing and paints

Question 11

what are general comments on the reactivity of G14

Answer 11

carbon doesn’t exhibit valency > 4
Si more reactive than C

going down group electropositive character and reactivity of elements increase

+4 OS dominant for C, Si and Ge compounds stability of +2 increases down (inert pair effect)

Question 12

what can Si combine with at high temperature

Answer 12

combines with O2, halides, S8, N2, P4, C and B

Question 13

what is catenation

how does this trend down G14

Answer 13

ability to form bonds with its self

decreases down group C»Si>Ge>Sn>Pb

Question 14

how are the hydrides of Si, Sn and Ge formed

Answer 14

SiCl4 + LiAlH4 -> SiH4 + LiCl + AlCl3

SnCl4 + LiAlH4 -> SnH4 + LiCl + AlCl3

GeO2 + Na[BH4] -> GeH4 + Na[BO2]

all analogues of CH4 and are very unstable compared to CH4

Question 15

how does SiH4 decompose

Answer 15

SiH4 + 2O2 -> SiO2 + 2H2O

SiH4 -> Si + H2 (add heat)

Question 16

which g14 tetrahalides are formed

Answer 16

known for all elements except for PbI4 as Pb-I bond is not strong enough to compensate for promotional energy required for hybridisation

All the halides are volatile covalent except SnF4 and PbF4 which are ionic

Question 17

how are G14 tetrahalides formed

Answer 17

CH4 + 4Cl2 -> CCl4 + 4HCl (add heat)

E + 2Cl2 -> ECl4
(E= Si , Ge, Sn)

GeO2 + 4HCl -> GeCl4 + 2H2O

Question 18

how does PbCl4 decompose

Answer 18

in contrast to others decomposes to PbCl2 and Cl2

due to inert pair effect which favours the +2 OS for Pb

Question 19

what makes EX4 lewis acids when E=Si Ge Sn Pb

Answer 19

use empty d orbitals to accept electrons
(G13 EX3 compounds used empty p orbitals to accept e)

eg:
SiF4 + F- -> SiF5- 10e-, trigonal bipyramidal, sp3d hybridised

Question 20

how does hydrolysis of tetra-halides occur

Answer 20

hydrolysis is thermodynamically favourable for all G14 elements
CX4 species are resistant to hydrolysis (kinetic reasons)
EX4+2H2O -> EO2 + 4HX

Si Ge Sn Pb can all extend coordination number beyond 4 which allows easy attack of EX4 by water.
C has no low-lying vacant orbitals and can’t do this

Question 21

describe G14 oxides

Answer 21

carbons forms CO and CO2 CO very strong due to efficient 2p-2p pi bonding

SiO2 forms polymeric 3D solids

dioxides of Ge Sn and Pb are all non-volatile solids
GeO2 display structure that closely resembles SiO2