Further fullerenes |
The point group of C24 is D6d which leads to the following cycle indices.
Z3(R(C24))= (1)/(12)( 2 e66 f12 f62 v64+ 2 e312 f12 f34 v38+ e218 f12 f26 v212+ 6 e12 e217 f27 v212+ e136 f114 v124)
Z3(S(C24))= (1)/(2)Z3(R(C24)) + (1)/(24)( 4 e123 f2 f12 v122+ 2 e49 f2 f43 v46 + 6 e14 e216 f14 f25 v14 v210 ).Since the C26 has a symmetry group of the form D3h we have
Z3(R(C26))= (1)/(6)( 3 e1 e219 f1 f27 v213+ 2 e313 f35 v12 v38 + e139 f115 v126 )
Z3(S(C26))= (1)/(2)Z3(R(C26)) + (1)/(12)( 2 e3 e66 f3 f62 v2 v32 v63 + 3 e15 e217 f15 f25 v14 v211 + e13 e218 f13 f26 v16 v210).The fullerene C28 is of tetrahedral symmetry Td so the cycle indices can be computed as
Z3(R(C28))= (1)/(12)( 8 e314 f1 f35 v1 v39 + 3 e12 e220 f28 v214 + e142 f116 v128 )
Z3(S(C28))= (1)/(2)Z3(R(C28)) + (1)/(24)( 6 e2 e410 f44 v47 + 6 e14 e219 f14 f26 v16 v211 ).The cycle index of R(C28) acting on its set of vertices can be found in [4] as well.
The C30 has D5h symmetry and the cycle indices for the action on the sets of edges, faces and vertices are
Z3(R(C30))= (1)/(10)( 4 e59 f12 f53 v56 + 5 e1 e222 f1 f28 v215 + e145 f117 v130)
Z3(S(C30))= (1)/(2)Z3(R(C30)) + (1)/(20)( 4 e5 e104 f2 f5 f10 v103 + 5 e15 e220 f15 f26 v16 v212 + e15 e220 f15 f26 v215 ).The symmetry group of C32 is D3, so it consists only of rotations and we have
Z3(R(C32))= (1)/(6)( 2 e316 f36 v12 v310 + 3 e12 e223 f29 v216 + e148 f118 v132 ).Since C34 has C3v symmetry the cycle indices are
Z3(R(C34))= (1)/(3)( 2 e317 f1 f36 v1 v311 + e151 f119 v134)
Z3(S(C34))= (1)/(2)(Z3(R(C34)) + e15 e223 f15 f27 v14 v215 ).The point group of C36 is D6h, from which the following cycle indices can be computed.
Z3(R(C36))= (1)/(12)( 2 e69 f2 f63 v66 + 2 e318 f12 f36 v312 + 4 e227 f12 f29 v218 + 3 e12 e226 f210 v218 + e154 f120 v136 )
Z3(S(C36))= (1)/(2)Z3(R(C36)) + (1)/(24)( 2 e32 e68 f2 f32 f62 v66 + e227 f210 v218 + 2 e69 f12 f63 v66 +
3 e14 e225 f14 f28 v18 v214 + 3 e16 e224 f16 f27 v14 v216 + e16 e224 f16 f27 v218 ).C38 is of C3v symmetry so we have
Z3(R(C38))= (1)/(3)( 2 e319 f37 v12 v312 + e157 f121 v138)
Z3(S(C38))= (1)/(6)( 2 e319 f37 v12 v312 + 3 e15 e226 f15 f28 v16 v216 + e157 f121 v138).For the C40 two possible symmetry groups are given, namely Td and D5d. In the first case the cycle indices are
Z3(R(C40))= (1)/(12)( 3 e230 f12 f210 v220 + 8 e320 f1 f37 v1 v313 + e160 f122 v140 )
Z3(S(C40))= (1)/(2)Z3(R(C40)) + (1)/(24)( 6 e415 f2 f45 v410 + 6 e16 e227 f16 f28 v14 v218 ).For the symmetry group D5d we compute
Z3(R(C40))= (1)/(10)( 4 e512 f12 f54 v58 + 5 e12 e229 f211 v220 + e160 f122 v140 )
Z3(S(C40))= (1)/(2)Z3(R(C40)) + (1)/(20)( 4 e106 f2 f102 v104 + e230 f211 v220 + 5 e16 e227 f16 f28 v14 v218 ).Since the point group of C42 is D3 we have
Z3(R(C42))= (1)/(6)( 2 e321 f12 f37 v314 + 3 e1 e231 f1 f211 v221 + e163 f123 v142 ).For the D3h symmetry group of C44 we can compute
Z3(R(C44))= (1)/(6)( 2 e322 f38 v12 v314 + 3 e12 e232 f212 v222 + e166 f124 v144)
Z3(S(C44))= (1)/(2)Z3(R(C44)) + (1)/(12)( 2 e32 e610 f32 f63 v2 v32 v66 + 4 e16 e230 f16 f29 v16 v219 ).A second form of C44 is chiral and has T as its symmetry group which gives the following cycle index:
Z3(R(C44))= (1)/(12)( 3 e12e232 f212 v222 + 8 e322 f38 v12v314 + e166 f124 v144 ).The symmetry group of C46 is C3 so we compute
Z3(R(C46))= (1)/(3)( 2 e323 f1 f38 v1 v315 + e169 f125 v146 ).The C48 has D3 as its symmetry group, so we have
Z3(R(C48))= (1)/(6)( 2 e324 f12f38 v316 + 3 e12 e235 f213 v224 + e172 f126 v148 ).For the D5h symmetry of C50 we derive
Z3(R(C50))= (1)/(10)( 4 e515 f12 f55 v510 + 5 e1 e237 f1 f213 v225 + e175 f127 v150 )
Z3(S(C50))= (1)/(2)Z3(R(C50)) + (1)/(20)( 4 e5 e107 f2 f5 f102 v52 v104 + 5 e17 e234 f17 f210 v14 v223 +
e15 e235 f15 f211 v110 v220 ) .The symmetry group of C52 is T, which consists only of rotational symmetries.
Z3(R(C52))= (1)/(12)( 3 e12e238 f214 v226 + 8 e326 f1 f39 v1 v317 + e178 f128 v152 ).C54 has point group D3, so we compute
Z3(R(C54))= (1)/(6)( 3 e1 e240 f1 f214 v227 + 2 e327 f12 f39 v318 + e181 f129 v154 ).For the D3d symmetry group of C56 the cycle indices are
Z3(R(C56))= (1)/(6)( 2 e328 f310 v12 v318 + 3 e12 e241 f215 v228 + e184 f130 v156 )
Z3(S(C56))= (1)/(2)Z3(R(C56)) + (1)/(12)( 2 e614 f65 v2 v69 + 3 e16 e239 f16 f212 v18 v224 +
e242 f215 v228 ) .The symmetry group of C58 is C3v so we have
Z3(R(C58))= (1)/(3)( 2 e329 f1 f310 v1 v319 + e187 f131 v158)
Z3(S(C58))= (1)/(2)(Z3(R(C58)) + e17 e240 f17 f212 v16 v226 ).C80 is the next fullerene with Ih symmetry. Its cycle index is
Z3(R(C80))=(1)/(60) ( 24 e524f12f58v516+ 20 e340f314v12 v326+ 15 e260f12 f220v240+ e1120f142v180)
Z3(S(C80))=(1)/(2) Z3(R(C80)) + (1)/(120) ( e260f221v240 + 20e620f67v2v613 + e18e256f18f21715v18v236 + 24e1012f2f104v108).The first fullerene with symmetry group I is the C140. Its cycle index is
Z3(R(C140))=(1)/(60) ( 24 e542f12f514v528+ 20 e370f324v12v346+ 15 e12e2104f236v270+ e1210f172v1140)
We could go on listing the cycle indices of many more fullerenes. In many cases it is possible to arrange the v vertices of Cv in several different ways leading to different symmetry groups and to different cycle indices. For instance for the fullerene C78 there are 4 possible isomers given in [9]. So from chemical properties we first have to determine the actual shape of the molecule. In [11] it is shown that C76 is of D2 symmetry and not of Td symmetry which would be possible as well.
Further fullerenes |