Example: Bond Orders¶
With the key BONDORDER a bond order analysis is performed based on SFOs. Note that symmetry used in the calculation should be NOSYM. Shown here is only the example for benzene, where the bond orders calculated are with respect to the atomic fragments.
$ADFBIN/adf <<eor
title benzene BP/SZ bondorders tol=0.05
define
cc=1.38476576
ccc=120.0
dih=0.0
hc=1.07212846
hcc= 120.0
dih2=180.0
end
atoms Z-matrix
C 0 0 0
C 1 0 0 cc
C 2 1 0 cc ccc
C 3 2 1 cc ccc dih
C 4 3 2 cc ccc dih
C 5 4 3 cc ccc dih
H 2 1 3 hc hcc dih2
H 3 2 4 hc hcc dih2
H 4 3 5 hc hcc dih2
H 5 4 3 hc hcc dih2
H 6 5 4 hc hcc dih2
H 1 2 3 hc hcc dih2
end
basis
Type SZ
Core None
end
symmetry NOSYM
xc
gga becke perdew
end
bondorder tol=0.05 printall
noprint sfo
eor