# B3LYP/6-31G* opt=addgic
Title Card Required
1 1
[coordinate part]
XCR(inactive)=XCntr(2-25)
YCR(inactive)=YCntr(2-25)
ZCR(inactive)=ZCntr(2-25)
scan(StepSize=0.4,NSteps=12)=sqrt[(X(1)-XCR)^2+(Y(1)-YCR)^2+(Z(1)-ZCR)^2]*0.529177
where Li is atom 1, and 12-crown-4 is assumed to be atoms 2~25. The relaxed scan will be performed between Li and center of 12-crown-4, the initial distance corresponds to the inputted coordinate, the stepsize is 0.4 Angstrom, and number of steps is 12.
If you can only use G09, unfortunately it is impossible to realize the relaxed scan as you expected, you can only perform rigid scan, namely scanning Z coordinate of Li while keep positions of atoms of 12-crown-4 fixed.
]]>%chk=C:\salt.chk
# hf/3-21g
Title Card Required
1 1
Li 0.00000000 0.00000000 0.00000000
O 1.80378214 -0.00000000 0.32927408
C 2.30761918 1.26176110 -0.19253296
C 2.30761918 -1.26176110 -0.19253296
C 1.26176110 2.30761918 0.19253296
C 1.26176110 -2.30761918 0.19253296
O 0.00000000 1.80378214 -0.32927408
O -0.00000000 -1.80378214 -0.32927408
C -1.26176110 2.30761918 0.19253296
C -1.26176110 -2.30761918 0.19253296
C -2.30761918 1.26176110 -0.19253296
C -2.30761918 -1.26176110 -0.19253296
O -1.80378214 -0.00000000 0.32927408
H 3.26664225 1.49572411 0.24426407
H 2.39390318 1.19789109 -1.26923404
H 3.26664225 -1.49572411 0.24426407
H 2.39390318 -1.19789109 -1.26923404
H 1.49572411 3.26664225 -0.24426407
H 1.19789109 2.39390318 1.26923404
H 1.49572411 -3.26664225 -0.24426407
H 1.19789109 -2.39390318 1.26923404
H -1.49572411 3.26664225 -0.24426407
H -1.19789109 2.39390318 1.26923404
H -1.49572411 -3.26664225 -0.24426407
H -1.19789109 -2.39390318 1.26923404
H -3.26664225 1.49572411 0.24426407
H -2.39390318 1.19789109 -1.26923404
H -3.26664225 -1.49572411 0.24426407
H -2.39390318 -1.19789109 -1.26923404
I also asked this question in researchgate but I can't find an answer:
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