Main calculation

Finally we compute PtCl₄H₂^2- from the fragments PtCl₄^2- and H₂.

$ADFBIN/adf <<eor

title PtCl4 H2

units

length bohr

end

integration 4.0

xc

LDA vwn

GGA becke perdew

end

relativistic scalar

corepotentials t12rel &

H 0

Cl 1

Pt 2

end

ATOMS

Pt 0 0 0 f=PtCl4

Cl 4.361580 0.000000 0.00000000 f=PtCl4

Cl 0.000000 4.361580 0.00000000 f=PtCl4

Cl -4.361580 0.000000 0.00000000 f=PtCl4

Cl 0.000000 -4.361580 0.00000000 f=PtCl4

H 0.0 0.0 5.58 f=H2

H 0.0 0.0 7.26 f=H2

end

fragments

PtCl4 t21PtCl4

H2 t21H2

end

end input

eor

Note that, although the key CHARGE is not supplied, the molecule is not neutral: the default charge (that is, omitting the keys CHARGE, OCCUPATIONS) is the sum-of-fragments: the fragments here are H₂ and PtCl₄^2-, yielding a net charge for the molecule of minus two.

Note the F= fragment specification in the ATOMS block. No fragment-numbering suffix (/n) is required because there is only one fragment of each fragment type.