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P 21/a symmetry not found

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  • job314
    job314
    job314
    wrote on last edited by
    #1

    I am trying to input the following cif file Mg, Acta Crystallographica 17 (1964) 1478-1479.cif which is P 21/a space group 14. However, in the manual space group 14 is P 21/c and my geometry ends up looking funky. I also tried entering manually, e.g. P 2 1a but that simply does not work. Can you help me enter the correct space group here?

    CRYSTAL
    1 0 0
    P 2 1a
    9.383 12.669 6.220 107.05
    20
    7 0.1321 0.3509 0.3611
    1 0.058 0.337 0.225
    1 0.208 0.305 0.394
    1 0.095 0.344 0.487
    1 0.174 0.421 0.346
    12 0 0 0
    8 0.1603 -0.1094 -0.0307
    8 0.1685 0.1042 0.1656
    8 -0.0017 -0.0687 0.2986
    1 0.2 0.091 0.317
    1 0.227 0.134 0.116
    1 0.252 -0.096 0.059
    1 0.143 -0.176 -0.008
    1 -0.097 -0.066 0.341
    1 0.027 -0.135 0.325
    16 0.0953 -0.3605 0.2575
    8 -0.0469 -0.4174 0.2116
    8 0.2185 -0.4328 0.3718
    8 0.1185 -0.3211 0.0456
    8 0.0951 -0.2702 0.4089

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    • job314
      job314
      job314
      wrote on last edited by
      #2

      I think I got it but I do not know how to explain it. I set it up as below after lots of trial and error

      MgNH4SO4*6H2O from Mg, Acta Crystallographica 17 (1964) 1478-1479
      CRYSTAL
      1 0 0
      P 1 21/a 1
      9.383 12.669 6.220 107.05
      20
      7 0.1321 0.3509 0.3611
      1 0.058 0.337 0.225
      1 0.208 0.305 0.394
      1 0.095 0.344 0.487
      1 0.174 0.421 0.346
      12 0 0 0
      8 0.1603 -0.1094 -0.0307
      8 0.1685 0.1042 0.1656
      8 -0.0017 -0.0687 0.2986
      1 0.2 0.091 0.317
      1 0.227 0.134 0.116
      1 0.252 -0.096 0.059
      1 0.143 -0.176 -0.008
      1 -0.097 -0.066 0.341
      1 0.027 -0.135 0.325
      16 0.0953 -0.3605 0.2575
      8 -0.0469 -0.4174 0.2116
      8 0.2185 -0.4328 0.3718
      8 0.1185 -0.3211 0.0456
      8 0.0951 -0.2702 0.4089
      EOS
      RANGE
      0.94 1.06 8
      PREOPTGEOM
      MAXCYCLE
      500
      END
      BASISSET
      pob-TZVP-rev2
      DFT
      B3LYP-D3
      XLGRID
      END
      TOLINTEG
      8 8 8 8 16
      SHRINK
      4 4
      BIPOSIZE
      41202400
      EXCHSIZE
      41202400
      END

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      0
      • aerba
        aerba
        aerba
        Developer
        wrote on last edited by aerba
        #3

        Hi,

        I think it was just a syntax problem on the way the symbol of the space group was inputted. This is how it should look like in this case:

        TEST
        CRYSTAL
        1 0 0
        P 21/A
        9.383 12.669 6.220 107.05
        20
        7 0.1321 0.3509 0.3611
        1 0.058 0.337 0.225
        1 0.208 0.305 0.394
        1 0.095 0.344 0.487
        1 0.174 0.421 0.346
        12 0 0 0
        8 0.1603 -0.1094 -0.0307
        8 0.1685 0.1042 0.1656
        8 -0.0017 -0.0687 0.2986
        1 0.2 0.091 0.317
        1 0.227 0.134 0.116
        1 0.252 -0.096 0.059
        1 0.143 -0.176 -0.008
        1 -0.097 -0.066 0.341
        1 0.027 -0.135 0.325
        16 0.0953 -0.3605 0.2575
        8 -0.0469 -0.4174 0.2116
        8 0.2185 -0.4328 0.3718
        8 0.1185 -0.3211 0.0456
        8 0.0951 -0.2702 0.4089
        CIFPRT
        TESTGEOM
        END
        END
        

        For future reference, the syntax for space group symbols in CRYSTAL is discussed at pages 22-23 of the CRYSTAL23 User's Manual (see screenshot below):

        Screenshot 2025-05-07 alle 14.29.01.png

        We ran a test and the structure built from CRYSTAL looks reasonable:

        image0 (4).jpeg

        Hope this helps,

        Alessandro Erba
        Professor of Physical Chemistry
        Department of Chemistry, University of Torino
        [email protected]

        1 Reply Last reply
        ♥
        0
        • job314
          job314
          job314
          wrote on last edited by
          #4

          My initial confusion was that I entered

          0 0 0
          14

          and that led to problems. Today I rectified, right before your email as

          CRYSTAL
          1 0 0
          P 1 21/a 1

          1 Reply Last reply
          0
          • aerba
            aerba
            aerba
            Developer
            wrote on last edited by
            #5

            Good! The clean way is:

            CRYSTAL
            1 0 0
            P 21/A

            I guess the extra "1"s you put are safe as they correspond to the identity operator.

            Alessandro Erba
            Professor of Physical Chemistry
            Department of Chemistry, University of Torino
            [email protected]

            1 Reply Last reply
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