Startseite    

Z-Matrix

Die Z-Matrix dient der numerischen Darstellung der geometrischen Struktur. Sie ist ein internes Koordinatensystem, bei dem ein molekularer Graph als Route durch das Molekül gewählt wird. Die dabei zur Anwendung kommenden Koordinaten sind Bindungslängen, Bindungswinkel und Diederwinkel. Sie soll hier am Beispiel des N2O4-Moleküls erläutert werden. Der $DATA-Input sieht wie folgt aus:

 $DATA
N2O4-Molekuel
DNH 2

N1
N2  1  1.800
O3  1  1.200  2  114.0
O4  1  1.200  3  132.0  2  180.0
O5  2  1.200  1  114.0  3    0.0
O6  2  1.200  5  132.0  1  180.0
 $END

Man hat hier N1 als Start-Atom gewählt. N2 ist mit diesem durch die Bindungslänge 1.800 A verbunden. O3 ist mit N1 mit der Bindungslänge 1.200 A verbunden und bildet mit N3 einen Bindungswinkel von 114° (blauer Graph)
O4 ist mit N1 durch die Bindungslänge von 1.20 A verbunden und bildet mit O3 einen Bindungswinkel von 132°. Mit N2 bildet O4 einen Diederwinkel von 180°. Man erkennt das am besten, wenn man die N1-N2-Bindung auf das O3-Atom verschiebt (grüner Graph). Der Diederwinkel ist durch den gekrümmten blauen Pfeil gekennzeichnet. Da O4 und N2 in in einer Ebene liegen und in "trans"-Stellung stehen, beträgt der Diederwinkel 180°. 
O5 ist mit N2 durch die Bindungslänge 1.20 A verbunden. Der Bindungswinkel 5-2-1 beträgt 114°. Der Diederwinkel 5-2-1-3 beträgt 0° (planare "cis"-Stellung)
O6 ist mit der Bindungslänge 6-2 (1.20 A) verbunden. Der Bindungswinkel 6-2-5 beträgt 132°. Der Diederwinkel 6-2-5-1 beträgt 180° (der grüne Graph entspricht der N2-N1-Bindung, die zum O5-Atom verschoben wurde.

Andere Beispiele: Butadien,

Die folgenden Ausführungen stammen von der Website "http://www.shodor.org/chemviz/"

Building a Z-Matrix:

When constructing a Z-matrix, you should follow these steps:

  1. Draw the molecule.
  2. Assign one atom to be #1.
  3. From the first atom, assign all other atoms a sequential number. When assigning atoms, you must be careful to assign them in an order that is easy to use. This will become clearer as you experiment with different molecules.
  4. Starting with atom #1, list the atoms you numbered, in order, down your paper, one right under the other.
  5. Place the atom designated as #1 at the origin of your coordinate system. The first atom does not have any defining measurements since it is at the origin.
  6. To identify the second atom, you must only define its bond length to the first atom. Use the reference charts given.
  7. For the third atom, you must define a bond length to atom #1 and a bond angle between atom #3 and atoms #1 and #2. (Bond angles are the angles between three atoms.)
  8. Remember that you can only use previously defined atoms when defining your current atom. This means that you can not reference atom #7 when defining atom #5.
  9. To identify atom #4 and all other atoms, you must include a bond length, bond angle and a dihedral angle. (Dihedral angles are the angles between an atom and the plane created by three other atoms.) This is done by using neighboring atoms to the atom you are describing. Again, the reference charts are helpful in locating bond lengths and angles.

Example 1: Water

This is the water molecule.
This is the numbered molecule. Use this for reference in this example.
  1. Label the oxygen as atom O1. Since the O1 atom is placed at the origin, you don't need to define it. Thus far, your z-matrix looks like this:

    O1

  2. Either hydrogen can be labelled as H2 and the other as H3.
  3. Starting with the H2 atom, we use the reference charts and find that the bond length between a single bond hydrogen-oxygen is 0.96 angstroms. Now your z-matrix should look like this:
    O1
    H2 1 0.96

  4. Finally, we want to describe atom H3. To do this, we first look at its bond length with atom O1. Again, using the reference charts, we find the bond length = 0.96 angstroms. We must also find the bond angle formed by the three atoms. Using the reference charts, we find this angle to equal 110.0 degrees. Your final z-matrix should look like this:
O1
H2 1 0.96
H3 1 0.96 2 110.0

Example 2: Acetaldehyde

This is acetaldehyde.
Notice the pyrimidal shape of the three hydrogens and the V-shape of the hydrogen and oxygen.
Here is the numbered molecule. Use this as a reference when following along in the example. Notice that the black arrow means that atom H7 is coming out of the screen and towards you. The shaded arrow H6 is going into the monitor.
These pictures show the five atoms that make up one plane. The included atoms are both carbons, the oxygen, and two hydrogens, each from opposite ends. This geometry is important when finding the dihedral angles.
  1. First we must decide which atom will be atom #1. Let's choose the carbon bonded to the oxygen and call it C1. Place this at the origin.
    C1

  2. Next we must order the rest of the atoms. Designate the oxygen O2, the hydrogen atached to C1 to be H3, the other carbon as C4, and the other hydrogens as atoms H5, H6, and H7.
  3. Now we must look at the oxygen. Using our reference charts, we find the double bond length equals, 1.22 angstroms.
    C1
    O2 1 1.22

  4. Next look at atom H3. Its bond length to atom C1 equals 1.09 angstroms. The bond angle formed between the first three atoms equals 120.0 degrees.
    C1
    O2 1 1.22
    H3 1 1.09 2 120.0

  5. Next we look at atom C4. Its bond length with atom C1 equals 1.54 angstroms. Its bond angle with atoms C1 and O2 equals 120.0 degrees. Its dihedral angle with atoms C1,O2,H3 equals 180.0 degrees.
    C1
    O2 1 1.22
    H3 1 1.09 2 120.0
    C4 1 1.54 2 120.0 3 180.0

  6. Now we must find a way to describe the atoms that are not bonded to atom C1. To do this, we will describe them with reference to their bond length with the carbon, atom C4, in addition to the angles they form with atoms C1,O2.
  7. Looking at atom H5, we find its bond length to atom C4 equals 1.09 angstroms. Its bond angle with atoms C1,C4 equals 110.0 degrees. Its dihedral angle with atoms C1,O2,C4 equals 000.0 degrees.
    C1
    O2 1 1.22
    H3 1 1.09 2 120.0
    C4 1 1.54 2 120.0 3 180.0
    H5 4 1.09 1 110.0 2 000.0

  8. We will do the same for atom H6. Its bond length to atom C4 equals 1.09 angstroms. Its bond angle with atoms C1,C4 equals 110.0 degrees. Its dihedral angle with atoms C1,O2,C4 equals 120.0 degrees.
    C1
    O2 1 1.22
    H3 1 1.09 2 120.0
    C4 1 1.54 2 120.0 3 180.0
    H5 4 1.09 1 110.0 2 000.0
    H6 4 1.09 1 110.0 2 120.0

  9. For atom H7, everything is the same as atom H6 except the dihedral angle formed with atoms C1,O2,C4. This angle will be -120.0 degrees since it is in the opposite direction as that of atom H6. Your final z-matrix for acetaldehyde is as follows:
    C1
    O2 1 1.22
    H3 1 1.09 2 120.0
    C4 1 1.54 2 120.0 3 180.0
    H5 4 1.09 1 110.0 2 000.0
    H6 4 1.09 1 110.0 2 120.0
    H7 4 1.09 1 110.0 2 -120.0

Example 4: Butane

This is Butane.

Steps:

Here is the numbered molecule. Use this as a reference when following along in the example. Notice that the black arrow means that the atom is coming out of the screen and towards you. The shaded arrows mean that the atom is going into the monitor.

 

  1. First we must decide which atom will be atom C1. We'll pick the carbon to the left and call it C1. Place this at the origin.
    C1

  2. Next we must order the rest of the atoms. Designate the other carbons C2, C3 and C4. The rest of the hydrogens will be H5 through H14.
  3. Now we must look at the C2 atom. Using our reference charts, we find the single bond length between two carbons equals, 1.54 angstroms.
    C1
    C2 1 1.54

  4. Next look at atom C3. Its bond length to atom C2 equals 1.54 angstroms. The bond angle formed between the first three atoms equals 110.0 degrees.
    C1
    C2 1 1.54
    C3 2 1.54 1 110.0

  5. Next we look at atom C4. As before, its bond length with atom C3 equals 1.54 angstroms. Its bond angle with atoms C2 and C3 equals 110.0 degrees. Its dihedral angle with atoms C1,C2,C3 equals 120.0 degrees.
    C1
    C2 1 1.54
    C3 2 1.54 1 110.0
    C4 3 1.54 2 110.0 1 120.0

  6. Atom H5 is the first hydrogen atom we will look at. We can see that it forms a 1.09 angstrom bond length with atom C1. It also forms a 110.0 bond angle with atoms C1 and C2 and a 120.0 degree dihedral angle with atoms C1,C2, and C3. Thus, the z-matrix now looks like this:
    C1
    C2 1 1.54
    C3 2 1.54 1 110.0
    C4 3 1.54 2 110.0 1 120.0
    H5 1 1.09 2 110.0 3 120.0

  7. All of the hydrogens will have the same measurements, but will reference different atoms. Remember to only use previously defined atoms when identifying your current atom. Your final z-matrix should look like this:
    C1
    C2 1 1.54
    C3 2 1.54 1 110.0
    C4 3 1.54 2 110.0 1 120.0
    H5 1 1.09 2 110.0 3 120.0
    H6 1 1.09 2 110.0 3 120.0
    H7 1 1.09 2 110.0 3 120.0
    H8 2 1.09 3 110.0 4 120.0
    H9 2 1.09 3 110.0 4 120.0
    H10 3 1.09 2 110.0 1 120.0
    H11 3 1.09 2 110.0 1 120.0
    H12 4 1.09 3 110.0 2 120.0
    H13 4 1.09 3 110.0 2 120.0
    H14 4 1.09 3 110.0 2 120.0

Example 3: Methyl Cyanide

This is Methyl Cyanide. Notice the pyrimidal shape of the three hydrogens and linear geometry of the two carbons and nitrogen. These three atoms form a 180 degree angle and will pose a problem when calculating the z-matrix.

There are two ways to build this z-matrix;

Using a Dummy Atom - (The preferred method)

Ignoring a connecting atom

Using a Dummy Atom:

Here is the numbered molecule. Use this as a reference when following along in the example. Notice that the black arrow means that atom H5 is coming out of the screen and towards you. The shaded arrow H4 is going into the monitor. The atom "X" is the dummy atom.

 

  1. First we must decide which atom will be atom #1. We'll pick the carbon attached to the hydrogens and call it C1. Place this at the origin.
    C1

  2. Next we must order the rest of the atoms. Designate the other carbon C2, the three hydrogens will be H3,H4,and H5, the dummy atom will be X6 and the nitrogen will be N7.
  3. Now we must look at the C2 atom. Using our reference charts, we find the single bond length between two carbons equals, 1.54 angstroms.
    C1
    C2 1 1.54

  4. Next look at atom H3. Its bond length to atom C1 equals 1.09 angstroms. The bond angle formed between the first three atoms equals 110.0 degrees.
    C1
    C2 1 1.54
    H3 1 1.09 2 110.0

  5. Next we look at atom H4. Its bond length with atom C1 equals 1.09 angstroms. Its bond angle with atoms C1 and C2 equals 110.0 degrees. Its dihedral angle with atoms C1,C2,H3 equals 120.0 degrees.
    C1
    C2 1 1.54
    H3 1 1.09 2 110.0
    H4 1 1.09 2 110.0 3 120.0

  6. Atom H5 is almost the same as atom H4 since their bond angles and bond lengths are identical. However, their dihedral angles are the negatives of each other since the atoms point in opposite directions. The new z-matrix is as follows:
    C1
    C2 1 1.54
    H3 1 1.09 2 110.0
    H4 1 1.09 2 110.0 3 120.0
    H5 1 1.09 2 110.0 3 -120.0

  7. Now, all we have left is the nitrogen. The nitrogen atom poses problems because it forms a 180 degree angle with the two carbons. This angle cannot be computed be many computer programs. So, scientists have learned to get around this problem by creating a "dummy atom." By placing a make-believe atom between the first carbon and the nitrogen, you break up the 180 degree angle into two 90 degree angles. Now you can solve the z-matrix. The X6 atom (the "X" atom) is given an arbitrary bond length of 1.00 angstroms. It forms a 90 degree bond angle with the atom C1. We will say that the dummy atom forms a 0.0 degree dihedral angle with atoms C1,C2,H3 in order to preserve the 180 degree angle for the nitrogen. The z-matrix is as follows:
    C1
    C2 1 1.54
    H3 1 1.09 2 110.0
    H4 1 1.09 2 110.0 3 120.0
    H5 1 1.09 2 110.0 3 -120.0
    X6 2 1.00 1 90.0 3 0.0

  8. Finally, we can identify the nitrogen atom (atom N7). We know that it creates a triple bond with the carbon of length 1.16 angstroms. It forms a 90 degree angle with atoms C2 and X6. Finally, we designate its dihedral angle with atoms C1,C2,X6 as 180 degrees. This is the final z-matrix:
    C1
    C2 1 1.54
    H3 1 1.09 2 110.0
    H4 1 1.09 2 110.0 3 120.0
    H5 1 1.09 2 110.0 3 -120.0
    X6 2 1.00 1 90.0 3 0.0
    N7 2 1.16 6 90.0 1 180.0

    Ignoring A Connecting Atom:

    This method is exactly like the method of using a dummy atom except for how we define the nitrogen atom.
    Here is the numbered molecule. Use this as a reference when following along in the example. Notice that the black arrow means that atom H5 is coming out of the screen and towards you. The shaded arrow H4 is going into the monitor.

     

    1. First we must decide which atom will be atom #1. We'll pick the carbon attached to the hydrogens and call it C1. Place this at the origin.
      C1

    2. Next we must order the rest of the atoms. Designate the other carbon C2, the three hydrogens will be H3,H4,and H5, and the nitrogen will be N6.
    3. Now we must look at the C2 atom. Using our reference charts, we find the single bond length between two carbons equals, 1.54 angstroms.
      C1
      C2 1 1.54

    4. Next look at atom H3. Its bond length to atom C1 equals 1.09 angstroms. The bond angle formed between the first three atoms equals 110.0 degrees.
      C1
      C2 1 1.54
      H3 1 1.09 2 110.0

    5. Next we look at atom H4. Its bond length with atom C1 equals 1.09 angstroms. Its bond angle with atoms C1 and C2 equals 110.0 degrees. Its dihedral angle with atoms C1,C2,H3 equals 120.0 degrees.
      C1
      C2 1 1.54
      H3 1 1.09 2 110.0
      H4 1 1.09 2 110.0 3 120.0

    6. Atom H5 is almost the same as atom H4 since their bond angles and bond lengths are identical. However, their dihedral angles are the negatives of each other since the atoms point in opposite directions. The new z-matrix is as follows:
      C1
      C2 1 1.54
      H3 1 1.09 2 110.0
      H4 1 1.09 2 110.0 3 120.0
      H5 1 1.09 2 110.0 3 -120.0

    7. Now we are once again left with the problem of defining the nitrogen atom. The problem is with the bond angle of 180 degrees. One way to get around this, other than using dummy atoms, is to ignore the second carbon(atom C2). When you do this, you act as if there is one huge bond between atom C1 and atom N6. You find the bond length by summing the bond length of the two carbon single bond and the carbon-nitrogen triple bond. This gives you 1.54 angstroms +1.16 angstroms = 2.70 angstroms.
    8. Next you look at the bond angle between atoms N6, C1, and H3. This is equal to 110.0 degrees. (When finding this, remember to treat the bond between C1 and N6 as a single bond.)
    9. The only thing left is the dihedral angle. However, this method does not allow for you to assign this angle. We say that it is undefined. Thus, you enter a dihedral angle equal to 0.00 degrees in reference to atom C2. Now your final z-matrix should look like this:
      C1
      C2 1 1.54
      H3 1 1.09 2 110.0
      H4 1 1.09 2 110.0 3 120.0
      H5 1 1.09 2 110.0 3 -120.0
      N6 1 2.70 3 110.0 2 0.00

Reference Charts and Hints:

When constructing a Z-matrix, you should follow these steps:
  1. Draw the molecule.
  2. Assign one atom to be #1.
  3. From the first atom, assign all other atoms a sequential number. When assigning atoms, you must be careful to assign them in an order that is easy to use. This will become clearer as you experiment with different molecules.
  4. Starting with atom #1, list the atoms you numbered, in order, down your paper, one right under the other.
  5. Place the atom designated as #1 at the origin of your coordinate system. The first atom does not have any defining measurements since it is at the origin.
  6. To identify the second atom, you must only define its bond length to the first atom. Use the reference charts given.
  7. For the third atom, you must define a bond length to atom #1 and a bond angle between atom #3 and atoms #1 and #2. (Bond angles are the angles between three atoms.)
  8. Remember that you can only use previously defined atoms when defining your current atom. This means that you can not reference atom #7 when defining atom #5.
  9. To identify atom #4 and all other atoms, you must include a bond length, bond angle and a dihedral angle. (Dihedral angles are the angles between an atom and the plane created by three other atoms.) This is done by using neighboring atoms to the atom you are describing. Again, the reference charts are helpful in locating bond lengths and angles.

Bond Length Hints:

  1. The bond length of each kind of bond varies very little from one particular compound to another.
  2. Single bonds of first-row elements (C, N, O, F) to hydrogen are all about 1 Å.
  3. Single bonds between first-row atoms are all about 1.5 Å.
  4. Double and triple bonds are shorter: 1.2 to 1.3 Å in first-row elements.
  5. Second-row, and higher, atoms (S, P, Cl, etc.) form correspondingly longer bonds.

Normal Bond Lengths (in angstroms)

H-C     1.09
H-N     1.00
H-O     0.96
C-C     1.54
C-N     1.47
C-O     1.43
C-Cl     1.76
C-Br     1.94
C-I     2.14
C=C     1.35
C=N     1.30
C=O     1.22
C (triple)C     1.20
C(triple)N     1.16
N-F		1.33
H-P		1.40

Bond Angles:

  1. Angles with all single bonds: 110 degrees
  2. Angles with a double bond: 120 degrees
  3. Angles with a triple bond: 180 degrees

Dihedral Angles:

  1. Angles with all single bonds: 120 degrees
  2. Angles with a double bond: 180 degrees

More Hints:

  • Remember that you can ONLY use previously defined atoms to identify the atom you are working on.
  • Angles can be positive and negative to represent directions. If one atom is going into the screen and another is coming out of the screen, one angle should be defined as negative and the other as positive. It does not matter which you chose to be positive or negative.

Z-Matrix Converter

http://www.shodor.org/chemviz/zmatrices/babel/html.