Dear community,
I would like to model NH4PF6 using COMPASS II. I created two separated sketch files one for NH4 and the other one for PF6, then optimized the ions using Module/Forcite/Calculation/Geometry Optimization with different bond types to see the charge of each atom.
1. For the case of NH4: I consider three cases: (1) all N-H bonds are single bond, (2) three N-H bonds are single bond and one N-H bond is partial double bond, and (3) all N-H bonds are partial double bonds. Then, the atomic charge is independent of the selection of bond type as follow:
Case 1: All N-H bonds are single bond
XXXX_1:N1 N n4+ ? 0 1+ -0.1200 0 0 8 1.0000 0.0000 H1 H2 H3 H4
XXXX_1:H1 H h1+ ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
XXXX_1:H2 H h1+ ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
XXXX_1:H3 H h1+ ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
XXXX_1:H4 H h1+ ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
Case 2: Three N-H bonds are single bond and one N-H bond is partial double bond
XXXX_1:N1 N n4+ ? 0 1+ -0.1200 0 0 8 1.0000 0.0000 H1 H2/1.5 H3 H4
XXXX_1:H1 H h1n ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
XXXX_1:H2 H h1 ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1/1.5
XXXX_1:H3 H h1n ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
XXXX_1:H4 H h1n ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1
Case 3: All N-H bonds are partial double bonds
XXXX_1:N1 N n4+ ? 0 1+ -0.1200 0 0 8 1.0000 0.0000 H1/1.5 H2/1.5 H3/1.5 H4/1.5
XXXX_1:H1 H h1 ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1/1.5
XXXX_1:H2 H h1 ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1/1.5
XXXX_1:H3 H h1 ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1/1.5
XXXX_1:H4 H h1 ? 0 0 0.2800 0 0 8 1.0000 0.0000 N1/1.5
2. For the case of PF4: I also consider three cases: (1) all P-F bonds are single bond, (2) five P-F bonds are single bond and one P-F bond is partial double bond, and (3) all P-F bonds are partial double bonds. The atomic charge is totally different. Please see below:
Case 1: All P-F bonds are single bond
XXXX_1:P1 P p6- ? 0 1- 1.3796 0 0 8 1.0000 0.0000 F1 F2 F3 F4 F5 F6
XXXX_1:F1 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F2 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F3 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F4 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F5 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F6 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
Case 2: Five P-F bonds are single bond and one P-F bond is partial double bond
XXXX_1:P1 P p6- ? 0 1- 0.9830 0 0 8 1.0000 0.0000 F1/1.5 F2 F3 F4 F5 F6
XXXX_1:F1 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
XXXX_1:F2 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F3 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F4 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F5 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
XXXX_1:F6 F f1p ? 0 0 -0.3966 0 0 8 1.0000 0.0000 P1
Case 3: All N-H bonds are partial double bonds
XXXX_1:P1 P p6- ? 0 1- -1.0000 0 0 8 1.0000 0.0000 F1/1.5 F2/1.5 F3/1.5 F4/1.5 F5/1.5 F6/1.5
XXXX_1:F1 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
XXXX_1:F2 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
XXXX_1:F3 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
XXXX_1:F4 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
XXXX_1:F5 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
XXXX_1:F6 F f1 ? 0 0 0.0000 0 0 8 1.0000 0.0000 P1/1.5
Can you please explain why the above difference? And, what should I choose the bond type?
Thank you and best regards,
Trang
