In the 2D representation of the ammonia molecule, we see one solid line, one solid triangle and one dashed triangle

enter image description here

A plane can be made to pass through any three given points. In other words, we can always find a plane which will pass through any three given points in space. So the central N atom and two H atoms will lie in a plane. Only one H will be out of plane. So I think, in the 2D representation, we need to show two solid lines and one solid triangle. Please give your opinion about this. Thanks

  • $\begingroup$ The image is fine. In the image the H with the straight line is in the plane with the N. The H with the dashed triangle is behind the plane, and the H with the solid triangle is in front of the plane. $\endgroup$ – MaxW May 21 at 9:30
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    $\begingroup$ You can draw the way you proposed too if you omit the lone pair (l.p.). Currently, the plane is defined by $\ce{\text{l.p.}-N-H}$ and this way the graphical depiction is also consistent with the one of tetrahedral environment of a carbon atom with $\mathrm{sp^3}$ hybridized orbitals. $\endgroup$ – andselisk May 21 at 10:21

According to the rules, yes, Ammonia can be drawn with three of it's atoms in one plane and the other outside the plane. Here's an example, along with a 3D projection:

2D view 3d projection

However, what do you comprehend from this structure? This structure does not convey the trigonal pyramidal geometry of the molecule very convincingly. It might be mistaken for trigonal planar at a glance, and this defeats the purpose of using the wedge/dash notation. Ammonia (and several other tetrahedral molecules) are represented in that manner since it is easiest to interpret their 3d structure from a 2d image.

enter image description here

| improve this answer | |
  • $\begingroup$ The choice is between trigonal pyramidal and trigonal planar. $\endgroup$ – Karsten Theis May 21 at 10:19

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