Numeric Hessian for a non-planar H2NO molecule with a Cs Point Group

Both H atoms are bonded to N; the point group is Cs; the symmetry plane contains N-O bond.

$force temp(1) = 10.00, 20.00, 30.00, 40.00, 50.00,
60.00, 70.00, 80.00, 90.00, 100.00
method=fullnum projct=.t.
$end
$contrl maxit=200 mult=2 scftyp=rohf runtyp=hessian coord=unique $end
$contrl ispher=+1 $end
$contrl cctyp=cr-ccl $end
$basis gbasis=cct $end
$system timlim=1000000 mwords=400 memddi=0 $end
$scf soscf=.t. dirscf=.t. $end
$statpt nstep=100 opttol=0.00001 $end
$data
H2NO
Cs

N 7.0 -0.1357158331 0.0890286127 0.0000000000
H 1.0 -0.3822861363 -0.3680309994 0.8687014208
O 8.0 0.0165139267 1.3587083668 0.0000000000
$end

The following 2nd derivatives are evaluated:

Between atoms 1 and 1 9 (2x-x 4x-y 2y-y 1z-z)
Between atoms 1 and 2 (not calculated because 1-2 = 1-3)
Between atoms 1 and 3 (4*3*3=36)
Between atoms 2 and 2 (not calculated because 2-2 = 3-3)
Between atoms 2 and 3 21
Between atoms 3 and 3 18
Between atoms 4 and 1 18
Between atoms 4 and 2 (not calculated because 4-2 = 4-3)
Between atoms 4 and 3 (4*3*3=36)
Between atoms 4 and 4 9

Total = 147 + 1 = 148

Note that due to the Cs symmetry, E(z+dz) = E(z-dz) at z=0


E"(z=0) with respect to z = [E(z+dz) + E(z-dz) - 2E(z)]/dz^2


E"(z=0) = 2[E(z=dz) - E(z=0)]/dz^2


Also, at z=0, E(x+dx,z+dz) = E(x+dx,z-dz);
E(x-dx,z-dz) = E(x-dx,z+dz) due to the Cs symmetry.

Therefore,
E"(z=0) with respect to x and z
= [E(x+dx,z+dz) + E(x-dx,z-dz) - E(x-dx, z+dz) - E(x+dx, z-dz)]/4dxdz = 0