MM2

NH3

Basis set: 6-31G(d,p)

calculation method: RB3LYP

point group: C3V

RMS Gradient Norm:0.00000485 a.u

the unique file

Maximum Force            0.000004     0.000450     YES
RMS     Force            0.000004     0.000300     YES
Maximum Displacement     0.000072     0.001800     YES
RMS     Displacement     0.000035     0.001200     YES
Predicted change in Energy=-5.986276D-10
Optimization completed.
   -- Stationary point found.

NH3 molecule

The first is a wagging bending; this is the umbrella mode. The second and third are scissoring bending and degenerate. This is shown by the bending having the same frequency and infrared absorption values. The fourth is a symmetric stretching. The fifth and sixth one are both an asymmetric stretch. there are 4 bands in the IR spectrum.

The nitrogen is more electronegative than the hydrogen. Hence the negative charge on the nitrogen. The negative charges cancel out the 3 positive charges on the hydrogen this shows it is correct.

H₂

Calculation Method:RB3LYP

Basis set:6-31G(d,p)

Point Group:D*H

RMS Gradient Norm:0.00000017 a.u.

the unique file

Maximum Force            0.000000     0.000450     YES
RMS     Force            0.000000     0.000300     YES
Maximum Displacement     0.000000     0.001800     YES
RMS     Displacement     0.000001     0.001200     YES
Predicted change in Energy=-1.164081D-13
Optimization completed.
   -- Stationary point found.

HYDROGEN MOLECULE

Since they are two identical hydrogen molecules there is no net charge.

N₂

Calculation Method:RB3LYP

Basis Set:6-31G(d,p)

RMS Gradient Norm:0.00000060 a.u.

Point Group:D*H

the unique file

Item Value Threshold Converged?

Maximum Force            0.000001     0.000450     YES
RMS     Force            0.000001     0.000300     YES
Maximum Displacement     0.000000     0.001800     YES
RMS     Displacement     0.000000     0.001200     YES
Predicted change in Energy=-3.400959D-13
Optimization completed.
   -- Stationary point found.

NITROGEN MOLECULE

Due to the identical nature of the nitrogen atoms each they both have a net charge of zero.

NH3 enthapy of formation

E(NH3)=-56.55776873 a.u.

2*E(NH3)=-113.1155374 a.u.

E(N2)=-109.52412868 a.u.

E(H2)=-1.17853936 a.u.

3*E(H2)=-3.53561808 a.u.

ΔE=2*E(NH3)-[E(N2)+3*E(H2)]=-146.47848285 kj/mol

The product is more stable because it is exothermic reaction: energy is released from the reaction. The products are of a lower energy level and so are more stable.

Chlorine

Calculation Method: RB3LYP

Basis Set:6-31G(d,p)

RMS Gradient Norm:0.00002510

2s orbital that do not contribute to bonding.

2p pi bonding

this is another 2p pi bond: this is a degenerate bond to the one above.

this is an antibonding orbital. That is a sigma bond.

this is a antibonding orbital that is degenerate immediately above.