1. Using Spartan:
(a) Calculate energies of the C and H atoms in their ground states (pay attention to the correct spin multiplicity you enter, i.e. 2S+1 - see “Lecture 7+2” for proper names);
(b) Obtain the equilibrium distance Re and the dissociation energy De of the CH molecule in its ground state;
(c) Gradually reduce (with step 0.1 Å) the C-H distance R (making it less than Re) and calculate the energies E until they cross the value Elimit for the dissociation limit (i.e. C + H), and record the distance R0 at which it happens;
(d) Now gradually increase the C-H distance R above the Re value and calculate energies E until they cross the value Elimit – (De/2);
(e) Put the distances R and the corresponding energies E you get in a table, and plot the graph of E(R), i.e. E as function of R, with the origin of energy at Elimit.
2. (a) Write down the VB wave function of CH, including explicitly only valence electrons of C and H, and neglecting hybridization and ionic contribution.
The valence bond wave function of hydrogen bonded to carbon. Using the valence bond model, the electrons of the molecule CH are located in the atomic orbitals of carbon and hydrogen. For hydrogen the electron configuration is 1s having one valence electron with the principle quantum number 1 part of the s orbital. The carbon’s electron configuration is 1s22s22p2 having four valence electrons on the principle quantum number of 2 from the s and p orbital. For the Valence bond wave function the principle quantum number is not needed. Neglecting hybridization of the s and p orbitals and ionic contributions from the carbon and hydrogen, the valence bond wave function will look as the following. Keeping in mind that neglecting hybridization and explicitly showing valences of carbon would mean the valence is equal to four. For hydrogen nothing has been affected since there is no hybridization but only a sigma bond formed. Though the type of valence bond theory used are for heteronuclear diatoms since CH is not a homonuclear diatom.
C=1s22s22p2=1s22s22px12py12pz0
-Currently there are 4 valence electrons that need to be filled to have a full p orbital.
H=1s1
ΨVB=1s22s22px12py1[1sH(1)2pzC(0) + 1sH(0)2pzC(1)]
Neglecting the hybridization of the s and p orbitals the electron on the hydrogen is shown on the Pz orbital on the CH molecule.
(b) Sketch the MO energy level diagram for CH (with valence electrons only), and show electrons by spin-arrows.
*the Pi and sigma 2 show that the sigma has one lone pair and one unpaired electron on the Carbon
*in sigma 3 shows that there is one bonding pair.
(c) Determine HOMO and LUMO, write down their wave functions in terms of appropriate AO’s, and draw them (with factor 1 for both C and H) using the Orbital Viewer.
(d) Specify the bond order of CH. Would you predict the molecule to stretch or shrink upon (1) ionization, (2) electron attachment, and in each case briefly explain why.
3. Using Spartan, test the transferability of covalent radii (rcov) of atoms:
(a) Obtain the Re values for H2 and C2 (in their ground states) and get the rcov(H) and rcov(C).
(b) Using these radii, predict the the R¬e value for CH and compare it with your result from part 1 above.
(c) Check how the same procedure works for CH+ and CH– (here pay attention to which atom is more likely to carry the charge – briefly explain why, and choose the corresponding homonuclear neutral and ionic diatoms accordingly).
is your teacher named fedor??? hhaahhaha I have the same assignment at UOIT