1. ' ) Draw all bonding orbitals (i.e. “balloons" for s, sp, sp², sp³ and л) for the following compounds: a) b) c) H H C... H C=C=CH H H H -CEN

Consider the following flat drawing of methane (CH4) . a. What is HCH bond angle implied by this drawing if you assume it is flat? b. Are the electron domains of this flat CH4 spread out as much as possible? c. Use model materials to make a model of CH4 (methane). If you assembled it correctly, thefour bonds (bonding electron domains) of your model will be 109.5° apart. d. In which representation, the drawing above or the model in your hand (circle one) are theH’s of CH4 more spread out around the central carbon? e. Confirm that your model looks like the following drawing. The wedgebond represents a bond coming out of the page, and the dash bondrepresents a bond going into the page f. You will often see methane drawn as if it were flat (like on the previous page). Why is thismisleading, and what is left to the viewer’s imagination when looking at such a drawing?

Indicate whether each statement is true or false. (a) s orbitalscan only make σ or σ* molecular orbitals. (b) The probabilityis 100% for finding an electron at the nucleus in a π* orbital.(c) Antibonding orbitals are higher in energy than bondingorbitals (if all orbitals are created from the same atomic orbitals).(d) Electrons cannot occupy an antibonding orbital.

Indicate whether each statement is true or false. (a) s orbitalscan only make s or s* molecular orbitals. (b) The probabilityis 100% for finding an electron at the nucleus in a p* orbital.(c) Antibonding orbitals are higher in energy than bondingorbitals (if all orbitals are created from the same atomic orbitals).(d) Electrons cannot occupy an antibonding orbital.