For each molecule below, draw in all implied lone pairs and/or protons (hydrogens) based on the formal charge shown.
Being able to determine implied lone pairs and/or hydrogens is a good skill to have, as its likely that much of the time your professor or TA will not write in all lone pairs, and will almost never draw in the implicit hydrogens. Until you are very comfortable with formal charges, you should always draw in all lone pairs and hydrogens on each atom.
MendelSet practice problem # 311 submitted by Matt on June 7, 2011.
For each molecule, determine the formal charge of the indicated atom.
Remember that when calculating formal charge, you count both electrons in a lone pair but only half of the electrons in a bonding pair. This is why a helpful formula is:
For example, in the first compound (the protonated oxygen), the oxygen has one lone pair ("2 dots") and three bonding pairs ("3 sticks"). Oxygen has a valence of 6, so its formal charge in this species is 6 - 5 = 1 or +1.
MendelSet practice problem # 310 submitted by Matt on June 7, 2011.
For the molecule shown below, indicate the hybridization (sp3, sp2, sp, etc.) of atoms A through H, and the bond angles of X, Y, and Z.
The trick to this problem is to look for implicit hydrogens- hydrogens (protons) that are not drawn in, but assumed to be there because there are no charges.
Also remember these rules:
If an atom has no multiple bonds, the total number of "things" (bonds and lone pairs) surrounding the atom is probably four, so it's sp3 hybridized (s1 + p3 = 4).
If an atom has one double bond, the total number of "things"surrounding the atom is probably three, so it's sp2 hybridized (s1 + p2 = 3).
If an atom has one triple bond or two double bonds, the total number of "things" surrounding the atom is probably two, so it's sp hybridized (s1 + p1 = 2).
Asp2 This oxygen has one double bond and two lone pairs. Three things total = sp2.
Bsp2 This carbon has one double bond and two single bonds, to a carbon and an oxygen (so two bonds). Three things total = sp2.
Csp3 This nitrogen has one long pair, two implicit hydrogens (two bonds), and one single bond to a carbo (another bond). Four things total = sp3.
Dsp3 This carbon has one implicit hydrogen and three single bonds. Four things total = sp3.
Esp3 This oxygen has two lone pairs and two bonding pairs. Four things total = sp3.
Fsp2 This carbon has one double bond (one bond), one implicit hydrogen (another bond), and is bonded to one other carbon (a third bond). So three bonds in total. Three things total = sp2.
Gsp This carbon has a triple bond (one bond) and a single bond (another bond). So two things in total = sp.
Hsp This nitrogen has a lone pair and a triple bond. Two things total = sp.
For angles, remember that electrons repel each other, and so will try to be as far apart from each other as possible:
Best way to separate two things is a straight line (180°).
Best way to separate three things is a "trifecta" (120°).
Best way to separate four things in three dimensional space is a tetrahedron (109.5°).
X 120° The carbon between the two oxygens is sp2 hybridized, so the shape is trigonal planar and the bond angle is 120°.
Y 180° The carbon bonded to the nitrogen is sp hybridized, so the shape is linear and the bond angle is 180°.
Z 109.5° The carbon is sp3 hybridized (the most common type of carbon). So the shape is tetrahedral and the bond angle is 109.5°.
MendelSet practice problem # 1286 submitted by Matt on October 2, 2011.
The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.
Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.
When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).
Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.
MendelSet practice problem # 755 submitted by Matt on July 27, 2011.
Draw the conjugate base forms of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).
Explain your reasoning.
Electronegativity increases as you travel from left to right along the periodic table, so the fluorine anion is more stable than a negative oxygen, which is more stable than a negative nitrogen, etc.
Because F- is the most stable, it is the weakest base, and its conjugate acid (HF) is the most acidic.
MendelSet practice problem # 286 submitted by Matt on June 6, 2011.
Draw the conjugate base form of each acid listed below, then rank the acids in order or decreasing acidity (1 = most acidic).
Explain your reasoning.
Perchlorate (ClO4-) has the most resonance forms and therefore has the most electron delocalization, so it is the most stable and weakest base, which makes its conjugate acid (HClO4) the strongest acid.
MendelSet practice problem # 303 submitted by Matt on June 7, 2011.
Rank each group of acids in order of decreasing acidity. (1 = most acidic)
Explain your reasoning. You will have to use more than one rule in your explanation (resonance, electronegativity, atomic radius, etc.).
Compounds 1 and 2 are more acidic than compound 3 because their conjugate bases have more resonance forms than that of compound 3.
Compound 1 is more acidic than compound 2 because the resonance form of its conjugate base has two oxygen atoms and is more stable than that of compound 2, which has one oxygen and one nitrogen; a negative oxygen atom is more stable than a negative nitrogen atom.
MendelSet practice problem # 306 submitted by Matt on June 7, 2011.