Test: Dot Structures - MCAT MCQ

The HONC (pronounced hunk or honk) shortcut, which correspond to 1, 2, 3, or 4 bonds, will be employed for all questions in this set. When these four atoms have the corresponding bonds, then the atoms are neutral. One less bond to that atom makes its more negative by 1, and one more bond more positive by 1.
Here is a chart that summarizes the shortcut:

Start by drawing a skeleton of C−O and consider the double bond. Looking at the formal charges, carbon would have no formal charge, as well as the oxygen. This would seem correct, but remember the octet rule. It is not much of a rule since C, N, O, and F are the only four elements that must adhere to the rule. Unless there is a carbene (:CH2), carbon usually needs to fill its octet even though C=O seems to be a feasible Lewis dot structure.

n analyzing the triple bond C≡O, carbon has a formal charge of -1, and oxygen a formal charge of +1. Both have their octets filled, and so here is the Lewis dot structure:

To find the formal charge of the oxygen, start with the number of valence electrons, which is 6. Subtract any unbonded electrons, which is 2, and then subtract one-half of any bonded electrons, which is 3.

Generally in determining structures, the more electronegative atom tends to have a negative formal charge, and the more electropositive a positive one. Despite being more electronegative, oxygen does not get assigned the automatic negative charge but has instead a formal charge of +1.

Treat the sulfur as if it were an oxygen since they are both in the same group, remembering of course that sulfur can expand its octet.

Start with the skeleton S−C−N and quickly give carbon four bonds to complete its octet. There are only two options: S=C=N and S−C≡N. S≡C−N can be quickly eliminated since N cannot have only 1 bond.

Looking at S=C=N, when sulfur like oxygen has 2 bonds and carbon has 4 bonds, they both have no formal charge. When nitrogen has 2 bonds, it has a formal charge of -1.

Looking at S-C≡N, when sulfur like oxygen has 1 bond, it has a formal charge of -1. When nitrogen has 3 bonds and carbon has 4 bonds, they both have no formal charge.

From the Lewis dot structures above, sulfur can have a formal charge of -1 or 0.

Formal charge is the charge assigned to an atom in a molecule assuming that electrons are shared equally between the atoms regardless of electronegativity. Generally formal charges are +1, 0, or -1, but can range from -4 to +4.

Oxidation state is the number assigned to an atom based on a set of rules. The number represents the number of electrons an atom can lose, gain, or share when it is chemically bonded to another atom.

The atoms are assigned oxidation states based on a prescribed set of rules and range in value from -4 to +7.

Lastly, in assigning oxidation states, we treat the bonds as if they are ionic bonds where the more electronegative atom is awarded the electrons.

We have to determine that the resonance structure of nitrate, which is NO3⁻. Nitrogen seems to be the central atom since it can form 4 bonds, while oxygen cannot. We start with each oxygen single bonded.

When oxygen is single bonded, it has a formal charge of -1, which means in total -3 with the three oxygens. This is not a feasible Lewis dot structure, so let’s move on by making one of the bonds a double bond:

When oxygen is double bonded, it is uncharged. When nitrogen has four bonds, it takes on a charge of +1. When oxygen is single bonded, it takes on a charge of -1, and we have 2 for -2. In total, we have +1 and -2, giving us the -1 for the overall charge we are looking for:

The bond order is the total number of bonds divided by the number of attaching atoms, which gives 4/3 or 1.33

The fractional charge is the total charge on the attaching atoms divided by the number of attaching ions, which is -⅔.

The correct answer is a bond order of 1⅓ and a fractional charge of -⅔ .

Applying the HONC strategy, when nitrogen has 4 bonds, it takes on a charge of +1.

To have a complete octet, carbon must have at least 3 bonds and when it does it takes on a charge of -1.

R is representative of any alkyl group, and R can be single, double, or triple bonded depending on whether it is -CH, -CH₂ or -CH₃.

The correct resonance structure is:

Sulfur dioxide SO₂ is a classic trap for linear molecular geometry due to its likeness to CO₂, which is definitely linear. SO₂ is bent due to the single lone pair:

H₂O is bent due to the two lone pairs on the oxygen. Here is the Lewis dot structure:

The chlorite ion ClO₂^- is also bent due to the two lone pairs on the chlorine. Here is the Lewis dot structure:

The triiodide ion I₃^- is not bent because it has 3 lone pairs, but it has a linear molecular geometry with a trigonal bipyramidal electronic geometry.

Starting with periodic acid, iodine has to be the central atom since it is the only atom that can expand its octet.

Since there are 5 hydrogens and 6 oxygens, it seems likely there are 5 hydroxyl (-OH) groups and a lone bonded oxygen. Since iodine has 7 valence electrons, it can expand its octet to form 7 bonds.
since each hydroxyl group is single bonded, then the lone oxygen has to be double bonded to form this Lewis dot structure:

Since iodine could have 7 bonds, each of the 3 oxygens has to be double bonded and the 1 hydroxyl group single bonded to form this Lewis dot structure:

Periodic acid has an octahedral molecular geometry and metaperiodic acid has tetrahedral molecular geometry.

In cis-platin, platinum is in a Pt(II) form, which means that platinum still has a d⁸ configuration. Additionally, each of the ligands donates both electrons to form the coordinate covalent bond, so platinum has a 16 electron configuration. That is not necessary to answer the question since we are looking for a molecular geometry.

What determines the molecular geometry is the number of atoms attached, and here the ligands. Cis-platin has four ligands attached, 2 NH₃ and 2 Cl‾.

We need a molecular geometry that allows for only 4 attached atoms or ligands, and tetrahedral and square planar fits that bill. Square pyramidal and trigonal pyramidal are geometries with 5 and 3 attached atoms, respectively.

In order for the molecule to have a stereocenter, only tetrahedral, square pyramidal, and square planar would work as the molecular geometry. Tetrahedral would give a stereocenter if all four substituents or attached atoms were different, and square pyramidal once again needs 5 attached atoms or ligands.

Additionally, there are 4 lone pairs as well as four ligands. With four lone pairs, square planar and square pyramidal are the only configurations that can accommodate four electron pairs.

Only in the square planar configuration will the following geometric isomers form:

The hypochlorite ion ClO₃^- has 2 double-bonded oxygens and 1 single-bonded oxygen to the chlorine central atom. Since chlorine can have 7 bonds, with 5 bonds to oxygen, there are 2 electrons remaining as an unbonded pair. Hypochlorite has the trigonal pyramidal molecular geometry.

The phosphite ion PO₃^3- has a central phosphorus atom with 3 single-bonded oxygens each with a negative charge. Since phosphorus has 5 valence electrons, there are 2 electrons remaining for a lone pair. Phosphite has the trigonal pyramidal molecular geometry.

Ammonia NH₃ , has a central nitrogen with 3 single-bonded hydrogens. Since nitrogen has 5 valence electrons, there are 2 electrons remaining for an unbonded pair.

Sulfur trioxide SO₃ has a central sulfur atom attached to 3 oxygens with double bonds. The molecular geometry is trigonal planar, and it has no molecular dipole.

Chlorine has seven valence electrons and can expand its octet to form 7 bonds. Chlorine has to be the central atom.

Fluorine has to be single bonded since it cannot expand its octet. Chlorine has 3 single bonds to fluorine, and there are 4 electrons left to make 2 lone pairs.

Count the number of attaching atoms and lone pairs to determine the electronic geometry. For a count of 5, the geometry is trigonal bipyramidal.

The molecular geometries for trigonal bipyramidal electronic geometry are linear, T-shaped, see-saw, and trigonal bipyramidal.

With 3 attached atoms, the molecular geometry is T-shaped. Here is the Lewis dot structure: