Draw a Lewis structure for OCl-, Ethane, C2H4. Predict the Geometry of NH4+. State the hybridization of Nitrogen in NH3. Give the number of pi and sigma bonds in NH3, NO2, N2. Draw structural formula of 3-isobutylhexane, 2,3,3-trimethyl-2-hexene.

Question One

a) Draw a Lewis structure for:

i) OCl⁻

Step 1: Count total valence electrons. Oxygen has 6, Chlorine has 7, and the negative charge adds 1 electron.
Total valence electrons = $6 + 7 + 1 = 14$.

Step 2: Draw a single bond between O and Cl. Distribute the remaining electrons to satisfy the octet rule for both atoms.

$$\begin{align*} &\text{[:}\ddot{\text{O}}-\ddot{\text{Cl}}: ]^{-} \end{align*}$$

The oxygen atom has 3 lone pairs and 1 bonding pair (8 electrons). The chlorine atom has 3 lone pairs and 1 bonding pair (8 electrons). The overall charge is -1.

ii) Ethane, C₂H₄
Assumption: The formula C₂H₄ refers to ethene, not ethane (which is C₂H₆).

Step 1: Count total valence electrons. Each Carbon has 4, each Hydrogen has 1.
Total valence electrons = $2(4) + 4(1) = 8 + 4 = 12$.

Step 2: Connect the two carbon atoms and then connect each carbon to two hydrogen atoms.

Step 3: Form a double bond between the carbon atoms to satisfy the octet rule for both carbons, as each carbon needs 8 electrons.

$$\begin{align*} &\text{H}_{2}\text{C}=\text{CH}_{2} \end{align*}$$

Each carbon atom forms two single bonds with hydrogen atoms and one double bond with the other carbon atom, satisfying the octet rule. Each hydrogen atom has 2 electrons.

b) Predict the Geometry of NH₄⁺

Step 1: Determine the central atom (Nitrogen) and count its valence electrons (5). Each Hydrogen contributes 1 electron. The positive charge means one electron is removed.
Total valence electrons = $5 + 4(1) - 1 = 8$.

Step 2: Nitrogen forms single bonds with four hydrogen atoms. This uses $4 \times 2 = 8$ electrons.

Step 3: Nitrogen has no lone pairs. There are 4 bonding pairs around the central nitrogen atom.

Step 4: According to VSEPR theory, 4 electron domains (all bonding pairs) result in a tetrahedral electron geometry and molecular geometry.
The geometry of NH₄⁺ is tetrahedral.

c) State the hybridization of Nitrogen in NH₃

Step 1: Determine the central atom (Nitrogen) and count its valence electrons (5). Each Hydrogen contributes 1 electron.
Total valence electrons = $5 + 3(1) = 8$.

Step 2: Nitrogen forms single bonds with three hydrogen atoms, using 6 electrons. The remaining $8 - 6 = 2$ electrons form one lone pair on the nitrogen atom.

Step 3: Count the number of electron domains around the central nitrogen atom: 3 bonding pairs (to H atoms) + 1 lone pair = 4 electron domains.

Step 4: A central atom with 4 electron domains is sp³ hybridized.
The hybridization of Nitrogen in NH₃ is sp³.

d) Give the number of pi ($\pi$) and sigma ($\sigma$) bonds in:

i) NH₃

Step 1: Draw the Lewis structure for NH₃. Nitrogen forms three single bonds with hydrogen atoms and has one lone pair.

Step 2: Each single bond is a sigma ($\sigma$) bond. There are no double or triple bonds.
Number of $\sigma$ bonds = 3
Number of $\pi$ bonds = 0

ii) NO₂
Assumption: This refers to the neutral nitrogen dioxide molecule, NO₂.

Step 1: Draw the Lewis structure for NO₂. Nitrogen is the central atom, forming one double bond with an oxygen atom, one single bond with another oxygen atom, and has one unpaired electron.

Step 2: A single bond contains one $\sigma$ bond. A double bond contains one $\sigma$ bond and one $\pi$ bond.
Number of $\sigma$ bonds = 2 (one for N-O, one for N=O)
Number of $\pi$ bonds = 1 (from the N=O double bond)

iii) N₂

Step 1: Draw the Lewis structure for N₂. The two nitrogen atoms are connected by a triple bond, and each nitrogen has one lone pair.

Step 2: A triple bond consists of one $\sigma$ bond and two $\pi$ bonds.
Number of $\sigma$ bonds = 1
Number of $\pi$ bonds = 2

Question Two

a) Draw structural formula of 3-isobutylhexane

Step 1: Identify the parent chain: hexane, which means 6 carbon atoms in a straight chain.

Step 2: Identify the substituent: isobutyl group, which is $-\text{CH}_{2}\text{CH}(\text{CH}_{3})_{2}$.

Step 3: Attach the isobutyl group to the 3rd carbon of the hexane chain.

$$\begin{align*} &\text{CH}_{3}-\text{CH}_{2}-\text{CH}(\text{CH}_{2}\text{CH}(\text{CH}_{3})_{2})-\text{CH}_{2}-\text{CH}_{2}-\text{CH}_{3} \end{align*}$$

Or, drawn out:

$$$$