Complete the following problem, clearly showing all steps taken

Complete the following problem, clearly showing all steps taken: Draw a Lewis structure and calculate the formal charge for each element for ICl4-. Provide the name of the electron geometry and molecular geometry, and state whether the molecule is overall polar or nonpolar Explain, in complete sentences and in your own words, all of the steps and considerations used to solve the problem above C. Select a covalent compound that contains at least four groups and at least one lone pair. Draw a Lewis structure and calculate the formal charge for each element for this compound-. Provide the name of the electron geometry and molecular geometry, and state whether the molecule is overall polar or nonpolar

The Correct Answer and Explanation is:

ICl₄⁻ Lewis Structure and Formal Charge

Step 1: Count the Total Number of Valence Electrons

Iodine (I) belongs to group 17 (halogens), so it has 7 valence electrons. Each chlorine (Cl) atom is also in group 17 and has 7 valence electrons. Since there are four chlorine atoms, this gives:

• 1 iodine: 7 electrons
• 4 chlorines: 4 × 7 = 28 electrons

Additionally, the ion has a -1 charge, which means it has gained one electron. Thus, the total number of electrons is:

• 7 (I) + 28 (Cl) + 1 (charge) = 36 valence electrons

Step 2: Draw the Skeleton Structure

• Place iodine (I) in the center because it is less electronegative than chlorine (Cl).
• Connect each chlorine (Cl) atom to iodine (I) with a single bond. This uses up 4 pairs of electrons (4 × 2 = 8 electrons).

Step 3: Distribute the Remaining Electrons

• After placing the single bonds, we have 36 – 8 = 28 electrons remaining.
• Place 3 lone pairs of electrons around each chlorine atom (3 × 4 = 12 electrons). This uses up 12 of the remaining electrons.
• The remaining 16 electrons (28 – 12) will be placed as lone pairs on iodine (I). Since iodine is in period 5, it can accommodate more than 8 electrons, allowing it to have lone pairs.

Step 4: Check the Octet Rule

• All chlorine atoms are surrounded by 8 electrons (6 from lone pairs and 2 from the bond), satisfying the octet rule.
• Iodine has 4 bonds and 2 lone pairs, so it has 12 electrons in its valence shell. This is acceptable for iodine because it can expand its valence shell beyond 8 electrons.

Step 5: Formal Charge Calculation

The formal charge for an atom is calculated using the formula:Formal Charge=(Valence Electrons)−(Lone Electrons)−12(Bond Electrons)\text{Formal Charge} = (\text{Valence Electrons}) – (\text{Lone Electrons}) – \frac{1}{2} (\text{Bond Electrons})Formal Charge=(Valence Electrons)−(Lone Electrons)−21​(Bond Electrons)

• Chlorine (Cl):
  • Valence electrons = 7
  • Lone electrons = 6
  • Bond electrons = 2
  • Formal charge = 7 – 6 – 1 = 0 (for each chlorine)
• Iodine (I):
  • Valence electrons = 7
  • Lone electrons = 2
  • Bond electrons = 8 (4 bonds)
  • Formal charge = 7 – 2 – 4 = +1

So, the formal charge for iodine is +1, and for each chlorine atom, it is 0.

Step 6: Electron Geometry and Molecular Geometry

• Electron Geometry: The iodine atom has four bonding regions and two lone pairs, so the electron geometry is octahedral.
• Molecular Geometry: The molecular geometry is determined by the positions of the atoms, and since there are four chlorine atoms and two lone pairs on iodine, the geometry is square planar.

Step 7: Polarity

• Since the iodine atom is in the center with symmetrically arranged chlorine atoms, the dipoles cancel out, making the molecule nonpolar.

Covalent Compound with Four Groups and One Lone Pair

Let’s consider NH₃ (ammonia).

Step 1: Count the Total Number of Valence Electrons

• Nitrogen (N) has 5 valence electrons.
• Each hydrogen (H) atom has 1 valence electron, and there are 3 hydrogens, so:
  • 3 hydrogens: 3 × 1 = 3 electrons
• Total valence electrons = 5 (N) + 3 (H) = 8 electrons.

Step 2: Draw the Skeleton Structure

• Place nitrogen (N) in the center and connect it to three hydrogen (H) atoms with single bonds. This uses up 6 electrons (3 × 2 = 6).

Step 3: Place the Remaining Electrons

• After placing the bonds, we have 8 – 6 = 2 electrons left.
• These 2 electrons will form a lone pair on nitrogen.

Step 4: Formal Charge Calculation

• Hydrogen (H):
  • Valence electrons = 1
  • Lone electrons = 0
  • Bond electrons = 2
  • Formal charge = 1 – 0 – 1 = 0 (for each hydrogen)
• Nitrogen (N):
  • Valence electrons = 5
  • Lone electrons = 2
  • Bond electrons = 6 (3 bonds)
  • Formal charge = 5 – 2 – 3 = 0

Step 5: Electron Geometry and Molecular Geometry

• Electron Geometry: The nitrogen atom has four regions of electron density (three bonds and one lone pair), so the electron geometry is tetrahedral.
• Molecular Geometry: The geometry considering only the atoms (ignoring lone pairs) is trigonal pyramidal.

Step 6: Polarity

• Ammonia (NH₃) has a trigonal pyramidal shape, which creates an asymmetric distribution of electron density, so the molecule is polar.

In summary:

• ICl₄⁻ is nonpolar, with square planar geometry and a formal charge of +1 on iodine.
• NH₃ is polar, with trigonal pyramidal molecular geometry and a formal charge of 0 on all atoms.