What is true about the CH2O molecule? Carbon utilizes sp2 hybridization. The H-C-H bond angle is expected to be slightly less than the H-C-O bond angle. The molecule contains 3 sigma bonds. The formal charge of all atoms equals zero in the best Lewis structure. All of the responses above are true. QUESTION 25 In the ICl2- ion, the lone pairs of electrons on the central atom occupy sp2 hybrid orbitals and are oriented 109.59° from each other.
The Correct Answer and Explanation is:
Correct Answer:
All of the responses above are true (for CH₂O).
For the ICl₂⁻ ion statement: It is false – the lone pairs occupy sp³d hybrid orbitals and are oriented 120° and 180°, not 109.5° from each other.
Explanation:
Part 1: CH₂O Molecule (Formaldehyde)
The carbon atom in formaldehyde (CH₂O) forms one double bond with oxygen and two single bonds with hydrogen. This results in three regions of electron density around carbon, giving it a trigonal planar geometry. To accommodate this shape, carbon undergoes sp² hybridization, where one s orbital and two p orbitals mix to form three sp² orbitals. These hybrid orbitals form three sigma (σ) bonds: two with hydrogen and one with oxygen. The second bond in the carbon–oxygen double bond is a pi (π) bond, which does not involve hybrid orbitals.
The ideal bond angle in trigonal planar geometry is 120°. However, due to the presence of a double bond (which has more electron density and repulsion), the H–C–H angle becomes slightly less than 120°, and the H–C=O angles become slightly more than 120°.
The formal charges in the best Lewis structure of CH₂O are all zero. Carbon has four bonds (zero formal charge), oxygen has two bonds and two lone pairs (zero formal charge), and each hydrogen has one bond (zero formal charge).
Therefore, all the given statements about CH₂O are true.
Part 2: ICl₂⁻ Ion
In the ICl₂⁻ ion, iodine is the central atom, surrounded by two bonding pairs and three lone pairs, making a total of five electron domains. This corresponds to a trigonal bipyramidal electron geometry, where the three lone pairs occupy the equatorial positions to minimize repulsion. The molecule itself has a linear shape, with the two bonding pairs 180° apart.
The hybridization of iodine in ICl₂⁻ is sp³d, not sp². Also, the angle between lone pairs is 120° in the equatorial plane, not 109.5°, which is associated with sp³ hybridization.
Thus, the statement about ICl₂⁻ is false.
