QQuestionChemistry
QuestionChemistry
"The following five species have a bent molecular shape. The central atom is underlined.
[H^2F]+ [IO^2]+ SeH^2 [NO^2]‾ SO^2"
10 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1:
Identify the central atom and its surrounding atoms (also known as ligands) in each molecule: - [H^2F]+: The central atom is F, and it is surrounded by two H atoms. - [IO^2]+: The central atom is I, and it is surrounded by two O atoms. - SeH^2: The central atom is Se, and it is surrounded by two H atoms. - [NO^2]‾: The central atom is N, and it is surrounded by two O atoms. - SO^2: The central atom is S, and it is surrounded by two O atoms.
Step 2:
Recall the valence electron pairs (VEP) theory, which states that the shape of a molecule is determined by the number of electron pairs (both bonding and nonbonding) around the central atom. The central atom tends to minimize repulsion between these electron pairs, leading to specific molecular shapes.
Step 3:
Determine the number of VEPs for each molecule: - [H^2F]+: F has 7 valence electrons (2 from bonds with H and 5 from the positive charge). Adding one electron from each H results in a total of 9 valence electrons, which are all bonding electrons (no lone pairs). So, there are 3 VEPs. - [IO^2]+: I has 7 valence electrons (2 from bonds with O and 5 from the positive charge). Adding one electron from each O results in a total of 9 valence electrons, which are all bonding electrons (no lone pairs). So, there are 3 VEPs. - SeH^2: Se has 6 valence electrons. Adding 2 electrons from each H results in a total of 10 valence electrons. Subtracting 2 electrons for the single bond between the two H atoms gives 8 valence electrons. These 8 electrons are distributed as 4 bonding pairs, resulting in 4 VEPs. - [NO^2]‾: N has 5 valence electrons. Adding 5 electrons from each O and one electron from the negative charge results in a total of 11 valence electrons. Subtracting 2 electrons for the double bond between N and O leaves 9 valence electrons. These 9 electrons are distributed as 3 bonding pairs and 2 lone pairs, resulting in 5 VEPs. - SO^2: S has 6 valence electrons. Adding 6 electrons from each O results in a total of 12 valence electrons. These 12 electrons are distributed as 2 bonding pairs and 4 lone pairs, resulting in 6 VEPs.
Step 4:
Analyze the VEP distribution and predict the molecular shape for each molecule using the table below (adapted from Molecular Shape and VSEPR Theory): | Number of VEPs | Molecular Shape | |----------------|-----------------| | 2 | Linear | | 3 | Trigonal Planar | | 4 | Tetrahedral | | 5 | Bent (Trigonal Bipyramidal) | | 6 | Octahedral | - [H^2F]+: 3 VEPs, Trigonal Planar - [IO^2]+: 3 VEPs, Trigonal Planar - SeH^2: 4 VEPs, Tetrahedral - [NO^2]‾: 5 VEPs, Bent (Trigonal Bipyramidal) - SO^2: 6 VEPs, Octahedral (but with 2 lone pairs, leading to a bent shape)
Step 5:
Final Answer
The bent molecular shapes are [H^2F]+, [IO^2]+, [NO^2]‾, and SO^2.
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students