Problem 63
Question
One of the following ions has a trigonal-planar shape: \(\mathrm{SO}_{3}^{2-} ; \mathrm{PO}_{4}^{3-} ; \mathrm{PF}_{6}^{-} ; \mathrm{CO}_{3}^{2-} .\) Which ion is it? Explain.
Step-by-Step Solution
Verified Answer
The ion that has a trigonal planar shape is \(CO_{3}^{2-}\).
1Step 1: Drawing Lewis Structure 1
Draw the Lewis structure of \(SO_{3}^{2-}\). It can be observed that sulphur (S) is the central atom bonded to three oxygen atoms and also carries a lone pair of electrons. This gives it a tetrahedral electronic geometry.
2Step 2: Drawing Lewis Structure 2
Next, draw the Lewis structure of \(PO_{4}^{3-}\). With phosphorus (P) as the central atom, it is bonded to four oxygen atoms, giving it a tetrahedral electronic geometry as well.
3Step 3: Drawing Lewis Structure 3
After that, draw the Lewis structure of \(PF_{6}^{-}\). Here, phosphorus (P) is the central atom and it is bonded to six fluorine atoms, resulting in an octahedral electronic geometry.
4Step 4: Drawing Lewis Structure 4
Finally, draw the Lewis structure of \(CO_{3}^{2-}\). In this case, carbon (C) is the central atom and is bonded to three oxygen atoms. This gives it three electron domains, meaning it has a trigonal planar electronic geometry.
Key Concepts
Lewis StructureTrigonal PlanarElectron DomainsTetrahedral Geometry
Lewis Structure
To understand molecular geometry and determine the shape of a molecule, drawing the Lewis structure is an essential first step. It provides a visual representation of the atoms in a molecule and their bonding.
Think of Lewis structures as blueprints for molecules. Here's how they work:
Think of Lewis structures as blueprints for molecules. Here's how they work:
- Identify the total number of valence electrons in the molecule or ion.
- Sketch the arrangement of atoms, starting by placing the least electronegative atom at the center (except for hydrogen).
- Connect the atoms with single bonds and distribute remaining electrons to satisfy the octet rule, or duet for hydrogen.
- Adjust electron pairs to manage formal charges and double check octet completion.
Trigonal Planar
Trigonal planar geometry is characteristic of molecules or ions with three electron domains around the central atom, all consisting of bonding pairs. This arrangement ensures the regions of electron density are spread as far apart as possible, minimizing repulsion among them.
In a trigonal planar shape, all atoms lie in a single plane:
In a trigonal planar shape, all atoms lie in a single plane:
- The central atom forms 120° bond angles with the surrounding atoms.
- Molecules like carbonate ion \((CO_{3}^{2-})\) commonly exhibit this geometry.
Electron Domains
An electron domain refers to a region around a central atom where electrons are likely to be found. This includes both bonding and non-bonding (lone) pairs.
Here's how to count electron domains:
Here's how to count electron domains:
- Each bond to an atom represents one domain, regardless of whether it's a single, double, or triple bond.
- Lone pairs are also considered as one domain each.
Tetrahedral Geometry
Tetrahedral geometry arises in molecules where the central atom is surrounded by four electron domains, typically all bonding pairs, arranged symmetrically around it.
Key characteristics of tetrahedral shapes:
Key characteristics of tetrahedral shapes:
- Identifiable by bond angles of about 109.5° between any two bonds.
- Common in molecules like methane (\(CH_4\)), where the central atom is spanned symmetrically by bonded atoms.
Other exercises in this chapter
Problem 61
Each of the following is either linear, angular (bent), planar, tetrahedral, or octahedral. Indicate the correct shape of (a) \(\mathrm{H}_{2} \mathrm{S} ;\) (b
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Predict the geometric shapes of (a) \(\mathrm{CO} ;\) (b) \(\mathrm{SiCl}_{4}\); (c) \(\mathrm{PH}_{3} ;\) (d) \(\mathrm{ICl}_{3} ;\) (e) \(\mathrm{SbCl}_{5} ;\
View solution Problem 65
Each of the following molecules contains one or more multiple covalent bonds. Draw plausible Lewis structures to represent this fact, and predict the shape of e
View solution Problem 66
Sketch the probable geometric shape of a molecule of (a) \(\mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{O}_{2} \mathrm{NNO}_{2}\right) ;\) (b) \(\mathrm{C}_{2} \m
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