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Up until now, you've been drawing first a Lewis dot diagram (two-dimensional, shows lone electron pairs) and then a separate diagram showing molecule shape (three-dimensional, no lone pairs shown). Now begin combining the two, drawing one structure showing both threedimensional shape and all lone pairs. Draw such a structure for each of the following polyatomic ions. Name each shape, and indicate whether the ion has an overall dipole moment. If so, draw the dipole moment vector. (Hint: It's a good idea to continue to draw a regular Iewis diagram first, even though you do not show it in your final answer.) (a) \(\mathrm{CN}^{-}\) (b) \(\mathrm{ClO}_{4}^{-}\) (c) \(\mathrm{PCl}_{4}^{+}\) (d) \(\mathrm{NO}_{2}^{-}\)

Short Answer

Expert verified
(a) \(\mathrm{CN}^{-}\): Linear geometry with an overall dipole moment. The dipole moment vector runs from carbon to nitrogen. (b) \(\mathrm{ClO}_{4}^{-}\): Tetrahedral geometry with no overall dipole moment. (c) \(\mathrm{PCl}_{4}^{+}\): Tetrahedral geometry with no overall dipole moment. (d) \(\mathrm{NO}_{2}^{-}\): Bent geometry with an overall dipole moment. The dipole moment vector runs from nitrogen to oxygen.

Step by step solution

01

(a) Draw the Lewis structure and determine geometry for \(\mathrm{CN}^{-}\)

Step 1: Draw the Lewis dot structure for \(\mathrm{CN}^{-}\), which will look like this: C is triple-bonded to N with one lone pair on C and two lone pairs on N, and there is one extra electron on N due to the negative charge. C≡N: Step 2: Determine the electron pair geometry. The central carbon atom is surrounded by 2 electron groups: one triple bond and one lone pair. This will result in a linear electron pair geometry. Step 3: Determine the molecular geometry. The molecular geometry will also be linear due to the presence of only two electron groups. Step 4: Dipole moment identification. The ion consists of carbon and nitrogen, both of which have different electronegativities. In linear molecules, the dipole moments of the opposite ends are not balanced; thus, we have an overall dipole moment. Step 5: Draw the vector. The dipole moment vector runs from the less electronegative atom (carbon) to the more electronegative atom (nitrogen).
02

(b) Draw the Lewis structure and determine geometry for \(\mathrm{ClO}_{4}^{-}\)

Step 1: Draw the Lewis dot structure for \(\mathrm{ClO}_{4}^{-}\), which will look like this: Cl is single-bonded to four O atoms, and each O atom has three lone pairs, and there is one extra electron on Cl due to the negative charge. Cl - O: | | O O Step 2: Determine the electron pair geometry. Cl is surrounded by 4 electron groups (4 single bonds), giving a tetrahedral electron pair geometry. Step 3: Determine the molecular geometry. The molecular geometry will be tetrahedral as well since there are no lone pairs on the central atom. Step 4: Dipole moment identification. Although there are differences in electronegativity between Cl and O, the dipoles cancel each other out due to the symmetrical tetrahedral shape. Thus, there is no overall dipole moment.
03

(c) Draw the Lewis structure and determine geometry for \(\mathrm{PCl}_{4}^{+}\)

Step 1: Draw the Lewis dot structure for \(\mathrm{PCl}_{4}^{+}\), which will look like this: P is single-bonded to four Cl atoms and there is one less electron on P due to the positive charge. P - Cl | | Cl Cl Step 2: Determine the electron pair geometry. P is surrounded by 4 electron groups (4 single bonds), resulting in a tetrahedral electron pair geometry. Step 3: Determine the molecular geometry. The molecular geometry will be tetrahedral since there are no lone pairs on the central atom. Step 4: Dipole moment identification. All bonds are P-Cl, and there are no lone pairs on the central atom. Due to the symmetrical tetrahedral shape, there is no overall dipole moment.
04

(d) Draw the Lewis structure and determine geometry for \(\mathrm{NO}_{2}^{-}\)

Step 1: Draw the Lewis dot structure for \(\mathrm{NO}_{2}^{-}\), which will look like this: N is double-bonded to one O atom and single-bonded to another O atom. The double-bonded O has two lone pairs while the single-bonded O has three lone pairs. There is one extra electron on N due to the negative charge. - O // N \ O Step 2: Determine the electron pair geometry. N is surrounded by 3 electron groups: one double bond, one single bond, and one lone pair. This results in a trigonal planar electron pair geometry. Step 3: Determine the molecular geometry. With one lone pair, the molecular geometry will be bent. Step 4: Dipole moment identification. The molecule consists of nitrogen and oxygen atoms with different electronegativities, and the bent shape doesn't cancel out the dipole moments. Therefore, there is an overall dipole moment. Step 5: Draw the vector. The dipole moment vector runs from the less electronegative atom (nitrogen) to the more electronegative atom (oxygen).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry is a key aspect of understanding how molecules form shapes in space. It is determined by the arrangement of atoms around a central atom in a molecule or ion. This shape is crucial because it influences the molecule’s physical and chemical properties, such as polarity and reactivity.

To start, we need to first use the Lewis structure to identify the arrangement of electron pairs around the central atom. Once we know the electron pair geometry, we can deduce the molecular geometry by considering how these pairs repel each other in three-dimensional space.

The geometry can be linear, bent, trigonal planar, tetrahedral, and more depending on the number of bonding and lone pairs. For instance, the molecule \(\mathrm{CN}^{-}\) is linear because it only forms one triple bond between carbon and nitrogen. However, in \(\mathrm{NO}_{2}^{-}\), we see a bent geometry because of the presence of a lone electron pair affecting the angles between atoms.
Dipole Moment
Dipole moment is a measure of the overall separation of positive and negative charges in a molecule. It is an important concept as it helps predict the behavior of molecules in electric fields and is central to understanding molecular polarity.

A molecule has a dipole moment if there is an imbalance in the distribution of charges, often due to differences in electronegativity between atoms and the molecular geometry. In a linear molecule like \(\mathrm{CN}^{-}\), the difference in electronegativity between carbon and nitrogen creates a dipole moment, pointing from carbon to nitrogen.

Meanwhile, molecules such as \(\mathrm{ClO}_{4}^{-}\) and \(\mathrm{PCl}_{4}^{+}\) do not have dipole moments due to their symmetrical tetrahedral shape that allow for the individual bond dipoles to cancel out. In contrast, \(\mathrm{NO}_{2}^{-}\) has an overall dipole moment because of its bent shape, causing a net dipole from nitrogen towards the oxygens.
Electron Pair Geometry
Electron pair geometry refers to the spatial arrangement of electron pairs (lone and bond pairs) around the central atom. It's a more general concept than molecular geometry, which only considers the positions of the atoms.

Using the VSEPR (Valence Shell Electron Pair Repulsion) theory, we predict that electron pairs will arrange themselves to be as far apart as possible to minimize repulsion. As a result, different numbers of electron pairs lead to distinct electron geometries like linear, trigonal planar, or tetrahedral.

For example, \(\mathrm{ClO}_{4}^{-}\) has a tetrahedral electron pair geometry, meaning all four pairs of electrons (all are bonding pairs in this case) are symmetrically arranged. But \(\mathrm{NO}_{2}^{-}\) has a trigonal planar electron pair geometry since there are three groups, two bonds and one lone pair around nitrogen.
Chemical Bonding
Chemical bonding is the foundation for understanding how atoms form molecules and compounds. The nature of the bonds depends on how electrons are shared or transferred between atoms to achieve stability by fulfilling the octet rule.

There are mainly three types of bonding: ionic, covalent, and metallic. In the context of the provided examples, covalent bonding is predominant, where atoms share electrons.

For instance, in \(\mathrm{CN}^{-}\), a triple covalent bond is formed between carbon and nitrogen, sharing electrons for stability. Similarly, \(\mathrm{PCl}_{4}^{+}\) involves single bonds between phosphorus and chlorine atoms, distributing electrons to satisfy the octet rule. Understanding these bonds is essential in predicting how these molecules will behave in different environments and reactions.

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Most popular questions from this chapter

What is a dipole-dipole force? Give an example.

Which of each pair is most polar? Explain. (a) Chlorine and phosphorus trichloride (b) Carbon disulfide and sulfur difluoride (c) Nitrogen trifluoride and phosphorus trifluoride (d) Silicon tetrabromide and hydrogen cyanide \((\mathrm{HCN})\) (e) Nitrogen trichloride and oxygen dichloride

The atoms in the molecule HSCN are connected in the order given in the formula. (a) Draw the dot diagram. (b) Draw the molecule's three-dimensional shape, and label the numeric value of all bond angles. (c) Draw in the individual bond dipole moments. (d) Is the molecule polar? If yes, draw the molecular dipole moment vector.

There are exceptions to the predictions of VSEPR. Consider \(\mathrm{CH}_{3}\), known as a methyl radical. (a) Create a dot diagram for the methyl radical. How is it fundamentally different from other dot diagrams you have done? (b) Use VSEPR to predict the shape of the methyl radical and draw it with that shape (treat the odd electron as a single electron group). (c) The methyl radical is known to be planar with \(120^{\circ} \mathrm{H}-\mathrm{C}-\mathrm{H}\) angles. What steric number is being employed here, and what is the carbon atom doing with respect to the odd electron in determining molecular shape? (d) The \(\mathrm{CF}_{3}\) radical does obey VSEPR. Draw it according to its VSEPR-predicted shape. What steric number is being employed here? (e) The \(\mathrm{C}-\mathrm{H}\) bond is shorter than the \(\mathrm{C}-\mathrm{F}\) bond. When bonds are short, the atoms at the ends of the bonds can bang into each other (this is called steric congestion) unless a geometry is adopted to get around this. Use this knowledge to explain why \(\mathrm{CH}_{3}\) violates VSEPR, but \(C F_{3}\) does not.

Consider the phosphonium ion, \(\mathrm{PH}_{4}^{+}\). (a) Draw the dot diagram. (b) Draw the ion's three-dimensional shape, and label the numeric value of all bond angles. (c) What is the shape of this polyatomic ion? (d) Draw in the individual bond dipole moments.

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