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(a) Does SCl\(_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does BeCl\(_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Short Answer

Expert verified
(a) Yes, SCl$_{2}$ has a dipole moment, with the net dipole pointing towards the more electronegative chlorine atoms. (b) No, BeCl$_{2}$ does not have a dipole moment due to its linear geometry causing symmetrical bond dipoles that cancel each other.

Step by step solution

01

Draw the Lewis structure for SCl2

Start by drawing the Lewis structure for SCl2: 1. Count the total number of valence electrons: S (6 valence electrons) + 2 Cl (2 x 7 valence electrons) = 20 valence electrons. 2. Write the central atom (S) followed by its surrounding atoms (2 Cl). 3. Distribute the valence electrons between the atoms to form a stable octet (fulfilling the octet rule for S and each Cl atom). 4. The Lewis structure for SCl2 is: Cl - S - Cl | | . . where "-" represents a single bond, and "." represents a lone pair of electrons.
02

Determine the molecular geometry using VSEPR

The number of electron regions around sulfur is 4 (2 bonding regions from the 2 Cl atoms and 2 lone pairs of electrons). According to the VSEPR theory, this gives the molecular geometry a bent/angular shape with a bond angle of approximately 104.5°, similar to that of water (H2O).
03

Determine the polarity and dipole moment of SCl2

SCl2 has a bent molecular geometry, and the electronegativity difference between sulfur (S) and chlorine (Cl) atoms indicates polar bonds. As a result, there is an uneven distribution of electron density that creates a net dipole moment pointing towards the more electronegative chlorine atoms.
04

Draw the Lewis structure for BeCl2

Now, draw the Lewis structure for BeCl2: 1. Count the total number of valence electrons: Be (2 valence electrons) + 2 Cl (2 x 7 valence electrons) = 16 valence electrons. 2. Write the central atom (Be) followed by its surrounding atoms (2 Cl). 3. Distribute the valence electrons between the atoms to fulfill the octet rule (Be can have less than 8 electrons though). 4. The Lewis structure for BeCl2 is: Cl - Be - Cl where "-" represents a single bond.
05

Determine the molecular geometry using VSEPR

There are only 2 electron regions around beryllium, which gives the molecular geometry a linear shape with a bond angle of 180°.
06

Determine the polarity and dipole moment of BeCl2

BeCl2 has a linear molecular geometry. Even though there is an electronegativity difference between beryllium (Be) and chlorine (Cl) atoms, indicating polar bonds, the linear geometry of the molecule causes the bond dipoles to be symmetrical and cancel each other out. Therefore, BeCl2 does not have a net dipole moment. To summarize the results: (a) SCl2 has a dipole moment, with the net dipole pointing towards the more electronegative chlorine atoms. (b) BeCl2 does not have a dipole moment due to its linear geometry which causes symmetrical bond dipoles that cancel each other.

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Key Concepts

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

Lewis structure
A Lewis structure is a simple diagrammatic way to represent molecules and their bonding. It shows how valence electrons are distributed around the atoms in a molecule. The main aim is to visualize the arrangement of atoms, the pairs of electrons that form bonds (bonded pairs), and the lone pairs or non-bonding electrons.
To create a Lewis structure, follow these steps:
  • Count the total number of valence electrons available for bonding.
  • Select a central atom, usually the one that needs the most bonds or is least electronegative (except hydrogen).
  • Distribute electrons to fulfill the octet rule, meaning each atom should have eight electrons around it, except hydrogen, which needs only two.
For example, in SCl\( _{2} \), we start by counting the total valence electrons (20). Sulfur is central, bonded to two chlorine atoms, forming a structure with single bonds and two pairs of lone electrons on sulfur.
VSEPR theory
The VSEPR (Valence Shell Electron Pair Repulsion) theory helps predict molecular shapes based on electron pair interactions. It is founded on the idea that electron pairs around a central atom will arrange themselves as far apart as possible to minimize repulsion.
This theory allows us to determine the three-dimensional shape of a molecule:
  • Identify the number of electron regions around the central atom, including both bonded atoms and lone pairs of electrons.
  • The shape of the molecule is dictated by these regions, where different numbers lead to distinct geometries like linear, trigonal planar, tetrahedral, etc.
In SCl\(_{2} \), the presence of two bonded pairs and two lone pairs on sulfur results in a bent or angular shape. In contrast, BeCl\(_{2} \) has only two bonded pairs, resulting in a linear shape.
Dipole moment
A dipole moment occurs in molecules where there is an uneven distribution of electrons across the bond, often due to differences in electronegativity between bonded atoms. It provides an insight into the polarity of the molecule.
Key points about dipole moments:
  • Directional: Dipole moments point towards the more electronegative atom.
  • Magnitude depends on the charge difference and the distance between the atoms.
For SCl\(_{2} \), the asymmetrical bent shape combined with polar bonds due to the electronegativity difference results in a net dipole moment, pointing towards the chlorine atoms. On the other hand, BeCl\(_{2} \) with its linear, symmetrical shape has its individual bond dipoles cancel out, leading to no net dipole moment.
Molecular geometry
Molecular geometry refers to the three-dimensional shape of a molecule, determined by the spatial arrangement of its atoms.
  • It helps in understanding reactivity, polarity, phase of matter, and how molecules interact with each other.
  • Different geometries include linear, bent, trigonal planar, and tetrahedral, among others.
For instance, SCl\(_{2} \) has a bent geometry due to the repulsion of the two lone pairs on sulfur, leading to a structure similar to H\(_{2}O \). Its bond angle is approximately 104.5°. BeCl\(_{2} \), however, is linear, resulting from the two chlorine atoms placed symmetrically around the beryllium, forming straight 180° angles. Both molecules exemplify how different electron configurations and arrangements significantly impact molecular geometry.

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Most popular questions from this chapter

Place the following molecules and ions in order from smallest to largest bond order: \(\mathrm{H}_{2}^{+}, \mathrm{B}_{2}, \mathrm{N}_{2}^{+}, \mathrm{F}_{2}^{+},\) and \(\mathrm{Ne}_{2}\) .

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_ \(_{4} )\) according to the following unbalanced reaction: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ in which the formal charges of all atoms are zero.(c) Use average bond enthalpies (Table 8.3 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

There are two compounds of the formula Pt \(\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} :\) The compound on the right is called cisplatin, and the compound on the left is called transplatin. (a) Which compound has a nonzero dipole moment? (b) One of these compounds is an anticancer drug, and one is inactive. The anticancer drug works by its chloride ions undergoing a substitution reaction with nitrogen atoms in DNA that are close together, forming a \(\mathrm{N}-\mathrm{Pt}-\mathrm{N}\) angle of about \(90^{\circ} .\) Which compound would you predict to be the anticancer drug?

Describe the bond angles to be found in each of the following molecular structures: (a) trigonal planar, (b) tetrahedral, (c) octahedral, (d) linear.

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 p}-\pi_{2 p}^{\star}\) ,\(\sigma_{25}-\sigma_{25}^{*},\) and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(N_{2} .\) (b) Is \(N_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(N_{2}\) molecule has the lowest energy peak at 170 nm. To what orbital transition does this corre- spond? (a) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of kJ/mol. (e) Is the N-N bondin the first excited state stronger or weaker compared to that in the ground state?

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