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Consider the \(\mathrm{SCl}_{2}\) molecule. (a) What is the electron configuration of an isolated S atom? (b) What is the electron configuration of an isolated Cl atom? (c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in \(\mathrm{SCl}_{2} ?\) (d) What valence orbitals, if any, remain unhybridized on the S atom in \(\mathrm{SCl}_{2} ?\)

Short Answer

Expert verified
(a) The electron configuration of an isolated S atom is \(1s^2 2s^2 2p^6 3s^2 3p^4\). (b) The electron configuration of an isolated Cl atom is \(1s^2 2s^2 2p^6 3s^2 3p^5\). (c) The hybrid orbitals needed for S-Cl bonds in \(\mathrm{SCl}_{2}\) are sp3 hybrid orbitals. (d) There are no unhybridized valence orbitals remaining on the S atom in \(\mathrm{SCl}_{2}\) as all four valence orbitals participate in sp3 hybridization.

Step by step solution

01

Part (a) - Electron Configuration of Sulfur

First, we need to determine the electron configuration of an isolated sulfur (S) atom. Sulfur has 16 electrons, and its electron configuration follows the Aufbau principle: S: \(1s^2 2s^2 2p^6 3s^2 3p^4\)
02

Part (b) - Electron Configuration of Chlorine

Next, we need to find the electron configuration of an isolated chlorine (Cl) atom. Chlorine has 17 electrons, and its electron configuration is as follows: Cl: \(1s^2 2s^2 2p^6 3s^2 3p^5\)
03

Part (c) - Hybrid Orbitals for S-Cl Bonds

To determine the hybrid orbitals needed for S-Cl bonds in \(\mathrm{SCl}_{2}\), we take into account the valence shell electron pair repulsion (VSEPR) theory. The central sulfur atom forms two sigma bonds with the two chlorine atoms and has two lone pairs. In VSEPR theory, this arrangement corresponds to a bent molecular geometry with a bond angle of about 104.5 degrees. Thus, we need a hybridization compatible with four orbitals arranging themselves in a bent geometry. This requires the sulfur atom to undergo sp3 hybridization, forming four sp3 hybrid orbitals – two to form sigma bonds with the chlorine atoms and two to hold the lone pairs.
04

Part (d) - Unhybridized Valence Orbitals on Sulfur

As we determined in part (c), sulfur undergoes sp3 hybridization for the \(\mathrm{SCl}_{2}\) molecule. Since all four valence orbitals (one 3s and three 3p) on sulfur are used in the hybridization process, there are no unhybridized valence orbitals remaining on the sulfur atom in \(\mathrm{SCl}_{2}\).

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Key Concepts

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

Understanding Hybrid Orbitals
Hybrid orbitals are a fascinating concept that helps explain the shapes of molecules and bonding behavior.
When atoms form covalent bonds, their atomic orbitals can blend to create hybrid orbitals. This blending happens to achieve lower energy and more effective overlapping needed for bonding.
  • Hybrid orbitals are formed by combining the standard atomic orbitals (s, p, d, etc.).
  • This process is essential for explaining molecular geometries observed in real-world compounds.
In the molecule we’re examining, \(\mathrm{SCl}_{2}\), the sulfur atom uses hybrid orbitals to bond with chlorine. These hybrid orbitals arrange in specific geometries to minimize repulsion between electron pairs. These arrangements are not random but follow predictable patterns influenced by the type of hybridization.
Exploring sp3 Hybridization
The concept of \(\text{sp}^3\) hybridization refers to the mixing of one s orbital and three p orbitals from the same atom. This results in the formation of four equivalent \(\text{sp}^3\) hybrid orbitals.
These orbitals are oriented in a way that they point towards the corners of a tetrahedron, allowing for an angle of 109.5 degrees between them.
  • In \(\mathrm{SCl}_{2}\), sulfur undergoes \(\text{sp}^3\) hybridization, even though its visible molecular shape is bent.
  • The reason sulfur uses \(\text{sp}^3\) hybridization is to accommodate its two sigma bonds with chlorine and two lone pairs.
Given this hybridization, the lone pairs and bonded atoms around sulfur push for maximum spatial separation, leading to the observed bent shape with an angle close to 104.5 degrees.
Applying VSEPR Theory
Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the shape of individual molecules based on electron-pair electrostatic repulsion.
This theory helps us understand the three-dimensional geometry that arises from molecular bonding.
  • The core idea is that electron pairs around a central atom will arrange themselves to be as far apart as possible.
  • In \(\mathrm{SCl}_{2}\), sulfur acts as the central atom with two bonded pairs and two lone pairs.
According to VSEPR theory, this arrangement results in a bent shape. The repulsion between lone pairs is greater than the repulsion between bonding pairs, which accounts for the smaller bond angle compared to a perfect tetrahedron. Understanding VSEPR theory allows chemists to predict shapes and reactivity patterns of molecules effectively.

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Most popular questions from this chapter

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{Fbond}\) angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

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.

The nitrogen atoms in \(\mathrm{N}_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has the stronger \(N-N\) bond?

In ozone, \(\mathrm{O}_{3}\) , the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

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