Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{SF}_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of these hybrid orbitals. (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4}\) ? Explain.

Short Answer

Expert verified
The sulfur atom's electron configuration is \(1s^{2}\, 2s^{2}\, 2p^{6}\, 3s^{2}\, 3p^{4}\). For SF2, one 3s and one 3p orbital combine to form two sp hybrid orbitals, which allow sulfur to form sigma bonds with two fluorine atoms. The large lobes of these sp hybrid orbitals are linear and have an angle of 180 degrees between them. This hybridization scheme is not appropriate for SF4 because it has four fluorine atoms that require four hybrid orbitals for bonding.

Step by step solution

01

Identify the sulfur atom's electronic configuration

Sulfur is in group 16 of the periodic table, and its atomic number is 16. Therefore, its electron configuration is \[1s^{2}\, 2s^{2}\, 2p^{6}\, 3s^{2}\, 3p^{4}\].
02

Construct the orbital diagram for sulfur

Based on the electron configuration, we can create the orbital diagram for sulfur. In the 3s subshell, there are two electrons, and in the 3p subshell, there are four electrons. The orbital diagram can be represented as: 3s: \(\uparrow\downarrow\) 3p: \(\uparrow\downarrow\), \(\uparrow\downarrow\), \(\uparrow\)
03

Hybridization for SF2

To describe the bonding in SF2, we need hybrid orbitals that can have two bonding pairs of electrons. One of the 3s and one of the 3p orbitals can combine to form two hybrid orbitals. This will allow the sulfur atom to form two sigma bonds with two fluorine atoms. The remaining two 3p orbitals and the unhybridized 3s orbital will contain lone electron pairs.
04

Name the hybrid orbitals

Since we are combining one 3s orbital and one 3p orbital, the hybrid orbitals are called "sp" orbitals. The two hybrid orbitals resulting from the combination of 3s and 3p orbitals are denoted by sp and sp.
05

Sketch the large lobes of sp hybrid orbitals

The sp hybrid orbitals result in two large lobes on opposite sides of the sulfur atom. The structure appears linear, with an angle of 180 degrees between the two lobes. The two large lobes are responsible for bonding with the fluorine atoms in SF2.
06

Determine if the hybridization scheme is appropriate for SF4

The hybridization scheme described in part (a) is not appropriate for SF4 because SF4 has four fluorine atoms that need to bond with the central sulfur atom. Thus, it requires four hybrid orbitals instead of two.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Orbital Diagram
The orbital diagram is a visual representation of the arrangement of electrons in an atom's orbitals. It helps to understand how electrons are distributed in an atom and is crucial when predicting how atoms will bond with each other. In our example, we're looking at sulfur, which has the electron configuration electronic configuration is \[1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{4}\]. The orbital diagram for sulfur shows us the spin direction of electrons in each orbital and how many electrons are present. Specifically, for sulfur's outer shell, the 3s and 3p subshells, we see the paired electrons in 3s and two pairs plus two unpaired electrons in the 3p orbitals. Understanding this diagram is essential to grasp the concept of hybrid orbitals and chemical bonding.

Electronic Configuration
Electronic configuration describes the distribution of electrons in an atom's orbitals. For example, sulfur's electronic configuration tells us that it has 16 electrons which fill the 1s, 2s, 2p, 3s, and 3p orbitals successively according to the rules of electron filling, such as Hund's Rule and the Pauli Exclusion Principle. This configuration is foundational for understanding how sulfur atoms will interact with other atoms and participate in bond formations. With this information, students can predict the type of bonds sulfur can form and how its orbitals will hybridize to accommodate bonding with other atoms, such as fluorine in \(\text{SF}_{2}\).

Chemical Bonding
Chemical bonding is the force that holds atoms together in compounds. There are several types of chemical bonds, including ionic, covalent, and metallic bonds. In the case of \(\text{SF}_{2}\), the bond between sulfur and fluorine is covalent, which means that electrons are shared between the atoms. Understanding the principles of chemical bonding helps in deciphering the behavior of elements when they form compounds. In our exercise, it is necessary to explore how sulfur's electrons can be rearranged through hybridization to create the optimal bonding conditions with fluorine.

sp Hybridization
sp hybridization occurs when one s orbital mixes with one p orbital to form two equivalent sp hybrid orbitals. This type of hybridization is seen in molecules where the central atom, like sulfur in \(\text{SF}_{2}\), forms two sigma bonds. Each hybrid orbital is made up of 50% s character and 50% p character, resulting in a linear arrangement due to the two hybrid orbitals aligning at 180 degrees to each other. Understanding sp hybridization is crucial for predicting the molecular geometry and bonding properties of molecules such as \(\text{SF}_{2}\), as it explains why sulfur can form two strong bonds with fluorine atoms.

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. The geometry of a molecule can be predicted by the Valence Shell Electron Pair Repulsion (VSEPR) theory, which shows that electron pairs around a central atom will arrange themselves to minimize repulsion. In \(\text{SF}_{2}\), the sp hybridization of the sulfur atom leads to a bent molecular shape, despite the linear arrangement of the sp orbitals. This is due to the lone pairs of electrons that are not involved in bonding, which exert repulsion forces on the bonding pairs, leading to the bent structure. An accurate understanding of molecular geometry is vital for grasping the physical properties and reactivity of molecules.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) 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) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

Consider the \(\mathrm{H}_{2}^{+}\)ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\)ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\)? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higherenergy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\)ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

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 charges 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 \(B\) 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 \(\mathrm{B}\) and \(\mathrm{N}\) 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 \(1.44 \AA\). Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(1.51 \AA \mathrm{A}\) 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?

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2}\) ? (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free