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Chapter 9: Problem 50

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

Short Answer

Expert verified
(a) The electron configuration of an isolated Sulfur (S) atom is: \(1s^2 2s^2 2p^6 3s^2 3p^4\). (b) The electron configuration of an isolated Chlorine (Cl) atom is: \(1s^2 2s^2 2p^6 3s^2 3p^5\). (c) In SCl₂, the Sulfur atom undergoes hybridization to form two \(sp^2\) hybrid orbitals that participate in bond formation with two Cl atoms. (d) After the \(sp^2\) hybridization, the remaining unhybridized valence orbitals on the Sulfur atom are one 3p orbital and one 3d orbital.

Step by step solution

01

(a) Electron configuration of an isolated Sulfur (S) atom

To determine the electron configuration of an isolated Sulfur atom, we need to know its atomic number, which is 16. The electron configuration can be represented as: \[1s^2 2s^2 2p^6 3s^2 3p^4\]
02

(b) Electron configuration of an isolated Chlorine (Cl) atom

For a Chlorine atom, the atomic number is 17. Its electron configuration is: \[1s^2 2s^2 2p^6 3s^2 3p^5\]
03

(c) Hybrid orbitals for S-Cl bonds in SCl₂

In SCl₂, Sulfur is the central atom which bonds with two Chlorine atoms. Sulfur has 6 valence electrons (3s², 3p⁴), and Chlorine has 7 valence electrons (3s², 3p⁵). To form two S-Cl bonds, the Sulfur atom needs two unpaired electrons that can participate in bonding. This can be achieved through hybridization. The Sulfur atom undergoes hybridization to form two \(sp^2\) hybrid orbitals in SCl₂. The 3s and two of the 3p orbitals on the S atom combine to form the \(sp^2\) hybrid orbitals that participate in bond formation with two Cl atoms.
04

(d) Unhybridized valence orbitals on the Sulfur atom in SCl₂

After the \(sp^2\) hybridization, the remaining valence orbitals on the Sulfur atom are one 3p orbital and one 3d orbital. The 3p orbital remains unhybridized as it does not participate in the bond formation. The 3d orbital is also unhybridized, as there are no d-electrons in the valence shell of Sulfur.

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Most popular questions from this chapter

(a) Which geometry and central atom hybridization would you expect in the series $\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+} ?(\mathbf{b})$ What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Consider the Lewis structure for acetic acid, which is known as vinegar: CCC(=O)O (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygen atoms, and what are the approximate bond angles at the oxygen that is connected to carbon and hydrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals combine to form a \(\pi\) bond? Explain.

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}:\) 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 \(90^{*} .\) Which compound would you predict to be the anticancer drug?

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: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(O\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central 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 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of 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.
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