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Chapter 9: Problem 25

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN}\), (b) \(\mathrm{SO}_{3}^{2-}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{PF}_{6}^{-}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).

Short Answer

Expert verified
In summary, the electron-domain and molecular geometries for the given molecules and ions are as follows: (a) HCN - electron-domain geometry: linear, molecular geometry: linear (b) SO3^2- - electron-domain geometry: trigonal planar, molecular geometry: trigonal planar (c) SF4 - electron-domain geometry: trigonal bipyramidal, molecular geometry: see-saw (d) PF6^- - electron-domain geometry: octahedral, molecular geometry: octahedral (e) NH3Cl^+ - electron-domain geometry: trigonal bipyramidal, molecular geometry: T-shaped (f) N3^- - electron-domain geometry: trigonal planar, molecular geometry: bent

Step by step solution

01

(a) HCN

Step 1: Find the central atom The central atom in HCN is Carbon (C). Step 2: Calculate the total number of valence electrons H: 1 valence electron C: 4 valence electrons N: 5 valence electrons Total: 10 valence electrons Step 3: Apply the VSEPR theory As there are 3 atoms and no lone pair of electrons on Carbon, the electron-domain geometry is linear. The molecular geometry is also linear.
02

(b) SO3^2-

Step 1: Find the central atom The central atom in SO3^2- is Sulfur (S). Step 2: Calculate the total number of valence electrons S: 6 valence electrons 3 O: 6(3) = 18 valence electrons Charge: 2 additional electrons Total: 26 valence electrons Step 3: Apply the VSEPR theory As there are 3 oxygen atoms bonded to the central sulfur atom and no lone pairs on the sulfur, the electron-domain geometry is trigonal planar. The molecular geometry is also trigonal planar.
03

(c) SF4

Step 1: Find the central atom The central atom in SF4 is Sulfur (S). Step 2: Calculate the total number of valence electrons S: 6 valence electrons 4 F: 7(4) = 28 valence electrons Total: 34 valence electrons Step 3: Apply the VSEPR theory As there are 4 fluorine atoms bonded to the central sulfur atom and one lone pair on the sulfur, the electron-domain geometry is trigonal bipyramidal. The molecular geometry is see-saw (also known as distorted tetrahedral).
04

(d) PF6^-

Step 1: Find the central atom The central atom in PF6^- is Phosphorus (P). Step 2: Calculate the total number of valence electrons P: 5 valence electrons 6 F: 7(6) = 42 valence electrons Charge: 1 additional electron Total: 48 valence electrons Step 3: Apply the VSEPR theory As there are 6 fluorine atoms bonded to the central phosphorus atom and no lone pairs on the phosphorus, the electron-domain geometry is octahedral. The molecular geometry is also octahedral.
05

(e) NH3Cl+

Step 1: Find the central atom The central atom in NH3Cl^+ is Nitrogen (N). Step 2: Calculate the total number of valence electrons N: 5 valence electrons 3 H: 3 valence electrons Cl: 7 valence electrons Charge: 1 less electron Total: 14 valence electrons Step 3: Apply the VSEPR theory As there are 3 hydrogen atoms, 1 chlorine atom, and one lone pair on the nitrogen, the electron-domain geometry is trigonal bipyramidal. The molecular geometry is T-shaped.
06

(f) N3^-

Step 1: Find the central atom The central atom in N3^- is Nitrogen (N). Step 2: Calculate the total number of valence electrons 3 N: 5x(3) = 15 valence electrons Charge: 1 additional electron Total: 16 valence electrons Step 3: Apply the VSEPR theory As there are 2 nitrogen atoms bonded to the central nitrogen atom and one lone pair on the central nitrogen, the electron-domain geometry is trigonal planar. The molecular geometry is bent (also known as angular or V-shaped).

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Most popular questions from this chapter

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}{ }^{-}\)ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\)ion to be stable? (e) Which of the following statements about part (d) 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?

(a) What conditions must be met if a molecule with polar bonds is nonpolar? (b) What geometries will signify nonpolar molecules for \(\mathrm{AB}_{2}, \mathrm{AB}_{3}\), and \(\mathrm{AB}_{4}\) geometries?

The orbital diagram that follows presents the final step in the formation of hybrid orbitals by a silicon atom. (a) Which of the following best describes what took place before the step pictured in the diagram: (i) Two 3 p electrons became unpaired, (ii) An electron was promoted from the \(2 p\) orbital to the \(3 s\) orbital, or (iii) An electron was promoted from the \(3 s\) orbital to the \(3 p\) orbital? (b) What type of hybrid orbital is produced in this hybridization? [Section 9.5]

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}{ }^{-}\)localized or delocalized?

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?
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