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Chapter 1: Problem 38

Which of the following substances has an asymmetrical molecular structure? (A) \(\mathrm{SF}_{4}\) (B) \(\mathrm{PCl}_{5}\) (C) \(\mathrm{BF}_{3}\) (D) \(\mathrm{CO}_{2}\)

Short Answer

Expert verified
The molecule with an asymmetrical molecular structure is \(\mathrm{SF}_{4}\).

Step by step solution

01

- Analyze \(\mathrm{SF}_{4}\)

Sulphur in \(\mathrm{SF}_{4}\) has 6 valence electrons. Four of these are used for bonding with fluorine atoms, and one lone pair remains. According to VSEPR theory, this molecule adopts a 'see-saw' shape, which is asymmetrical.
02

- Analyze \(\mathrm{PCl}_{5}\)

Phosphorus in \(\mathrm{PCl}_{5}\) has 5 valence electrons, all of which are used for bonding with the chlorine atoms. So, there are no lone pairs on the phosphorus atom. As per the VSEPR theory, it forms a trigonal bipyramidal structure which is symmetrical.
03

- Analyze \(\mathrm{BF}_{3}\)

Boron in \(\mathrm{BF}_{3}\) has 3 valence electrons which are used for bonding with the fluorine atoms. There are no lone pairs on boron, so according to VSEPR theory, the molecule is trigonal planar, which is a symmetrical structure.
04

- Analyze \(\mathrm{CO}_{2}\)

Carbon in \(\mathrm{CO}_{2}\) has 4 valence electrons. Two oxygen atoms double bond with the carbon, allowing for a linear structure without any lone pairs. This is also symmetrical according to the VSEPR theory.

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Key Concepts

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

VSEPR theory
VSEPR theory, which stands for Valence Shell Electron Pair Repulsion theory, is foundational in predicting the shape of molecules. It helps determine the 3D structure based on the repulsion between electron pairs around a central atom. The basic principle is that electron pairs, whether bonding or non-bonding (lone pairs), will position themselves as far apart as possible to minimize repulsion. This impacts molecular geometry, resulting in distinct shapes and angles.

The VSEPR theory considers not only bonding pairs but also lone pairs, which occupy more space and can thus affect the arrangement of atoms. For instance, in molecules like
  • oXyLoNiaTeX For example, in oXyLoNiaTeX in __lem ,molecules like SFAjp4 < / % >... ,sulfur tetrafluoride ( X(y.NiH( ( AGLEC2 o2 ) ,s the assew with very scy asymmetrically whereas in PCIL y / 124 . h phosphorous pentachloride ( B.F pd+ 5 remains symmetrically trigonal bipyramidal. Understanding how VSEPR theory applies to molecules helps anticipate their physical and chemical properties effectively.
Molecular geometry
Molecular geometry refers to the 3D arrangement of atoms within a molecule. It is crucial because it affects everything from boiling point to reactivity. Taking examples from the original problem:
  • In S F Mg 4, li... Send tetrafluorosulfach% the shape is described as a 'See-Saw,' canine: a asymmetrical due to an unsideryllic lone pair on the sulfidallynmelon. This structure makes SFM4 sulfur tetrafluoride ( Bz SF %j asymmetrical. + sine other hand_ In PHJ series of pentachloride ( P->CIM4aa:
    br=< is a symmetrical bocohagonal bipyramidal structure, as all over five to the pier _HHHTTS LCI< = valence electrons are shared.Link disk BBC solitarypair.
  • In ( C2 OM3 , boron trihalighth fuffyroune brackclra - tDDitional bonds are present, ! III! lettracaptioning its form to a planar triangle. That makes it symmetrical.
  • Lastly, in CO pg2 _carbondisufurlone H - OC ) O- JNMLL thorium one/| II I linearl mec! this symmetrical form comes from double bonds between carbon and oxygen.
By understanding these geometries, we can predict a molecule's behavior in different contexts, such as reactions and interactions with other molecules.
Valence electrons
Valence electrons play an essential role in chemical bonding and shape determination. They are the electrons found in the outermost shell of an atom, and they are crucial in forming chemical bonds. Based on the number of valence electrons, different molecular geometries are realized.

Let's look at the cases:
  • In SF 4, sulfur has six valence electrons. Out of these, four are used for sigma bonds with fluorine atoms, while two remain as a lone pair.
  • In P Cl licht 5, phosphorous has five valence electrons; each contributes to a bond with a chlorine atom, resulting in a plane-spanning trigonal bipyramidal shape.
  • In BF trikuorophosphisus 3, boron has three valence electrons, each engages one fluorine atom, forming a _symmetrical trigonal plane.
    Sustain_NE_FS _
  • For C O2, four of carbon's valence electrons are used in double bonds with oxygen atoms to create its linear form.
Knowing the count and configuration of valence electrons enables predictions around molecular geometry and characteristics.
Lone pairs
Lone pairs refer to valence electrons that are not involved in bonding. Their presence significantly affects molecular shape because they tend to occupy more space around the central atom compared to bonding pairs. This is primarily because they are attracted solely to the nucleus of the central atom and not shared with other atoms.

Here's how lone pairs work in different molecules:
  • In SF 4, the central sulfur atom has one lone pair. This causes an asymmetrical 'see-saw' shape, since the lone pair exerts a repulsive force on the other bonds.
  • In P Cl 5 and BF 3^, there are no lone pairs on the central atom. This allows them to maintain symmetrical trigonal bipyramidal and trigonal planar shapes, respectively.
  • Similarly, CO 2 doesn't have any lone pairs on its central carbon atom. This creates a straight-lived and symmetric linear shape.
Lone pairs, therefore, are pivotal in explaining why certain molecules deviate from standard geometries predicted by merely considering bonded atoms.

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Most popular questions from this chapter

A sample of a compound known to consist of only carbon, hydrogen, and oxygen is found to have a total mass of 29.05 g. If the mass of the carbon is 18.02 g and the mass of the hydrogen is 3.03 g, what is the empirical formula of the compound? (A) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (B) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\) (C) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{3}\) (D) \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\)

How many liters of distilled water must be added to 1 liter of an aqueous solution of HCl with a pH of 1 to create a solution with a pH of 2? (A) 0.1 L (B) 0.9 L (C) 2 L (D) 9 L

Starting with a stock solution of 18.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) , what is the proper procedure to create a 1.00 \(\mathrm{L}\) sample of a 3.0 \(\mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in a volumetric flask? (A) Add 167 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (B) Add 600 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (C) Fill the flask partway with water, then add 167 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water. (D) Fill the flask partway with water, then add 600 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water.

Molten \(\mathrm{AlCl}_{3}\) is electrolyzed with a constant current of 5.00 amperes over a period of 600.0 seconds. Which of the following expressions is equal to the maximum mass of Al(s) that plates out? (1 faraday = 96,500 coulombs) (A) \(\frac{(600)(5.00)}{(96,500)(3)(27.0)}\) grams (B) \(\frac{(600)(5.00)(3)(27.0)}{(96,500)}\) grams (C) \(\frac{(600)(5.00)(27.0)}{(96,500)(3)}\) grams (D) \(\frac{(96,500)(3)(27.0)}{(600)(5.00)}\) grams

A student added 1 liter of a 1.0\(M\) \(\mathrm{KCl}\) solution to 1 liter of a 1.0 \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution. A lead chloride precipitate formed, and nearly all of the lead ions disappeared from the solution. Which of the following lists the ions remaining in the solution in order of decreasing concentration? (A) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]\) (B) \(\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{K}^{+}\right]\) (C) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{Pb}^{2+}\right]>\left[\mathrm{NO}_{3}^{-}\right]\) (D) \(\left[\mathrm{K}^{+}\right]>\left[\mathrm{NO}_{3}^{-}\right]>\left[\mathrm{Pb}^{2+}\right]\)
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