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Chapter 1: Problem 113

Ir terms of polat character, which of the following order is correct? (a) \(\mathrm{NH}_{3}<\mathrm{H}_{z} \mathrm{O}<\mathrm{HF}<\mathrm{H}_{2} \mathrm{~S}\) (b) \(11_{2} \mathrm{~S}<\mathrm{NH}_{3}<\mathrm{H}_{2} \mathrm{O}<11 \mathrm{~F}\) (c) \(\mathrm{H}_{2} \mathrm{O}<\mathrm{N} 11_{5}<11_{2} \mathrm{~S}<\mathrm{H} \mathrm{F}\) (d) \(\mathrm{H} \mathrm{F}<\mathrm{H}_{2} \mathrm{O}<\mathrm{NH}_{3}<\mathrm{H}_{2} \mathrm{~S}\)

Short Answer

Expert verified
Option (b) H_2S < NH_3 < H_2O < HF is correct.

Step by step solution

01

Define Polarity

Polarity in molecules arises due to differences in electronegativity between atoms forming a molecule. As electronegativity differences increase, polarity also increases. A highly electronegative atom bonded with a less electronegative atom results in a stronger dipole moment.
02

Analyze Hydrogen Fluoride (HF)

Among the given options, HF is highly polar due to the large electronegativity difference between hydrogen and fluorine. HF should appear at the higher end of the polarity spectrum.
03

Compare Water (H_2O)

Water (H_2O) is highly polar due to the electronegative oxygen atom creating a significant dipole with hydrogen atoms. However, HF is still more polar than water because fluorine is more electronegative than oxygen.
04

Evaluate Ammonia (NH_3)

Ammonia is less polar than water and HF, due to the relatively smaller electronegativity difference between nitrogen and hydrogen.
05

Assess Hydrogen Sulfide (H_2S)

Hydrogen sulfide is the least polar among the group because sulfur has a relatively lower electronegativity compared to oxygen and nitrogen, creating a weaker dipole moment with hydrogen.
06

Determine Correct Order

Based on polarity, the correct order from least polar to most polar is H_2S, NH_3, H_2O, HF. Check which option matches this order.

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Key Concepts

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

Electronegativity
Electronegativity is a measure of an atom's ability to attract and hold onto electrons when forming a bond with another atom. The higher the electronegativity value, the stronger the atom pulls the shared electrons towards itself. This concept is critical when discussing molecular polarity because differences in electronegativity between atoms can lead to polarity.
Understanding electronegativity helps explain why some molecules are polar. For example, in a water molecule (H鈧侽), oxygen is more electronegative than hydrogen. This causes the electrons to be shared unequally, creating a dipole moment. Similarly, in hydrogen fluoride (HF), fluorine is much more electronegative than hydrogen, making HF a highly polar molecule.
Electronegativity values can be found on the periodic table, with fluorine being the most electronegative element. This attribute increases the polarity of bonds it forms, such as in HF. By comparing electronegativity values, we can predict the polarity order of molecules like NH鈧�, H鈧侽, HF, and H鈧係.
Dipole Moment
A dipole moment occurs when there is an uneven distribution of electrons between atoms bonded together. This happens because one atom attracts the bonded electrons more strongly than the other, creating a slight charge separation with one part being more negative and the other more positive.
The dipole moment is a vector quantity, having both magnitude and direction. It is measured in Debye units (D) and is a direct indicator of a molecule's polarity. A larger dipole moment implies a more polar molecule.
  • For example, in HF, the molecule has a high dipole moment due to the large electronegativity difference between hydrogen and fluorine.
  • Water (H鈧侽) also has a significant dipole moment due to its bent shape and the electronegativity of oxygen compared to hydrogen.
  • On the other hand, hydrogen sulfide (H鈧係) has a smaller dipole moment because sulfur is less electronegative, leading to weaker charge separation.
By examining dipole moments, you can understand which molecules are more polar, helping solve polarity-related problems.
Hydrogen Bonding
Hydrogen bonding is a specific type of intermolecular force that occurs when a hydrogen atom, covalently bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine, attracts neighboring electronegative atoms.
Hydrogen bonds are much stronger than regular van der Waals forces but weaker than covalent or ionic bonds. This unique bond affects the physical properties and molecular interactions of substances.
  • In water (H鈧侽), hydrogen bonds are responsible for the liquid's high boiling point and surface tension. Each water molecule can form up to four hydrogen bonds, contributing to its stable structure.
  • In HF, hydrogen bonding also plays a crucial role due to fluorine鈥檚 high electronegativity, making HF highly polar.
  • Ammonia (NH鈧�) exhibits hydrogen bonding as well, but it's less effective compared to water because ammonia molecules engage in fewer hydrogen bonds.
Understanding hydrogen bonding is essential for predicting molecular behavior and properties in various environmental and biological processes.

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Most popular questions from this chapter

Which of rhe following starcmenrs is ruc? (a) 'The dipole momenr of \(\mathrm{N} 1 \mathrm{t}_{3}\) is acro. (b) 1he dipole momenc of \(\mathrm{NT}_{3}\) is ecjual ro \(\mathrm{NH}_{s^{\circ}}\) (c) 'The dipole moment of \(\mathrm{NT}_{3}\) is zero. (d) "lhe dipole moment of \(\mathrm{NF}_{3}\) is less than \(\mathrm{NH}_{3}\).

In the following quesrions rwo statcmenrs (Asscrtion) (A) and Rcason (R) are givcn. Mark (a) If borh \(A\) and \(R\) arc corrcct and \(R\) is the corcct cxplanation of \(A\) (b) If both \(A\) and \(R\) arc corrccr bur \(R\) is nor the corrcer cxplanation of \(A\) (c) \(\mathrm{A}\) is rruc bur \(\mathrm{R}\) is false (d) \(\mathrm{A}\) is false bur \(\mathrm{R}\) is rruc. (c) \(\mathrm{A}\) and \(\mathrm{R}\) horh are false (A): Warcr is a good solvent for ionic compounds bur poor one for covalene compounds. (R): 11ydration cncrgy of ions relcased is sufficicnr to overcomc latrice cncrgy and ro brcak hydrogcn bonds in water while covalenrly bonded compounds inreracr so weakly rhar cven van der Waal's force hcrwecn molccules of covalent compounds cannor be brokcn.

'The strength of sigma bonds formed by axial overlap of s or porbitals of the 2nd shell of the participating atoms decrease as (a) \(\mathrm{s}-\mathrm{s}>\mathrm{p}-\mathrm{s}>\mathrm{p}-\mathrm{P}\) (b) \(s-s>p-p>s-p\) (c) \(p-p>s-p>s-s\) (d) \(\mathrm{p}-\mathrm{s}>\mathrm{s}-\mathrm{s}>\mathrm{p}-\mathrm{p}\)

Which of rhe following has rhe highcst bond orklcr? (a) \(\mathrm{N}_{2}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{He}_{2}\) (d) \(\mathrm{H}_{2}\)

The following compounds are ro be arrangcd in order of rheir increasing rhermal stabilicics. Idenrify rhe correcr order: \(\mathrm{I}, \mathrm{K}_{2} \mathrm{CO}_{3}\) \(\mathrm{II}, \mathrm{M}_{\mathrm{g}} \mathrm{CO}_{3}\) III. \(\mathrm{CaCO}_{3}\) IV. \(\mathrm{BcCO}_{3}\) (a) \(\mathbf{I}<\mathbb{I} I<\mathrm{III}<\mathrm{IV}\) (b) \(\Pi \mathrm{V}<\mathrm{II}<\mathrm{III}<\mathrm{I}\) (c) \(\mathrm{IV}<\mathrm{II}<\mathrm{I}<\mathrm{III}\) (d) II \(
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