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Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F} ;\) (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P}\) (c) \(\mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

Short Answer

Expert verified
(a) The order of increasing polarity is as follows: \(O-F < C-F < Be-F\) (b) The order of increasing polarity is as follows: \(O-Cl < C-P < S-Br\) (c) The order of increasing polarity is as follows: \(C-S < N-O < B-F\)

Step by step solution

01

Calculate electronegativity difference for each bond

C-F: \(|3.98 - 2.55| = 1.43\) O-F: \(|3.98 - 3.44| = 0.54\) Be-F: \(|3.98 - 1.57| = 2.41\)
02

Arrange the bonds in order

The order of increasing polarity is as follows: O-F < C-F < Be-F (b) Arrange the following bonds in order of increasing polarity: \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P}\)
03

Calculate electronegativity difference for each bond

O-Cl: \(|3.44 - 3.16| = 0.28\) S-Br: \(|2.58 - 2.96| = 0.38\) C-P: \(|2.55 - 2.19| = 0.36\)
04

Arrange the bonds in order

The order of increasing polarity is as follows: O-Cl < C-P < S-Br (c) Arrange the following bonds in order of increasing polarity: \(\mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)
05

Calculate electronegativity difference for each bond

C-S: \(|2.55 - 2.58| = 0.03\) B-F: \(|3.98 - 2.04| = 1.94\) N-O: \(|3.04 - 3.44| = 0.40\)
06

Arrange the bonds in order

The order of increasing polarity is as follows: C-S < N-O < B-F

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Key Concepts

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

Electronegativity
Electronegativity is a fundamental concept in chemistry that explains how strongly an atom can attract a shared pair of electrons towards itself in a chemical bond. Each element has an electronegativity value on the Pauling scale, which can range from 0 to about 4. For example, fluorine is the most electronegative element with a value of 3.98.
Understanding electronegativity is crucial when predicting the nature of chemical bonds. Here's why it matters:
  • When two different atoms form a bond, the atom with a higher electronegativity will attract the shared electrons more strongly.
  • This difference in electronegativity leads to bond polarity, where the electron density is higher around the more electronegative atom.
  • Electronegativity differences can predict whether a bond will be nonpolar covalent, polar covalent, or ionic.
When calculating the electronegativity difference, you simply subtract the smaller electronegativity value from the larger one for the two atoms involved in the bond. This difference can then indicate the degree of polarity in the bond.
Chemical Bonds
Chemical bonds are the glue that holds atoms together in molecules. They're formed because atoms have a tendency to achieve a more stable electron configuration. There are three main types of chemical bonds: ionic, covalent, and metallic, but we focus mainly on covalent bonds here, which involve the sharing of electron pairs between atoms.
Here's a quick breakdown:
  • Covalent Bonds: These involve the sharing of electron pairs between atoms. Depending on the atoms' electronegativities, a covalent bond can be nonpolar (equal sharing) or polar (unequal sharing).
  • Ionic Bonds: These occur when there is a complete transfer of electrons from one atom to another, typically between a metal and a non-metal. These are not the focus in polar covalent bonds but are at one end of the spectrum of electronegativity differences.
In covalent bonds, the behaviour of shared electrons is what dictates the bond's properties, including its polarity. Understanding chemical bonds helps predict molecular behavior and properties.
Polarity Order
The polarity order of chemical bonds is determined by the difference in electronegativity between the two bonding atoms. The greater the difference, the more polar the bond. This concept helps in determining how electrons are distributed across a molecule.
Consider these points when arranging bonds by polarity:
  • If the electronegativity difference is high, the bond is highly polar. A classic example is the hydrogen-fluoride (HF) bond.
  • If the electronegativity difference is very low, the bond tends to be nonpolar, like the bond between two carbon atoms (C-C).
In exercises involving polarity order, calculating the electronegativity differences first is key. Let's say we are evaluating bonds like C-F and Be-F from the exercise, you calculate: - Be-F has a higher electronegativity difference compared to C-F, making it more polar.
The sequence you're looking for with increasing polarity in such problems typically starts with the least difference in electronegativity, moving to the greatest. This organized arrangement helps understand which bonds have more "electron attraction inequality," leading to predictions about molecular properties like solubility and boiling points.

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Most popular questions from this chapter

(a) Determine the formal charge on the chlorine atom in the hypochlorite ion, \(\mathrm{ClO}^{-}\), and the perchlorate ion, \(\mathrm{ClO}_{4}^{-}\), using resonance structures where the \(\mathrm{Cl}\) atom has an octet. (b) What are the oxidation numbers of chlorine in \(\mathrm{ClO}^{-}\) and in \(\mathrm{ClO}_{4}^{-} ?(\mathrm{c})\) Is it uncommon for the formal charge and the oxidation state to be different? Explain. (d) Perchlorate is a much stronger oxidizing agent than hypochlorite. Would you expect there to be any relationship between the oxidizing power of the oxyanion and either the oxidation state or the formal charge of chlorine?

You and a partner are asked to complete a lab entitled "Fluorides of Group \(6 \mathrm{~B}\) Metals" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a colorless liquid and the other a green powder. You also find the following notes in your partner's notebook-Compound 1: \(47.7 \% \mathrm{Cr}\) and \(52.3 \% \mathrm{~F}\) (by mass), Compound 2: \(45.7 \% \mathrm{Mo}\) and \(54.3 \% \mathrm{~F}\) (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound 2? (c) Upon determining the melting points of these two compounds you find that the colorless liquid solidifies at \(18^{\circ} \mathrm{C}\), while the green powder does not melt up to the maximum temperature of your apparatus, \(1200{ }^{\circ} \mathrm{C}\). What is the identity of the colorless liquid? What is the identity of the green powder? Be sure to use the appropriate naming convention depending upon whether the compound is better described as a molecular or ionic compound.

(a) How does a polar molecule differ from a nonpolar one? (b) Atoms X and Y have different electronegativities. Will the diatomic molecule \(X-Y\) necessarily be polar? Explain. (c) What factors affect the size of the dipole moment of a diatomic molecule?

What is the Lewis symbol for each of the following atoms or ions: (a) \(\mathrm{Ca}\), (b) \(\mathrm{P}\), (c) \(\mathrm{Mg}^{2+}\), (d) \(S^{2-}\) ?

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}\), \(1.83 \% \mathrm{H}, 64.30 \% \mathrm{Cl}\), and \(19.35 \%\) O by mass and has a molarmass of \(165.4 \mathrm{~g} / \mathrm{mol} .\) (a) What is the empirical formula of this substance? (b) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the \(\mathrm{Cl}\) atoms bond to a single \(C\) atom and that there are a \(C-C\) bond and two C \(-\) O bonds in the compound.

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