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Chapter 8: Problem 32

Without using Fig. 8.3, predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb} \quad\) c. $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ b. \(\mathrm{B}, \mathrm{O}, \mathrm{Ga} \qquad\) d. $\mathrm{S}, \mathrm{O}, \mathrm{F}$

Short Answer

Expert verified
The order of increasing electronegativity for each group of elements is: a. \(Na > K > Rb \) b. \(Ga < B < O\) c. \(F > Cl > Br \) d. \(S < O < F\)

Step by step solution

01

a. Na, K, Rb

First, observe that all three elements, Sodium (Na), Potassium (K), and Rubidium (Rb) are in the same group (Group 1: Alkali metals) and are placed one below the other in the periodic table. As we move down a group, electronegativity decreases. Therefore, the order should follow this trend: \[Na > K > Rb \]
02

b. B, O, Ga

Boron (B), Oxygen (O), and Gallium (Ga) are elements in different groups and periods of the periodic table: B is in Group 13, O is in Group 16, and Ga is in Group 13, below B. First, we'll compare B and Ga. As we move down a group, electronegativity decreases: \[ B > Ga \] Now, we should compare these elements to oxygen. Moving across the periodic table from B to O, electronegativity increases, so we have: \[B < O\] and since \(B > Ga\) \[Ga < O\] Based on our comparisons, the order of increasing electronegativity is: \[Ga < B < O\]
03

c. F, Cl, Br

Fluorine (F), Chlorine (Cl), and Bromine (Br) are all in the same group (Group 17: Halogens) and are placed one below the other in the periodic table. As we move down a group, electronegativity decreases. Therefore, the order should follow this trend: \[F > Cl > Br \]
04

d. S, O, F

Sulfur (S), Oxygen (O), and Fluorine (F) are elements of different groups but from the same period in the periodic table (Period 2). When we move across the periodic table from left to right, electronegativity increases. So, the order of increasing electronegativity should follow this trend: \[S < O < F\] So, the order of increasing electronegativity for each group of elements is: a. \(Na > K > Rb \) b. \(Ga < B < O\) c. \(F > Cl > Br \) d. \(S < O < F\)

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Most popular questions from this chapter

List the bonds $\mathrm{P}-\mathrm{Cl}, \mathrm{P}-\mathrm{F}, \mathrm{O}-\mathrm{F},\( and \)\mathrm{Si}-\mathrm{F}$ from least polar to most polar.

Place the species below in order of the shortest to the longest nitrogen–oxygen bond. $$\mathrm{H}_{2} \mathrm{NOH}, \quad \mathrm{N}_{2} \mathrm{O}, \quad \mathrm{NO}^{+}, \quad \mathrm{NO}_{2}^{-}, \quad \mathrm{NO}_{3}^{-}$$ $\left(\mathrm{H}_{2} \mathrm{NOH} \text { exists as } \mathrm{H}_{2} \mathrm{N}-\mathrm{OH} .\right)$

Two different compounds exist having the formula $\mathrm{N}_{2} \mathrm{F}_{2}$ . One compound is polar whereas the other is nonpolar. Draw Lewis structures for \(\mathrm{N}_{2} \mathrm{F}_{2}\) consistent with these observations.

Use bond energies to estimate \(\Delta H\) for the combustion of one mole of acetylene: $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$

Consider the following bond lengths: $$\mathrm{C}-\mathrm{O} \quad 143 \mathrm{pm} \quad \mathrm{C}=\mathrm{O} \quad 123 \mathrm{pm} \quad \mathrm{C} \equiv \mathrm{O} \quad 109 \mathrm{pm}$$ In the \(\mathrm{CO}_{3}^{2-}\) ion, all three \(\mathrm{C}-\mathrm{O}\) bonds have identical bond lengths of 136 \(\mathrm{pm} .\) Why?
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