Chapter 4: Problem 54
Using the periodic table, arrange the atoms in each set in order of increasing electronegativity. a. \(\mathrm{Cl}, \mathrm{F}, \mathrm{Br}\) b. \(\mathrm{B}, \mathrm{O}, \mathrm{N}\) c. \(\mathrm{Mg}, \mathrm{F}, \mathrm{S}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Periodic Table
As you move from left to right across a period, the number of protons in the nucleus increases. This gives each element unique characteristics and allows scientists to predict chemical behaviors and trends.
Understanding the layout of the periodic table is crucial for determining trends like electronegativity, atomic radius, and ionization energy. Generally, elements in the same group (vertical column) share similar properties because they have the same number of valence electrons.
Electronegativity
On the periodic table, electronegativity exhibits clear trends:
- It increases across a period from left to right.
- It decreases down a group from top to bottom.
Understanding these trends helps in predicting the behavior of different elements during chemical reactions.
Chemical Elements Order
Let's look at how we order elements by their electronegativity:
- For \(\text{Cl, F, Br}\), we see that fluorine (F) is the most electronegative, followed by chlorine (Cl), and finally bromine (Br). Thus, the order is: \[ \text{Br < Cl < F} \].
- For \(\text{B, O, N}\), the sequence based on increasing electronegativity is boron (B), nitrogen (N), and then oxygen (O), i.e., \[ \text{B < N < O} \].
- For \(\text{Mg, F, S}\), magnesium (Mg) has the lowest electronegativity, followed by sulfur (S), with fluorine (F) being the highest: \[ \text{Mg < S < F} \].