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Chapter 2: Problem 153

Fluorine occurs naturally as a single isotope. How many protons, neutrons, and electrons are present in deuterium fluoride \(\left({ }^{2} \mathrm{HF}\right) ?\) (Deuterium is \({ }^{2} \mathrm{H}\).)

Short Answer

Expert verified
Deuterium fluoride has 10 protons, 11 neutrons, and 10 electrons.

Step by step solution

01

Identify Fluorine Details

Fluorine naturally appears as a single isotope \(^{19}\text{F}\). It has an atomic number of 9, which means it has 9 protons and, for a neutral atom, 9 electrons. The number of neutrons is determined by subtracting the atomic number from the atomic mass, i.e., \(19 - 9 = 10\) neutrons.
02

Identify Deuterium Details

Deuterium is a heavier isotope of hydrogen \(^{2}\text{H}\), consisting of 1 proton, 1 electron, and 1 neutron. The atomic number of hydrogen is 1, so deuterium has 1 proton, and its mass number is 2, thus it has \(2 - 1 = 1\) neutron.
03

Combine in Deuterium Fluoride

Deuterium fluoride \(^{2}\text{HF}\) is a compound consisting of one deuterium atom and one fluorine atom. Thus, the total count of subatomic particles is the sum of those in deuterium and fluorine. This results in 1 + 9 = 10 protons, 1 + 10 = 11 neutrons, and 1 + 9 = 10 electrons.

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Key Concepts

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

Deuterium and Its Unique Characteristics
Deuterium, often referred to as heavy hydrogen, distinguishes itself from regular hydrogen by having one extra neutron. This gives it a mass number of 2 since it contains 1 proton and 1 neutron, unlike typical hydrogen which only has 1 proton and no neutrons.

In the field of chemistry and physics, deuterium ( ^2 ext{H} ) is especially interesting because it creates heavier versions of compounds, like deuterium fluoride ( ^2 ext{HF} ).

This extra neutron hardly affects how atoms bond chemically, but it does make deuterium-containing compounds slightly denser. Understanding deuterium's composition helps in discerning its role in forming compounds like deuterium fluoride.
Fluorine: A Single-Isotope Element
Fluorine is a unique element in that it is primarily found in nature as just one isotope, ^{19} ext{F}. This characteristic simplifies its atomic structure analysis, as all naturally occurring fluorine atoms share the same number of subatomic particles.

Fluorine's atomic number is 9, which tells us it has 9 protons. For neutral atoms, the number of electrons equals the number of protons, so in neutral fluorine, there are also 9 electrons.

To find the number of neutrons in fluorine, we can subtract the atomic number from the atomic mass, which gives us: 19 - 9 = 10 neutrons.
  • Protons: 9
  • Neutrons: 10
  • Electrons: 9

This consistency makes studying fluorine's chemical behavior more straightforward and predictable.
Understanding the Atomic Structure of Compounds
The atomic structure of a compound is determined by the sum of the subatomic particles from each of its components.

In deuterium fluoride ( ^2 ext{HF} ), we combine the atomic structures of deuterium and fluorine to understand the whole picture.

From deuterium, we have:
  • 1 proton
  • 1 neutron
  • 1 electron

And from fluorine:
  • 9 protons
  • 10 neutrons
  • 9 electrons

When creating the compound, we add these values together:
  • Total Protons: 1 + 9 = 10
  • Total Neutrons: 1 + 10 = 11
  • Total Electrons: 1 + 9 = 10

Understanding the detailed atomic structure helps predict the compound's behavior, providing insights into its chemical reactions and properties.

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Most popular questions from this chapter

Write formulas for the following binary compounds: (a) Potassium chloride (b) Tin(II) bromide (c) Calcium oxide (d) Barium chloride (e) Aluminum hydride

In an alternate universe, the smallest negatively charged particle, analogous to our electron, is called a blorvek. To determine the charge on a single blorvek, an experiment like Millikan's with charged oil droplets was carried out and the following results were recorded: $$ \begin{array}{cl} \hline \text { Droplet Number } & \text { Charge(C) } \\ \hline 1 & 7.74 \times 10^{-16} \\ 2 & 4.42 \times 10^{-16} \\ 3 & 2.21 \times 10^{-16} \\ 4 & 4.98 \times 10^{-16} \\ 5 & 6.64 \times 10^{-16} \\ \hline \end{array} $$ (a) Based on these observations, what is the largest possible value for the charge on a blorvek? (b) Further experiments found a droplet with a charge of \(5.81 \times 10^{-16} \mathrm{C}\). Does this new result change your answer to part (a)? If so, what is the new largest value for the blorvek's charge?

In borane, one part hydrogen combines with \(3.6\) parts boron by mass. A compound containing only hydrogen and boron contains \(6.0 \mathrm{~g}\) of hydrogen and \(43.2 \mathrm{~g}\) of boron. Could this compound be borane? If it is not borane, show that the law of multiple proportions is followed for borane and this other substance.

There are two compounds of titanium and chlorine. One compound contains \(31.04 \%\) titanium by mass, and the other contains \(74.76 \%\) chlorine by mass. What are the ratios of titanium and chlorine atoms in the two compounds?

Look at the alphabetical list of elements inside the front cover. Give the names corresponding to the following symbols: (a) \(\mathrm{B}\) (b) \(\mathrm{Rh}\) (c) Cf (d) Os (e) \(\mathrm{Ga}\)
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