Iron is commonly found as Fe, Fe 2+ , and Fe 3+ . (b) What are the n and l quantum numbers of the electron removed on going from Fe 2 + to Fe 3+ ?

Hello everyone today. We are being given the falling problem. Mercury exists as nature as mercury. Mercury one plus and Mercury two plus, Mercury loses an electron when it goes from Mercury one plus two. Mercury two plus. Give the principle number and the angular momentum. Quantum numbers of the electron lost from this transition. So the first thing we want to do is we want to recall that electrons are removed from the outermost orbital first. So when an element loses an electron, it's always moved from the outermost orbital first. And so for our first one here we have just regular mercury. Its atomic number, so its atomic number is 80. and so therefore it has 80 electrons and is therefore considered neutral. It's regular condensed electron configuration per the periodic table is going to be xenon. Since we go we see where mercury is. And then we go backwards until we hit the first noble gas. We then go to R six S orbital which is in the sixth period six S to to be specific because we are filling in all of those orbital's all of those the entire orbital for the S orbital. We then dip into the four f orbital four F 14. And then we go back to our five D 10. And this is our condensed electron configuration. And so we have to note then our highest orbital here That isn't going to be 6 S2, essentially. Our six is the highest number here. So it's considered the highest orbital. So we're gonna just keep that number or answer safe for now. Next thing you wanna do is you wanna write the electron configuration for our Mercury one plus and The configuration for our Mercury two plus for our Mercury plus. The condensed formula is once again xenon because that's the closest noble gas. And then we have six S one, four F 14 and then five D 10. So notice that we are missing an electron in the six S orbital, as we said before, electrons are removed from the outermost orbital first and the outermost is synonymous with highest orbital. And so therefore, an electron was lost in The six S orbital. If we look at Mercury two plus, once again we have our Xenon, We simply just have four F 14, five D 10. And so we also note that we lost two electrons in R six S orbital because R six S orbital can hold a maximum of two electrons and the neutral form already had two electrons. So we simply removed one and then we removed too. So ultimately our N. or our principle quantum number is going to be six. And now we have to find our angular momentum and our angular momentum Is essentially going to be in -1. So you're already designated in as being six. So it ranges from 0 to N -6, sort of L is equal to zero. We have an S sub cell if L is equal to one, we have a p sub shell If L is equal to two, we have a. D. Sub shell, L can be equal to three, and we'll have an S. Sub shell and we'll stop there notice this was a six S. So we have an S. Orbital, meaning that our L. Is going to equal zero. So our answers here are going to be our principle number, or in is gonna be equal to six, and our angular momentum is going to be equal to zero. And ultimately we have solved the problem. I hope this helped, and until next time.

Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory

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Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory

Problem 104b

Chapter 6, Problem 104b

Iron is commonly found as Fe, Fe²⁺, and Fe³⁺. (b) What are the n and l quantum numbers of the electron removed on going from Fe²+ to Fe³⁺?

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Determine the electron configuration of Fe^{2+}.

Identify the electron configuration of Fe^{3+}.

Compare the electron configurations of Fe^{2+} and Fe^{3+} to find the electron that is removed.

Identify the quantum numbers n and l for the removed electron.

Conclude with the n and l values for the electron removed from Fe^{2+} to form Fe^{3+}.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Quantum Numbers

Quantum numbers are numerical values that describe the unique quantum state of an electron in an atom. The four quantum numbers include the principal quantum number (n), which indicates the energy level, and the azimuthal quantum number (l), which describes the shape of the orbital. Understanding these numbers is essential for determining the electron configuration and the specific electron being removed during ionization.

Electron Configuration

Electron configuration refers to the distribution of electrons in an atom's orbitals. For iron (Fe), the electron configuration is [Ar] 4s² 3d⁶. When iron loses electrons to form ions, the electrons are removed from the outermost energy levels first, which is crucial for identifying which electron is lost when transitioning from Fe²⁺ to Fe³⁺.

Ionization

Ionization is the process of removing one or more electrons from an atom or ion, resulting in a positively charged ion. In the case of Fe²⁺ to Fe³⁺, one electron is removed from the 3d subshell. Understanding the ionization process helps in determining the specific quantum numbers of the electron that is lost during the transition between these two oxidation states.

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Consider the electronic structure of the element bismuth.(d) Would you expect element 115 to have an ionization ene-rgy greater than, equal to, or less than that of bismuth? Explain.

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Consider the electronic structure of the element bismuth. (a) The first ionization energy of bismuth is Ei1 = +703 kJ/ mol. What is the longest possible wavelength of light that could ionize an atom of bismuth?

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Textbook Question

Iron is commonly found as Fe, Fe²⁺+, and Fe³⁺. (c) The third ionization energy of Fe is E_i3 = +2952 kJ/mol. What is the longest wavelength of light that could ionize Fe²⁺(g) to Fe³⁺(g)?

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Textbook Question

The ionization energy of an atom can be measured by photo-electron spectroscopy, in which light of wavelength l is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron 1EK2 is measured by determining its velocity, v since EK = 1/2 mv2. The Ei is then calculated using the relationship that the energy of the inci-dent light equals the sum of Ei plus EK. (a) What is the ionization energy of rubidium atoms in kilo-joules per mole if light with l = 58.4 nm produces elec-trons with a velocity of 2.450 * 106m/s? (The mass of an electron is 9.109 * 10-31 kg.)

Iron is commonly found as Fe, Fe2+, and Fe3+. (b) What are the n and l quantum numbers of the electron removed on going from Fe2+ to Fe3+?

Verified video answer for a similar problem:

Key Concepts

Quantum Numbers

Electron Configuration

Ionization

Iron is commonly found as Fe, Fe²⁺, and Fe³⁺. (b) What are the n and l quantum numbers of the electron removed on going from Fe²+ to Fe³⁺?