College Chemistry : Electron Configurations

Study concepts, example questions & explanations for College Chemistry

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Example Questions

Example Question #1 : Electron Configurations

What is the electron configuration for \displaystyle \text{Cd}^{2+}?

Possible Answers:

\displaystyle [Kr]5s^14d^9

\displaystyle [Kr]5s^24d^8

\displaystyle [Kr]5s^14d^{10}

\displaystyle [Kr]4d^{10}

Correct answer:

\displaystyle [Kr]4d^{10}

Explanation:

Cadmium normally has \displaystyle 48 electrons, but \displaystyle \text{Cd}^{2+} only has \displaystyle 46 electrons.

The normal electron configuration for \displaystyle \text{Cd} is as follows:

\displaystyle [Kr]4d^{10}5s^2

Since the cadmium is losing \displaystyle 2 electrons, it must lose them from the highest energy shell. In this case, this means the cadmium will be losing its electrons in \displaystyle 5s

Thus, the electron configuration for \displaystyle \text{Cd}^{2+} is \displaystyle [Kr]4d^{10}.

Example Question #2 : Electron Configurations

What is the electron configuration for \displaystyle \text{Co}^{3+}?

Possible Answers:

There is no valid electron configuration for \displaystyle \text{Co}^{3+}.

\displaystyle [Ar]3d^6

\displaystyle [Ar]4s^13d^5

\displaystyle [Ar]4s^23d^4

Correct answer:

\displaystyle [Ar]3d^6

Explanation:

The normal electron configuration for \displaystyle \text{Co} is as follows:

\displaystyle [Ar]3d^74s^2

Since it is losing \displaystyle 3 electrons, the element must lose the electrons from the highest energy shell first. Thus, the element loses \displaystyle 2 electrons from \displaystyle 4s and \displaystyle 1 electron for \displaystyle 3d.

The electron configuration for \displaystyle \text{Co}^{3+} is then \displaystyle [Ar]3d^6

Example Question #91 : Introductory Topics

What is the electron configuration for \displaystyle \text{W}^{3+}?

Possible Answers:

\displaystyle \text{[Xe]6s}^2\text{4f}^{11}\text{5d}^4

\displaystyle \text{[Xe]6s4f}^{13}\text{5d}^3

\displaystyle \text{[Xe]6s}^2\text{4f}^{14}\text{5d}

\displaystyle \text{[Xe]4f}^{14}\text{5d}^3

Correct answer:

\displaystyle \text{[Xe]4f}^{14}\text{5d}^3

Explanation:

Start by finding the noble gas core. For tungsten, this will be xenon as this is the noble gas that is closest to it.

The normal electron configuration for \displaystyle \text{W} is as follows:

\displaystyle \text{[Xe]}\text{4f}^{14}\text{5d}^{4}\text{6s}^2

Recall that electrons are lost in the highest energy level subshell first.

\displaystyle \text{W}^{3+} has lost \displaystyle 3 electrons. It will lose the first two electrons from the \displaystyle \text{6s} shell, then it will lose \displaystyle 1 electron from the \displaystyle \text{5d} shell, giving it the following electron configuration:

\displaystyle \text{[Xe]4f}^{14}\text{5d}^3

Example Question #3 : Electron Configurations

What is the electron configuration for \displaystyle \text{Fe}^{3+}?

Possible Answers:

\displaystyle \text{[Ar]4s}^2\text{4d}^5

\displaystyle \text{[Ar]4s}^2\text{3d}^3

\displaystyle \text{[Ar]3d}^5

\displaystyle \text{[Ar]4s3d}^4

Correct answer:

\displaystyle \text{[Ar]3d}^5

Explanation:

Start by finding the noble gas core. For iron, this will be argon as this is the noble gas that is closest to it.

Next, recall that since the \displaystyle \text{4s} orbitals are higher in energy that the \displaystyle \text{3d} orbitals, electrons will be lost from the \displaystyle \text{4s} orbital first.

The normal electron configuration for \displaystyle \text{Fe} is as follows:

\displaystyle \text{[Ar]3d}^6\text{4s}^2

\displaystyle \text{Fe}^{3+} has lost \displaystyle 3 electrons. It will lose the first two electrons from the \displaystyle \text{4s} shell, then it will lose \displaystyle 1 electron from the \displaystyle \text{3d} shell, giving it the following electron configuration:

\displaystyle \text{[Ar]3d}^5

Example Question #1 : Electron Configurations

When an electron moves from a lower energy state to a higher energy state, the electron __________.

Possible Answers:

absorbs energy

None of these. An electron cannot move from a lower energy state to a higher energy state.

releases energy

neither absorbs nor releases energy

both absorbs and releases energy 

Correct answer:

absorbs energy

Explanation:

Electrons of an atom are located within electronic orbitals around a nucleus. The electrons of each atoms have their own specific energy level called principal energy level. When electrons are excited by absorbing energy the electrons can jump to a high energy level. Then when an electron drops back to a lower energy level the electron emits the energy. Therefore, when an atom moves from a lower energy state to a higher energy state. the electrons absorb energy. 

Example Question #281 : College Chemistry

What is the full electron configuration of sodium?

Possible Answers:

\displaystyle 1s^{2}2s^{2}2p^{6}3s^{1}

\displaystyle [He]3s^{1}

\displaystyle 2s^{2}2p^{6}3s^{1}

\displaystyle 2s^{2}2p^{6}3s^{1}

\displaystyle 1s^{2}2s^{2}3s^{1}

Correct answer:

\displaystyle 1s^{2}2s^{2}2p^{6}3s^{1}

Explanation:

Each element has a unique electron configuration that represents the arrangement of electrons in orbital shells and sub shells. There are four different orbitals, s, p, d, and f that each contain two electrons. The p, d, and f orbitals contain subshells that allow them to hold more electrons. The orbitals for an element can be determined using the periodic table. The s-block consists of group 1 and 2 (the alkali metals) and helium. The p-block consists of groups 3-18. The d-block consists of groups 3-12 (transition metals), and the f-block contains the lanthanides and actinides series. Using this information we can determine the full electron configuration of sodium.  

To do this, start at hydrogen \displaystyle (H) located at the top left of the periodic table. Hydrogen \displaystyle (H) and helium \displaystyle (He) are in the first s orbital and account for \displaystyle 1s^{2}. Next, we move to the second s-orbital that contains lithium (Li) and beryllium (Be), which accounts for \displaystyle 2s^{2}. Then we move to boron, carbon, nitrogen, oxygen, fluorine, and neon, which are all in the p-block and account for \displaystyle 2p^{6}. There is no 1p orbital. Finally, we are at sodium, which is in the s-block and accounts for \displaystyle 3p^{1}. Therefore the full electron configuration of sodium is \displaystyle 1s^{2}2s^{2}2p^{6}3s^{1}

Example Question #21 : Atoms And Elements

What is the electron configuration of iodine in nobel gas notation?

Possible Answers:

\displaystyle [Xe]5s^{2}6d^{10}5p^{5}

\displaystyle [Kr]5s^{2}6d^{10}5p^{5}

\displaystyle [Xe]5s^{2}6d^{6}5p^{2}

\displaystyle 5s^{2}6d^{10}5p^{5}

\displaystyle [Kr]5s^{2}5d^{10}5p^{5}

Correct answer:

\displaystyle [Kr]5s^{2}6d^{10}5p^{5}

Explanation:

Each element has a unique electron configuration that represents the arrangement of electrons in orbital shells and subshells. There are four different orbitals, s, p, d, and f that each contain two electrons. The p, d, and f orbitals contain subshells that allow them to hold more electrons. The orbitals for an element can be determined using the periodic table. The s-block consists of group 1 and 2 (the alkali metals) and helium. The p-block consists of groups 3-18. The d-block consists of groups 3-12 (transition metals), and the f-block contains the lanthanides and actinides series. Using this information we can determine the electron configuration of iodine in nobel gas configuration.  

The nobel gas configuration is a short hand to writing out the full electron configuration. To do this, start at the nobel gas that come before the element of interest. In the case of iodine, the nobel gas is krypton. Therefore, the electron configuration will begin with \displaystyle [Kr], and this will be the new starting place for the electron configuration.

After krypton comes the s-block, which contains elements with the atomic numbers 37 and 38 that account for \displaystyle 5s^{2}. Then comes the d-block containing elements 39-48 that account for \displaystyle 6d^{10}. Finally comes the p-block containing elements 49-53 that account for \displaystyle 5s^{5}. Therefore, the electron configuration of iodine in nobel gas configuration is \displaystyle [Kr]5s^{2}6d^{10}5p^{5}

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