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Here is a sketch of a 2px orbital: ▬▬▬▬▬▬▬▬▬▬ I X This sketch is about 800 pm wide. The coordinate (x, y, and z) axes are also shown. You can rotate the sketch for a better view of the orbital by dragging the slider with your mouse. Suppose an atom with its nucleus at the origin has an electron in a 2p, orbital. Complete each row of the table below by deciding how P4, the probability of finding the electron at point A, compares to P3, the probability of finding the electron at point B. point A 100 pm behind the nucleus, along the +y axis. 100 pm below the nucleus along the -z axis. 100 pm to the left of the nucleus, along the -x axis. point B 100 pm above the nucleus, along the +z axis. 100 pm in front of the nucleus, along the -y axis. 100 pm to the right of the nucleus, along the +x axis. compare PA to PB P, < P B A O O O P₁ = P₁ PB A P < P B PA P₁ = P, B P. > P. A PA PB

Here is a sketch of a 2px orbital: ▬▬▬▬▬▬▬▬▬▬ I X This sketch is about 800 pm wide. The coordinate (x, y, and z) axes are also shown. You can rotate the sketch for a better view of the orbital by dragging the slider with your mouse. Suppose an atom with its nucleus at the origin has an electron in a 2p, orbital. Complete each row of the table below by deciding how P4, the probability of finding the electron at point A, compares to P3, the probability of finding the electron at point B. point A 100 pm behind the nucleus, along the +y axis. 100 pm below the nucleus along the -z axis. 100 pm to the left of the nucleus, along the -x axis. point B 100 pm above the nucleus, along the +z axis. 100 pm in front of the nucleus, along the -y axis. 100 pm to the right of the nucleus, along the +x axis. compare PA to PB P, < P B A O O O P₁ = P₁ PB A P < P B PA P₁ = P, B P. > P. A PA PB

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter7: Atomic Structure And Periodicity
Section: Chapter Questions
Problem 180CP: Answer the following questions, assuming that ms, could have three values rather than two and that...
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Here is a sketch of a 2px orbital: ||||| | | This sketch is about 800 pm wide. The coordinate (x, y, and z) axes are also shown. You can rotate the sketch for a better view of the orbital by dragging the slider with your mouse. Suppose an atom with its nucleus at the origin has an electron in a 2px orbital. Complete each row of the table below by deciding how PA the probability of finding the electron at point 4, compares to P3, the probability of finding the electron at point B. point A 100 pm behind the nucleus, along the +y axis. 100 pm below the nucleus along the -z axis. 100 pm to the left of the nucleus, along the -x axis. point B 100 pm above the nucleus, along the +z axis. 100 pm in front of the nucleus, along the -y axis. 100 pm to the right of the nucleus, along the +x axis. compare PA to PB P, < PR B O O P₁ = PR O O O B P₁ = PR O PA P, > P. B B
Transcribed Image Text:Here is a sketch of a 2px orbital: ||||| | | This sketch is about 800 pm wide. The coordinate (x, y, and z) axes are also shown. You can rotate the sketch for a better view of the orbital by dragging the slider with your mouse. Suppose an atom with its nucleus at the origin has an electron in a 2px orbital. Complete each row of the table below by deciding how PA the probability of finding the electron at point 4, compares to P3, the probability of finding the electron at point B. point A 100 pm behind the nucleus, along the +y axis. 100 pm below the nucleus along the -z axis. 100 pm to the left of the nucleus, along the -x axis. point B 100 pm above the nucleus, along the +z axis. 100 pm in front of the nucleus, along the -y axis. 100 pm to the right of the nucleus, along the +x axis. compare PA to PB P, < PR B O O P₁ = PR O O O B P₁ = PR O PA P, > P. B B
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