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Experiment 5 Flame Tests and Electron Configuration

Experiment 10: Flame Tests of Metal Cations

Required reading: Ebbing, 11 th Edition Chapter 7. -Light waves, photons and the Bohr Theory -Quantum mechanics and quantum numbers

Learning Goals: - To understand the relationship between wavelength and energy of the visible light to the flame color of the metal cations. - To understand the relationship between atomic orbital concept and electron emission of energy. - To practice how to calculate energy, wavelength and frequency..

Background information and theory: When a metallic compound is heated in a flame a colored light is emitted. To understand why light is emitted by the metal cations, we have to look at the arrangement of electrons in an atom, the electron configuration. In the region outside the nucleus of an atom there are energy levels, called electron shells. Each energy level has a specific energy. An electron in an atom can have only these specific energy values, meaning the energy of an electron is quantized. Within the energy levels there are sublevels, called electron subshells. Within the sublevels there are regions called atomic orbitals, where the electron is most likely to be found. Although an electron can have only certain energy values, it can change in energy. The lowest energy arrangement or configuration of electrons in an atom is called the ground state electron configuration. In the ground state the electrons are in the orbitals closest to the nucleus. When an atom or ion absorbs energy (in the form of heat, electrical energy, or electromagnetic radiation), one or more of its electrons jump to an orbital of a higher energy level. The electron is said to be in "excited or promoted state". But the tendency of the electron is to not stay there, but to come back to the ground state. The electron does this in either one step or multiple steps. Each time the electron jumps to a lower energy level, it loses (releases) energy. This energy lost by the electron is equal to the difference in energy between the two energy levels, and is emitted as a particle of electromagnetic radiation, called "photon". This can be seen as emitted light (colored flame) in a flame test. The energy of the photon that is emitted is equal to the positive energy lost by the electron.

πΈπ‘β„Žπ‘œπ‘‘π‘œπ‘›=|Ξ”πΈπ‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘›|= β„Žπœˆ =

β„Žπ‘

πœ†

Where: h is the Planck's constant = 6.626 x 10-34 J.s c is the speed of light = 3.00 x 10 8 m/s  is the wavelength of light emitted  is the frequency of light emitted

A different colored flame is produced by each element because each element has a different number of electrons, and therefore different arrangement of electrons. So, the light emitted by each element can be used to identify the element.

When white light is passed through a prism it disperses into the seven colors of a rainbow, giving a continuous spectrum. But when the colored flame from a heated element is passed through a prism it does not disperse into all the colors of a rainbow, but only certain colors, giving a line spectrum. The colored flame produced by an element is a combination of all the colors in the line spectrum of the element. Each element produces a characteristic line spectrum, that can be used to identify that element. It is referred to as "Element Fingerprinting".

In this experiment we will use the color of a flame generated by some elements (ions) to identify a unknown samples.

Procedure:

Part 1:

1) Watch the following video: https://youtu.be/kkBFG1mTSBk 2) Answer the questions regarding safety in the data paper (answer according to the info provided on the video). 3) Observe the color of the flame produced by each metal in the video. (Colors will change depending on your device, do the best you can to adjust the colors of your screen). 4) Record your observation in your data table. 5) Find a color copy of the visible spectrum. Try to match the color of each flame with a color in the visible spectrum. 6) Record the wavelength corresponding to the flame color for that specific sample. 7) Determine the energy of the photons emitted by that sample. To understand this type

of calculation, watch this video: https://www.youtube.com/watch?v=heXRHrJK6sE

Part 2: Unknown sample:

1) Watch the following short video https://lonestar.techsmithrelay.com/uTzz

2) Observe the flame color for the 2 unknown solids indicated in the image. 3) Compare the color of the flame from the unknown metal cations with the flame color of the known solids in the video from part 1 to identify metal cations in the unknown samples. You may want to use a "split screen" and play both videos at the same time to compare the colors side to side. 4) Record your observations and results in the data table provided.

KCl pink 700

6.626 x 10-34(3.0 x 10 8 )/700 x10-9 = 2.84 x 10-

NaCl Yellow- orange

590

6.626 x 10-34(3.0 x 10 8 )/590 x 10-9 = 3.37 x 10-

SrCl 2 Orange-red 635

6.626 x 10-34(3.0 x 10 8 )/635 x 10-9 = 3.130 x 10-

Unknown Metal cations:

Test Samples Color Flame observed Identification of Metal Cation(s)

Unknown A Orange-yellow NaCl

Unknown B Pinkish-Red KCl

Name: ____Judy Pham______

Flame Tests Postlab questions: Your instructor may ask you to answer these in your lab notebook, or to answer directly on this page and turn it in, or to include these answers in a formal lab report. Follow your instructor's directions.

  1. Rank the following colors of visible light from highest to lowest energy , including their estimated energy ( show your calculations ): (Write their wavelength and from there calculate the energy for each color.)

Blue, Red, Green, Violet

Violet: 4.42 x 10-19, blue: 4.06 x 10-19, violet: 3.55 x 10-19 , green: 2.84 x 10-

490 700 560 450

  1. How many electrons are in the outermost, valence, level of these atoms?

Ga 3_ Cl 7_ Ne 8_ Sr 2_

  1. What is the frequency in Hertz of a photon emitted after an electron transitions from n=6 to n=2.? (Hint: You first need to find the energy and then the frequency) ( show your calculations )

E= 6.626 x 10-34(3 x 108 )/4.10 x 10-

E = 4.85 x 10-

F = 1/t = 4.85 x 10-19/

F = 1.2125 x 10-

Experiment 5 Flame Tests and Electron Configuration

Source: https://www.studocu.com/en-us/document/lone-star-college-system/general-chemistry-ii/experiment-10-flame-tests/14075894

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