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OXIDATION STATES

−3, −2, −1, +1, +2, +3, +4, +5 (a mildly acidic oxide)

IONISATION ENERGIES

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ISOTOPES

only bismuth 209 has a one hundred percent abundance on Earth as all the other isotopes are synthetic, however an exception to this is bismuth 210 which is a trace isotope meaning that it has a half life that is relatively short compared to the age of the Earth, in this case bismuth-210 only has a half life of five days which is significantly less than the half life of bismuth 209 which has a half life of ten raised to the power of nineteen!

Research: Research
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SPECTROMETER

Diagram

A spectrometer is a device that disperses different wavelengths of visible light onto an object and measures which wavelengths of light were reflected back.

Research: Welcome

FINDING THE IONISATION ENERGIES

FIRST IONISATION ENERGY

January 25, 2025

Using the IB chemistry data booklet, we can find the first ionization energy of bismuth which is 703 Kj/mol as seen in the region with the red square, remember that the first ionisation energy is the energy involved in removing one mole of electrons from one mole of atoms in the gaseous state.

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FINDING THE WAVELENGTH

January 25, 2025

The wavelength at which a photon was sent back can be found using the emission or absorption spectrum of bismuth and measuring how far the thin lines are from the origin.

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IB CHEMISTRY DATA BOOKLET

necessary formulas

In the formula section of the IB data chemistry data booklet, we can find two very useful equations as shown in the picture (shown to be circled in blue) to the right hand side:

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In the top equation, the letter ''c'' represents the speed of light which is given in the data booklet as 3 x 10^8 m/s, the upside down ''y'' is lambda which represents the wavelength of a wave in meters and the letter ''v'' represents frequency in hertz.

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In the bottom equation the capital ''E'' represents energy in joules and the letter ''v'' represents frequency in hertz while the letter ''h'' represents plank's constant which is given in the IB chemistry data booklet as 6.63 x 10^-34 j/s.

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USING THE FORMULAS AND THE GIVEN WAVELENGTHS FOR EACH PHOTON RELEASED BY THE ELECTRON RETURNING TO THE GROUND STATE FOR EACH IONIZATION STATE WE CAN FIND THE IONIZATION ENERGY.

formulas: Energy= frequency x Planck's constant (6.63 x 10^-34)
lightspeed (3 x 10^8 m/s)= frequency x wavelength

[Xe] 4f145d106s26p2

FIRST IONIZATION ENERGY (703KJ/MOL)

703 KJ/6.02(10^23) x 1000= 1.168x 10^-18 J

1.168 x 10^-18/ 6.63 x 10^-34= frequency (using Planck's equation)

frequency (hz)= 1.76 x 10^15 

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Now using c= f x lambda

3 x 10^8/ 1.76 x 10^15= wavelength

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wavelength

at 1st ionization energy= 1.705 x 10^-7 meters

[Xe] 4f145d106s26p1

SECOND IONIZATION ENERGY (1610 KJ/MOL)

1610 KJ/6.02(10^23) x 1000= 2.67x 10^-18 J

2.67 x 10^-18/ 6.63 x 10^-34= frequency (using Planck's equation)

frequency (hz)= 4.03 x 10^15 

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Now using c= f x lambda

3 x 10^8/ 4.03 x 10^15= wavelength

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wavelength

at 2nd ionization energy= 7.44 x 10^-7 meters

[Xe] 4f145d106s2

THIRD IONIZATION ENERGY (2466 KJ/MOL)

2466 KJ/6.02(10^23) x 1000= 4.1 x 10^-18 J

4.1 x 10^-18/ 6.63 x 10^-34= frequency (using Planck's equation)

frequency (hz)= 6.18 x 10^15 

​

Now using c= f x lambda

3 x 10^8/ 6.18 x 10^15= wavelength

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wavelength

at 3rd ionization energy= 4.85x 10^-8 meters

[Xe] 4f145d106s1

FOURTH IONIZATION ENERGY (4370 KJ/MOL)

4370 KJ/6.02(10^23) x 1000= 7.26 x 10^-18 J

7.26 x 10^-18/ 6.63 x 10^-34= frequency (using Planck's equation)

frequency (hz)= 1.1 x 10^16

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Now using c= f x lambda

3 x 10^8/ 1.1 x 10^16= wavelength

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wavelength

at 4th ionization energy= 2.73x 10^-8 meters

[Xe] 4f145d10

FIFTH IONIZATION ENERGY (5400 KJ/MOL)

5400 KJ/6.02(10^23) x 1000= 8.97 x 10^-18 J

8.97 x 10^-18/ 6.63 x 10^-34= frequency (using Planck's equation)

frequency (hz)= 1.53 x 10^16

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Now using c= f x lambda

3 x 10^8/ 1.35 x 10^16= wavelength

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wavelength

at 5th ionization energy= 2 x 10^-8 meters

[Xe] 4f145d9

SIXTH IONIZATION ENERGY (8520 KJ/MOL)

8520 KJ/6.02(10^23) x 1000= 1.42 x 10^-17 J

81.42 x 10^-17/ 6.63 x 10^-34= frequency (using Planck's equation)

frequency (hz)= 2.13 x 10^16

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Now using c= f x lambda

3 x 10^8/ 2.13 x 10^16= wavelength

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wavelength

at 5th ionization energy= 1.41 x 10^-8 meters

Research: Open Positions

WHAT DOES THE GRAPH TELL US?

If we look at the graph, the largest jump in energy is between the fifth and sixth ionization energy, this is because bismuth has an electronic configuration of [Xe] 4f145d106s26p3 where in the sixth ionization you are removing an electron from the d sub shell which requires far more energy than than the p and s sub shells.

The last sub shell of bismuth is p3 thus placing the element in group 15, and thus the first ionization energy is the lowest as electrons are taken from the lowest energy sub shell first.

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Ionization energy increases across a period and decreases down a group.  Down a group, the number of energy levels (n) increase and the distance is greater between the nucleus and highest-energy electron. 

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Successive ionization energy data for an element give information that shows relations to electronic configurations.

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Trends in first ionization energy across periods account for the existence of main energy levels and sub-levels in atoms.

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Research: Open Positions
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