


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!
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.

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.

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.

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
​
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
​
Now using c= f x lambda
3 x 10^8/ 4.03 x 10^15= wavelength
​
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
​
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
​
Now using c= f x lambda
3 x 10^8/ 1.1 x 10^16= wavelength
​
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
​
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
​
Now using c= f x lambda
3 x 10^8/ 2.13 x 10^16= wavelength
​
wavelength
at 5th ionization energy= 1.41 x 10^-8 meters