12.1 Electrons in Atoms HL

Successive ionization energy Mg+ (g) + 1451 kJ ---> Mg2+ (g) + e- IE is the energy required to remove one mole electron from the atom in the gaseous state. (kJmol-1) Successive ionization energy Mg (g) + 738 kJ ---> Mg+ (g) + e- This is called the FIRST ionization energy because we removed only the OUTERMOST electron Mg+ (g) + 1451 kJ ---> Mg2+ (g) + e- This is the SECOND IE.

Trends across the periodic table -decreases down the periods -increases up the group Li and Be have a greater atomic radius, but B have a slightly bigger radius so a slightly lower I.E. -two peaks because of a full filled ‘s’ orbital and a half filled ‘p’ orbital

FACTORS THAT AFFECT THE I.E. ATOMIC RADIUS NUCLEAR CHARGE ELECTRON SHIELDING OR SCREENING Li has higher I.E. than Na -smaller atomic radius F has higher I.E. than O -F has more protons than O, higher nuclear charge

Trend in ionization energy- ACROSS GROUPS I.E. increases across GROUPS because the NUCLEAR CHARGE increases. Metals lose electrons more easily than nonmetals. Nonmetals lose electrons with difficulty (they like to GAIN electrons).

Trend in ionization energy- ACROSS PERIODS I.E. increases DOWN PERIODS Because number of ENERGY LEVELS increases Higher electron shielding effect

Which has the highest first ioniztion energy? Mg or Ca ? Al or S ? Cs or Ba ?

In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy.

Energy levels in the hydrogen atom converge at higher energy. At the convergence limit, the lines merge to form a continuum. Beyond this point, the electron is free from any influence of the nucleus. We can calculate the ionization energy from the convergence limit. The transition from n=1 to n=∞ corresponds to ionization which is the removal of the electron from the 1s orbital.

EXPLANATION OF THE TRENDS AND DISCONTUITIES IN FIRST IONIZATION ENERGY First two I.E. of Ca are low Because the first two electron sin the outer most energy level -larger atomic radius AND higher electron screening effect =LOW I.E. Third I.E. of Ca is high Because the third electron is in a full energy level and the nuclear charge have a greater attraction to this electron than it did to the first two electrons =HIGH I.E. SAME PRINCIPLES APPLY TO TITANIUM

SUCESSIVE IONIZATION ENERGY Na (g)  Na+ (g) + e- 1st IE = + 496 kJmol-1 Na+ (g)  Na2+ (g) + e- 2nd IE = + 4563 kJmol-1 Na2+ (g)  Na3+ (g) + e- 3rd IE = + 6913 kJmol-1 SUCCESSIVE IONIZATION ENERGY OF Na

He Ne F N O H C Be B Na Li 11 1st IONISATION ENERGY / kJmol-1 1s 2s 2p 1s 2s 1s ATOMIC NUMBER 1st IONISATION ENERGY / kJmol-1 1s 2s 2p 3s 11 He H Li Be B C N F Ne Na O

E=hv E=hv = hc/λ. Where H = Planck’s constant (6.63 x 10-34 Js) C= λv can be rearranged to V=c/λ Wavelength is related to the frequency of the radiation Substitute this into e=hv E=hv = hc/λ. Where H = Planck’s constant (6.63 x 10-34 Js) v = frequency of light (Hz or 1/s) c = speed of light (3.00 x 108 m/s) λ = wavelength (m) (1x109nm = 1m) (You may have to convert from nanometers.)

E=hv The energy of a photon is related to the frequency of the electromagnetic radiation. This equation can be used to work out the differences in energy between various levels in the hydrogen atom.

Question 1 Determine the energy in joules of a photon of red light, correct to 4 significant figures, given that the wavelength is 650.0 nm. h=6.626x10-34 Js; c=2.998 x108 m/s

Answer E=hv = hc/λ convert nm to m E=(6.626x10-34 Js)(2.998 x108 m/s

Question 2 Calculate the first ionization energy in kJ/mol for hydrogen given that its shortest wavelength line in the Lyman series is 91.16 nm. NA = 6.022x1023 mol-1 h=6.626x10-34 Js; c=2.998 x108 m/s

Answer E=hv = hc/λ convert nm to m E=(6.626x10-34 Js)(2.998 x108 m/s) E=2.179x10-18 J This represents the energy for one atom of hydrogen. Multiply by Avogadro’s # to get energy per mole of hydrogen atoms. IE1 = (2.179x10-18J) x (6.022x1023 mol-1) =1.312x106 J/mol = 1312 kJ/mol  

THANK YOU