Early Quantum Theory In what may seem like an abrupt change of pace, we will now drop our periodic table with its focus on practical chemistry of elements and start working on the background needed to understand and advanced theory called Quantum theory. Don t worry, the two are related but you won t see the tie until the next chapter. 4-1 First ionization energies Let s do a quick review of atom structure again Nucleus - dense and tiny. Surrounded by cloud of electrons. Electrons have virtually no mass but they are what occupies the volume of the atom. They are also how one atom interacts chemically with another atoms SO if you want to understand chemistry, you have to understand electrons Let s start with ionization energy - The minimum energy needed to remove and electron from an atom. In particular I want to discuss the first ionization energy (I 1), the minimum energy required to remove an electron from a gaseous atom (A) to produce a positively charged gaseous cation A + + - A(g) A (g) + e (g) First ionization energy There are also second, third, fourth, etc ionization energies for the reactions until you run out of electrons: + 2+ - A (g) A (g) + e (g) 2nd ionization energy 2+ 3+ - A (g) A (g) + e (g) 3rd ionization energy 3+ 4+ - A (g) A (g) + e (g) 4th ionization energy With each ionization getting harder and harder to do, so the energy gets higher and higher (Can you explain why?) st Let s look at the 1 ionization energies of the elements and see if we can dicover any periodic trends Figure 4.1 Very distinct trend the group 1 metal has the lowest E 1 in any period, and the inert gas has the highest E, with atoms in the middle generally following along. 1 Thinking chemically the electron cloud of the inert gas must be very stable, and that of the 1A metal very unstable. The low ionization energy of the group 1 metals ties to their high reactivity So our first insight is that periodic properties of the elements are tied to the stability of the electron cloud around the atom
4-2 Ionization Energies and periodicity Now let s look at he successive ionization, what happens as you remove more electrons Table 4.1 (aj = attaj = 10 joules/atom) First let s get oriented. Flip back to Figure 4.1 Moderately hard ti ionize H, then really hard to do He, then really easy for Li then gets harder as you go to Ne, then easy to Na, etc (maybe sketch on board?) You can see that inthe first column of data in Table 4.1 No go across for Na Easy for the first Then a lot harder for the second, then slowly increases up to I 9 Then big jump to I (don t bother with figures, just rough sketch on board) 10 Figure 4.6 & 4.7 in powerpoint, but just as easy to sketch onboard Kind of like peeling an onion first few come off easy (outer skin) Then a big break and a bunch that come off ~ the same Then a bigger break and anoth bunch that com off the same Suggest Shells of electrons with similar properties Now look again at Table 4.1 Always 2 in innermost shell Then 8 in next shell The 8 in next Key Concept: Call the electrons in the outermost shell the valence electrons Call the electrons in inner shells the core electrons Another way to represent valence electrons is with a Lewis dot diagram Put the name of the atom in the middle and surround it with one dot for each valence electron Na. He: etc Similarly we can emphasize the core by writing the [core] See Table 4.2 Clicker question: Write Lewis diagram for some atoms
4-3 Electromagnetic Spectrum Now that we are focusing on electrons, we need to better understand how electrons and matter interact, and to do that we need to understand the electromagnetic spectrum EM spectrum - Radio, Microwave, infrared, visible, ultraviolet X-rays and gamma rays all part of EM spectrum Figure 4.8 1860's about the time the periodic table was getting nailed down Maxwell proposed electromagnetic theory of radiation. Radiation was oscillating electronic and magnetic fields moving through space Figure 4.9 Magnetic field and electronic fields are perpendicular to each other Key Concepts/Equations: Wavelength(ë) - the distance between consecutive peaks or troughs Frequency (í) - # of crests that pass a point in 1 sec 8 í ë = c Speed of light: 2.9979x10 m/s Exercise What is the frequency of green light (green ë= 500 nm)? -9-7 500 nm = 500x10 m = 5x10 m ëí=c í=c/ë 8-7 14 =3x10 /5x10 = 6x10 Hz What is the wavelength of an FM radio wave with a frequency of 91.1MHz? 6 +6-1 91.1 MHz = 91.1x10 Hz = 91.1x10 sec ëí=c ë=c/í 8 +6 =3x10 m/s / 91.1x10 /s = 3.29 meters Clicker question more of same? Maybe a set up rather than a calculation?
4-4 Line Spectra of atoms When you pass while light through a prism you the visible rainbow 400 nm violet 750 nm red is a continuous spectrum. No gaps When we excite atoms to emit light (in a flame or arc a spark through a gas) we get discrete wavelengths - line spectra If we use a single atom for the source we call this an atomic emission spectrum Simplest atomic emission spectrum is that of hydrogen Figure 4.13 & 4.14 Fore years Scientists tried to make sense of this and figure out why we the lines are where they are Rydberg figured ou an empirical equation that fit the data 2 1/ë = Constant (1/4-1/n ) where n = 3,4,5,... 7-1 The constant was called the Rydberg constant = 1.097 x 10 m ) Every element will give an atomic spectrum the number of lines and the ë of each line is different can be used as a fingerprint to identify an element 4-5 Photons If light is a continuum, why do atomic spectra consist of discrete line? At this point in time (late 1800's-early 1900's)classical physics, the view that light is a continuum was having all sorts of problems explaining experiment two key phenomena Black Body radiation - Refers to the fact that as heat something up it emits a continuum of radiation. The hotter it gets the more radiation. Practical example heating iron from red hot to white hot. It is emitting a continuum of radiation, so you would think that it would fit the classical (continuum ) theory. But it was emitting the wrong continuum! According to the continuum theory it should give off lots of light at high frequencies (short wavelengths) at all temperature. According to experiment it only did this at very high temperatures The Photo-electric effect - refers to light hitting a metal in a vacuum tube as shown in Figure 4.17 & 4.18. Here light had to have a certain minimum frequency (called threshold frequency) before electrons would pass from one electrode to the other
It took the brilliance of Einstein to figure out that the two things were related. Key Concept: The explanation was that the energy of light is NOT a continuum, but was tied up in small discrete packets called photons (just as matter is tied up in small discrete packets called atoms) Key Equation: The energy of a single photon of light energy is given by: E=hí (if you have the í of light) -or- E=hc/ë (if you have the ë of light) Practice calculation: Going back to our green light. What is the energy of a single photon of green light What is the frequency of green light (green ë= 500 nm)? In our previous work on this problem -9-7 500 nm = 500x10 m = 5x10 m ëí=c í=c/ë 8-7 14 =3x10 /5x10 = 6x10 Hz E=hí =6.63x10 Js x 6x10 sec -19 =3.98 x10 J -34 14-1 or E=hc/ë -34 8-7 =(6.63x10 Js 3x10 m/s) /5x10 m -19 =3.98 x10 J Clicker question, set up of more of the same
4-6 De Broglie Wavelength So the explanation seems to be that light displays wave-like properties under certain conditions, and particle like properties under other conditions. This is called the wave-particle duality of light In 1924 de Broglie proposed a really radical idea. If the wave of light can display particle like properties under certain conditions, then why can t particles of matter display wavelike properties? Using Einstein s theory of relativity he proposed that both light and matter obey the equation Key Equation: ë=h/mv Where ë is the wavelength, m is the mass of the particle at rest and v is the velocity of the particle Practice problems What is the debroglie wavelength of an electron traveling at 1% of the speed of light -31 mass of electron 9.11x10 kg 8 6 v =.01 x c =.01 x 3x10 m/s = 3x10 m/s -34 6 ë=h/mv = 6.626x10 J sec/(9.11 kg x 3x10 m/s) 2 2 Note 1 J = 1 kg m /s -10 so ë = 2.43x10 = =.243 nm this is the wavelength of an X-ray What is the mass of a photon of UV light (ë=200 nm) ë=h/mv më= h/v m = h/ëv ë in m = 2x10-7 8 Photons travel at the speed of light so v = c =3x10 m/s -34-7 8 =6.626x10 / (2x10 x 3x10 ) -35 =1.10x10 Kg which is about 10,000 less mass than an electron! Clicker question on De broglie equation
4-7 Wave-Particle Duality Proof of de Broglie s hypothesis came just a few years latter when 2 Scientists showed that you could point a stream of electrons at a crystal, and the electrons would diffract, Figure 4.22 Something that only wave can do! So we can regard light as wave or a particle depending on the experiment Similarly we can regard matter as particles or waves depending on the experiment! This is called wave-particle duality. (Interesting aside in text: JJ Thomson (and others) got Nobel Prize in 1906 for demonstrating that electron was a particle. His Son GP Thomson (and others) got Nobel prize in 1937 for showing it was a wave!) 4-8 Quantization Getting back to our line spectra for hydrogen In 1913 Niels Bohr had looked at the line spectra for H and the Rydberg equations, and had proposed that the reason the E was quantized was that tha the electrons in H were restricted to certain circular orbitals 10 years later, when debroglie proposed that electrons could act like waves, it was possible to rationalize why the electrons were restricted to only certain orbitals. Figure 4.24 It was proposed that the orbit could only be stable if the wave took an integral number of periods to get around the nucleus so it would return to its starting point after 1 revolution, otherwise it would cancel itself out like waves do in a diffraction pattern Putting a little math together, saying the circumference of the orbit is 2ðr, and this has to equal në. Throwing in the force of attraction between the nucleus and the electron he came up with the equation E = -2.18x10 n J/n 2 This quantized the energy of the orbitals to distinct values or energy states These quantized states are called stationary states The lowest stationary state (n=1) is called the ground state All the higher states (n>1) are called excited states n=2 is called the first excited state
Note that the energy of an orbital is negative. This means that energy is released as an electron falls into this state. And then you have to supply energy (ionization energy) to release the electron from this orbital 4-9 Electronic Transitions When an electron is in a stationary state it cannot emit or absorb energy. This only happens when en electron moves from on e state to another So if energy is going to be released from an H atom to make an emission line, and electron has to drop from a high stationary state to a lower stationary state so some energy can be released E of photon = E i - Ef Where E = E of initial state and E = E of final state i f Key Equation: 2 2 E = 2.18x10 J (1/n - 1/n ) photon f i Practice problems What is the energy released in J, and the wavelength of the photon carrying that energy when an electron in a hydrogen atom falls from the n=5 state to the n=2 state? 2 2 E = 2.18x10 J (1/n f - 1/n i ) 2 2 =2.18x10 J (1/2-1/5 ) =2.18x10 J (1/4-1/25) =2.18x10 J (.25-.04) =2.18x10 J (.21) =.458x10 J E = hc/ë; ë=hc/e -34 8 =(6.626x10 J s x 3x10 m/s)/.458x10 J =4.34x10-7 m =434 nm Which checks with one of the line you can see! Clicker question: Set up for E photon type calculation
I mentioned earlier that you can get these line spectra from having an electrical spark arc through a gas (that is how glow tubes work) or by simply putting the atoms in a flame. Heat energy kicks electrons to higher states, and the atom cools light is emitted as electrons drop from highly excited states to lower states This is the basis for the flame test we use in the lab, where each element gives off a characteristic color in a flame ( demo?) The opposite can also happen. Light of the right wavelength can be absorbed by an atom to kick an electron from a low orbital to a higher orbital. (Figure 4.30) We can use the absorption of light in the chemistry lab, not only to identify compounds, but to measure the amount of a compound in a material. As good as the Bohr theory was at explaining the hydrogen spectrum. It failed miserably as explaining the spectrum of any other atom. Many people atried and tried, but this theory could never be extended to anything else. As a result inthe next chapter we will have to give up on the Bohr theory with its nice circular orbital that you see in all the picture books and try something new!