{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

test3 - Atoms and ions can be excited to very high states...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Atoms and ions can be excited to very high states called Rydberg states. Consider a He+ ion which has been excited to the n = 45 state. The nucleus of the one-electron ion is an alpha particle, e.g. 4He. Calculate the ionization energy of the ion. The helium cation is a one-electron species so the Bohr formula for the energy, E = -R(Z2/n2), is exact. IE(eV) = -E = (13.6 eV)(22/452) = 0.027 eV Calculate the average size of the ion. r = (0.529 )(452/2) = 540 The result calculated above is an estimate of the uncertainty of the position of the electron in the ion. Use the result to estimate the uncertainty of the electron's momentum. Use the Heisenberg Uncertainty Principle. DrDp = (h/2p) Dp = h/2pDr = (6.63 x 10-34 J-s)/(2)(3.14)(5.4 x 10-8 m) = 2.0 x 10-27 Kg-m/s Calculate the maximum possible value of the orbital angular momentum in units of h/2p. The quantum number l ranges from 0 to n-1 so the maximum value of l is 45-1 or 44. In units of h/2p, L equals the square root of l(l+1), i.e. [44(45)]0.5 The ion is moving with a speed of 2.0 km/s. Calculate the de Broglie wavelength of the ion. l = h/p = h/mv. We require the mass of the 4He ion in Kg. m = 0.0040 Kg/6.02 x 1023 = 6.64 x 10-27 Kg l = (6.63 x 10-27 J-s)/(6.64 x 10-27 Kg)(2000 m/s) = 5.0 x 10-11 m Suppose that a sample of 1000 He+ ions in the n = 45 state is prepared. When the excited ions decay to the ground state, will monochromatic or polychromatic radiation be emitted? Explain. Polychromatic. An atom can decay directly to the ground state or it can first decay to one of several excited states with an energy lower than that of the n=45 state. As it cascades down, several photons will be emitted. A figure is given below of a hydrogen-type orbital. The +'s and -'s indicate the sign of the wavefunction in the various regions. The z-axis is in the vertical direction. The information provided by the figure is sufficient to answer this question. Indicate the location of the radial node(s) on the figure with ........ The single radial node is marked on the gif file of the orbital. Indicate the location of the angular node(s) on the figure with -------- The two angular nodes which are planes are marked on the gif file of the orbital. The value of n is ________. There are a total of 3 nodes so n-1 = 3 and n = 4. The value of l is ________. There are 2 angular nodes so l = 2. The value of the absolute value of ml is ________. The answer is one as one of the angular nodes contains the z axis. What symmetry operation will convert the orbital shown above to its partner with the same value of n, l, and absolute value of ml? Rotation about the z axis of 90 degrees (90/1). Will an electron represented by this orbital ever be found at the nucleus? Give a succinct but complete explanation. No. First l is not zero so there is a signficant centrifugal force resulting from the orbital angular momentum which keeps the electron from the nucleus. Also there are two angular nodes which intersect at the nucleus so the wave function is zero there. A gif file of the orbital with the angular and radial nodes marked. For each property, arrange the species in order of increasing value (smallest to the LEFT, largest to the RIGHT). first ionization energy: Al, Cl, F, He-, K, K+, Mg The order is He-,K,Al,Mg.Cl,F,K+ The cation has the highest IE as ionization would remove a core electron. The anion is the easiest to ionize, inert gases in fact have negative electron affinities. Al, Mg, and Cl are third row elements and will be easier to ionize than F which is at the right of the 2nd. row. Within the third row, there is a general trend in IE increasing with atomic number except the order of Al and Mg are reversed as a 3p electron is involved rather than a 3s in the case of Al. atomic radius: Cl, Cs, Cs+, Cs-, F, He, P The order is He,F,Cl,P,Cs+,Cs,Cs- Cs is at the far left of the 6th row so the Cs species are by far the largest. For fixed atomic number, size increases with the number of electrons so the order is cation, neutral, anion. He is first row to the right so it is the smallest. F, in the second row is next. Cl is in the same family as F but in the third row and is hence larger than F. P is in the same row as Cl but has a lower atomic number and hence a larger size. The following questions deal with the chemistry of the transition metal titanium, atomic number 22. Predict the electronium configuration of a titanium atom. [Ar](3d)2(4s)2 Is a titanium atom magnetic or not? Briefly explain. Yes. The two 3d electrons can be in separate orbitals and will be unpaired. Predict the electronic configuration of a Ti+2 cation. The 4s electrons are shielded by the 3d electrons and are removed first. the configuration is [Ar] (3d)2. Is a Ti+2 cation magnetic or not? Briefly explain. Yes for the same reason as above. What is the maximum oxidation number of titanium achievable under normal chemical conditions. Briefly explain. +4 as four valence electrons, the two 4s's and the two 3d's, can be removed before ionization of the core is required. A student proposed [Ar] (5s)4 as a configuration for an excited state of titanium. Comment on the choice. A heretical choice. The configuration which places 4 electrons in a single orbital, the 5s, violates the Pauli Exclusion Principle. The experimentally determined electronic configuration of the V+1 cation which is isoelectronic to Ti is [Ar] (3d)4. Is this the expected result or not? Discuss. The actual result is a surprise. If one argues that Ti and V+ as isolectronic species should have the same configuration, [Ar](3d)2(4s)2. If one starts with the configuration of V, [Ar](3d)2(4s)2, one would expect that the 4s would be the first electron to go yielding [Ar](3d)3(4s) . Neither case agrees with the actual result. By the way, the configuration of V+2 is the expected [Ar](3d)3 . Briefly discuss an experimental method for the measuring the first ionization energy of titanium. Method 1. Employ the photoelectric effect. Irradiate the gaseous atoms with UV radiation and measure the kinetic energy of the liberated electron. Method 2. Insert two electrodes in the tube containing the gaseous atoms and measure the current between the electrodes as a function of the applied voltage. The graph of I versus V will show a discontinuity when the voltage equals the ionization energy in eV. This is the celebrated Franck-Hertz experiment. ...
View Full Document

{[ snackBarMessage ]}