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Question

My colleague and I disagree on the answer to a quiz question. "Identify the following element by its quantum numbers; 3, 2, -2, - I my the numhers identify the 3rd energy level, d orbital, 8th element, which is nickel. He says that for nickel, it should be 4, 2, -2, - ' I , , andso it isincorrectly written and can identify no element. Who is correct?

Answer

This question cannot be answered because a set of quantum numhers refers to an electron, not an element. The quantum numhers define the energy of the electron; that is, they identify the particular state of an electron in a definite atomic orbital. Three of the four quantum numbers; n, 1, and ml, define the atomic orbital. The fourth, called the spin quantum number, m,, identifies which of the two spin states possible in a given orbital is possessed by the electron. Two or more electrons in the same atom cannot have an identical set of quantum numbers.

However, to answer the question, we could assunw that the set of quantum numbers given in the question is that of the last electron added following the usual pro- cedure in the fictitious building up process. We shall also assume that the atom in question is in the ground state; that is, all its electrons exist in the lowest possible energy levels and each electron possesses a different set of quantum numhers to identify its state. Then, we could decide which element is referred to in the ques- tion. For example, this table identifies the atomic orbital and the spin state of the last electron in the fictictious building up process for a few elements:

Element n 1 mr ma

It should be noted that there are two ways in which the hydrogen atom can he in its ground state. These correspond to the two choices of the spin state. The quantum numhers would he either 1,0,0, or 1,0,0, -I/% for n, 1, ml and m,, respectively. However, by

We solicit questions for this column and reactions to the an- swers given. Correspondence should be addressed to Dr. J. G. Malik at San Diego State College, San Diego, California.

chemical queries

. . especially for

high school teachers

convention, the positive value is chosen to represent the most stable state. In a similar manner, when it be- comes necessary to specify individually the five states of the d orbitals as given by the magnetic quantum num- ber ml, the values are assigned in the order +2, +1, 0, -1, -2. Thus, 3, 2, +2, would represent the most stable ground state with 3, 2, +l, next. And, after five single electrons have been placed in these orbitals the sixth electron would pair up and have the quantum number designation 3,2, +2, - I / * .

From the table, and assuming that the question refers to atoms in their ground state, and that it specifies the configuration of the "last electron added" in the ficti- tious building up process, it can he seen that the quan- tum numbers 3,2, -2, '5dentify1' zinc.

Question What are same of the other substances besides water in which

the solid phase floats in the liquid phaie, as ice floats in water?

Answer

by William Hutton, Bowling Green State University, Bowling Green, Ohio.

There are a few elements which display a slight nega- tive slope to the solid-liquid equilibrium in their phase diagrams (temperature versus pressure). These suh- stances show an increase in volume as the change from the liquid to the solid state a t the melting point pro- gresses.

Elemental bismuth shows this property with solid bismuth floating on liquid bismuth at its melting point. The increase in volume when liquid bismuth solidifies is about 3.32%. Gallium increases in volume upon solidification (3.lY0) as does antimony. Some alloys containing these metals also show this property. Of these, type metal alloys (various proportions of lead, antimony, and tin, with a little copper occasionally) are the most common.

Question My students and I have run several esterilication reactions and

sn ionic reaction rtt 25" and 35°C and have caledated the mercies of activation. But now we can't find anv literature - vahlnes in tables anywhere for comparison. Where are energies of activation available?

Answer

by H. F. Shuruell, Queen's University, Kingston, Ontario, Canada

Unfortunately, it is true that energy of activation values are not readily available in many of the common

252 / Journal of Chemical Educofion

tables found in texts, laboratory manuals, or even chemistry handbooks. However, excellent summaries of available values for many common reactions can be found in the National Bureau of Standards Tables of Chemical Kinetics, N.B.S. Circular 510 ($4). Similar information for other reactions is listed in circular 510, supplements 1 ($3.25) and 2 ($0.35; an index) and in N.B.S. Monograph 34, volumes 1 ($2.75) and 2 ($2). These may be purchased from The Superintendent of Documents, U.S. Government Printing Office, Washing- ton, D.C. 20402.

Question

In Millikan's oil drop appsratus, how is the mass or radius of the drop of oil determined experimentally for nse in ealculcttion of the charge of the electron? Do not just state that one us- Stokes' Law.

Answer

by Norman Dessel, San Diego State College, San Diego, California

Neither the mass nor the radius of an oil drop is used. Let us examine the details: As a small drop of oil falls through a viscous fluid, such as air, under the inhence of gravity, it meets with a resistive viscous force which opposes the force of gravity. Stokes has described this resistive force F, as being proportional to the relative velocity v, of the spherical body,

where k is a proportionality constant. The question was, then, what physical variables determine the value of the proportionality constant, k? If the resisting gaseous medium is homogeneous, and if the radius r, of the spherical body is not too large, then Stokes reported that k is given by

k = 6mr

where 7 is the coefficient of viscosity. This property of viscosity can be thought of as a kind of friction manifest when a body moves through a fluid mass. The coeffi- cient is expressed in units of force times distance divided by area times velocity. The viscosity of air a t 20°C is found experimentally to be about 1.81 X 10W4 dyne- sec/cm2.

The falling drop of oil in Millikan's experiment is acted upon by a resultant force F, in the downward di- rection

where Fn is a buoyant force equal to the weight of air displaced by the drop and F, is the force of gravity. The values of F, and F, are unchanged during the fall. The value of F, is initially zero and as the drop acceler- ates, the resistive force increases in value until it, plus the constant buoyant force opposing the fall, equals the force of gravity. When this happens, the drop no longer accelerates since the net downward force now equals zero and a constant maximum velocity of the drop has been reached. Under this condition, F = 0 and F, - F, = F,.

If d is the oil density and d' is the air density, then, from Newton's Second Law,

where g is the acceleration due to gravity. Fo is related to the mass of air displaced by the oil drop and is given by

Substituting these values into the prior expression for the balanced forces; at constant, maximum, velocity (umaz)

The one quantity in this equation which is not known nor directly measurable is r, the oil drop radius. Solv- ing the above equation for r, one obtains

Substituting this value for the radius into the expression for the resistive force, F, = kv = 6~7rv,,,,,,, it is pos- sible to specify the value of the proportionality constant

since all quantities are known, or measurable. When the drop is given a charge q and placed within a

uniform electric field, E, the drop reaches an upward constant maximum velocity urn.,' when the electrical force, Eq, just balances the mechanical forces. Thus,

But for constant, maximum velocity fall under gravity, F, - F, = ku,.,. So

Eq = kv,.. + hum,,' Hence, the charge, q, is given as

Note that om., is the free fall terminal velocity under no electric field, and v,.,' is the terminal upward velocity under the influence of an electric field. The electric field is the applied voltage divided by the distance be- tween the charge plates.

Both terminal maximum velocities are constant and both are reached very quickly experimentally. They can be determined by measuring the distance a drop moves as a function of time.

But, particularly, note that from the expression we have for k and from the above expression for q, it is pos- sible to determine the charge on the oil droplet without ever determining the radius or the mass of the actual drop.

Thus, while it may be true that some explanations in- dicate that the mass or radius of the oil droplet is de- termined, the actual experiment and calculations aa performed by Millikm did not use these parameters.'

Far further information, see MILLIKAN, R. A,, Phys. Rev., 32, 349 (1911); or a. reproduction of this article in OMER, G. C., et al., "PhysicalSeienee: Menand Concepts," D. C. Heath, Boston, 1962.

Volume 45, Number 4, April 1968 / 253