2. (2 points) Calculate the Effective Nuclear Charge for each atom. Show all of your work for full credit. Se: Br: (1 point) Which atom is larger? (This is your claim.)

The question is incomplete. Here is the complete question.

Consider the selenium atom, Se, and the bromine atom, Br.

Write out the full electron configuration fro each atom.

Se:

Br:

Calculate the Effective Nuclear Charge for each atom. Show all of your work for full credit.

Se:

Br:

Which atom is larger?

Answer and Explanation: Electron Configuration of an atom demonstrates the shape and energy of its electrons. One of the rules used for it is given by using Madelung's Rule, in which the order of increased energy of the electrons is:

1s < 2s < 2p < 3s < 3p < 4s < 4p < 5s < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p

Atom of Selenium (Se) has 34 electrons. Its electron configuration is

Se:

Atom of Bromine has 35 electrons. Its electron configuration is

Br:

Effective Nuclear Charge ()is the net positive charge experienced by an electron in a multi-electron atom. In other words, it is the net force that helds nucleus and electrons together.

It is calculated by:

where

Z is the atomic number, i.e., number of protons of the atom

S is the nonvalence electrons, i.e., the number of electrons between the nucleus and the electron in question.

For Selenium (Se):

From the electron configuration, the valence shell is 4 with 6 valence electrons. Nonvalence electrons is the difference between valence and total electrons:

S = 34 - 6 = 28

= 6

The Effective Nuclear Charge of Se is 6

For Bromine (Br):

The valence shell, according to the configuration, is 4 and valence electrons are 7.

S = 35 - 7 = 28

= 7

The Effective Nuclear Charge fo Br is 7.

Bromine is larger than Selenium because it has bigger Effective Nuclear Charge, which means it held its electrons more loosely and, consequently, has a larger atomic radius.

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.