Atomic Structure and Properties - AP Chemistry

First, find how many moles of chlorine are present in the sample:

Because the ratio of moles of to moles of is 1:1, there are also 2 moles of . The next step is to find the mass of 2 moles of .

Find the number of moles of each element.

If the chemist has 2.0 moles of the compound, 1 mole would contain half as much. So 1mol of carbon, 4mol of hydrogen, and 2mol of oxygen, which forms acetic acid

This question is asking us to determine how many atoms there are in moles of . Rather than asking for the number of molecules, our final answer will need to tell us the number of atoms. Since there are atoms for every molecule of , we'll first need to find out how many molecules there are.

The peak at ~ 40 MJ binding energy corresponds to the 1s orbital. Since there are peaks at lower binding energy, it is implied that this orbital is filled (by 2 electrons). There are two peaks at lower binding energy corresponding to the 2s and 2p orbitals, respectively. The 2s peak has the same intensity as the 1s peak indicating it has the same number of electrons (2). The 2p peak is 3/2 as tall as the 2s peak and indicates that indicates it has 3 electrons in the orbital. The electron configuration of 1s22s22p3 makes the element N.

Mg has a Z=12 while Na is Z=11. The increased number of protons in Mg gives rise to a greater effective nuclear charge. This greater effective nuclear charge binds the electrons in their respective orbitals more tightly than an element with a lower effective nuclear charge. The peak height of Mg is 2x that of Na for the 3s orbital because it has twice the electrons present.

Oxygen has a Z=8 vs nitrogen’s Z=7. This greater effective nuclear charge binds the electrons in their respective orbitals more tightly than an element with a lower effective nuclear charge.

The binding energy at ~ 53 Mj represents the 1s orbital. With other peaks present, this 1s orbital is fully occupied by 2 electrons. The peaks at lower energy would correspond to the 2s and 2p, respectively. Since the 2p peak is twice the intensity of the 2s or 1_s_ peak, it has twice the electrons (4). This gives the electron configuration 1s22s22p4.

Step 1. Convert the Binding Energy from MJ/mol to J

x x

Step 2:

The sets of quantum numbers needs to follow the following rules: n (principal quantum
number) needs to be a positive integer, l can have any integral value from 0 to n – 1, and ml can range from –l to l. The only quantum numbers that follows these rules are (a).

The azimuthal quantum number is the second quantum number, designated by the letter l. It gives the shape and number of subshells in a principal energy level (shell).

There is not enough information given in the question stem to determine if the velocity of the electron is changed. We can definitively determine, however, the electron will lose energy and a photon will be emitted.

For any n, the energy level can hold 2n2 electrons, since there are two electrons for each orbital.

A fully filled orbital is generally more stable than a half-filled orbital. The splitting of the energy levels is also dependent on the geometry of the compound that one is analyzing. The 4s electrons are of higher energy than the 3d electrons, thus are lost first, leaving half filled d orbitals, which is the most stable configuration.

The energy of an electron is related to the quantum number by the equation E = -R/n2, where R is constant. The negative charge make it so that as n increases, the numerical value of the energy becomes less negative, approaching zero. Thus the energy increases as n increases.

For the 3d subshell, n (principal quantum number) = 3 and l (azimuthal quantum number) = 2, so n + l = 5. For the 4s subshell, n = 4 and l = 0, so n+l = 4. From this information, we can see that the 4s subshell has lower energy and will fill with electrons first.

The elements in copper's group will fill the d orbital at the expense of having a half filled s orbital. This is because this configuration is more stable than a full s orbital and d orbital with 9/10 spots filled.

6 more electrons are present in chromium than argon. It's more stable for d orbitals to be half-filled. Thus, one electron will fill each of the 5 d orbitals.

First, we will write out the electron configuration for the ground state Manganese atom, considering only the valence electrons:

The orbitals highest in energy will be filled last, so our highest energy orbitals are orbitals.

The angular momentum quantum number describes the shape of the orbital, with orbitals corresponding to , orbitals corresponding with , and orbitals corresponding to , and so on.

We are considering a orbital, so .

First, we will write down the electronic configuration for the ground state of Cobalt:

Remember that a half-filled orbital is very stable, and when it is possible for an ion to have a electron configuration it will. In order to attain this, two electrons will be removed from the orbitals and one will be removed from the orbital, yielding: