Atomic Structure and Properties - AP Chemistry

N has 5 valence electrons. In the ion, N has 4 bonds, and thus 8 bonding electrons and 0 nonbonding electrons. Thus, formal charge = 5- (1/2)*8 - 0 = 1.

Carbon-14 has 8 neutrons

Nitrogen-13 has 6 neutrons

Oxygen-15 has 7 neutrons

Fluorine-16 has 7 neutrons

Boron-10 has 5 neutrons

The denotation is element name-proton+neutron. With 7 protons and 8 neutrons, this necomes Nitrogen-15.

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.

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).

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 .

For the carbonate ion, there are 3 resonance structures, and each of the 3 oxygen atoms carries a formal charge of –1 in two out of the three structures. This means that there is a –2/3 charge on each O atom. There are no formal charges in H2O or O2. and in O3, the central O has a positive charge and each outer O has a negative charge, so there is an average charge of –1/2 on the two outer ones.

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:

Since this is a monovalent cation, the non-ionic electron configuration will be (note that the electron was removed from the s shell and not the d shell, as this is a transition metal). This electron configuration corresponds to iron, so the cation must be .

We discover that element X is sodium, due to the masive jump after the first ionization energy. Sodium has filled 1s and 2s orbitals, and a half filled 3s orbital. We know that sodium will have the base configuation of \[Ne\], with one additional electron in the third shell, due to its position in the third period. This leads us to our answer of 1s22s22p63s1.

Electron configurations can easily be figured out by using the periodic table. It is important to memorize the order that the orbitals become filled; then, you can simply follow the periodic table, adding one electron as you move from one element to the next. To make things easier, you can also remember that the first period ends with 1s, the second ends with 2p, and the third ends with 3p. It gets a little more complicated after that, but knowing those first few periods will let you work quicker.

When given an electronic configuration, you can also add the number of electrons to determine the element's atomic number.

shows a total of electrons. Phosphorous corresponds to atomic number 15.

To find the answer, we would write out the electron configuration for each given choice.

Arsenic has an electron configuration of , with the fourth shell representing the valence electrons. The valence includes both the 4s and 4p, so it has a total of 5 (). We could also look at the periodic table and see that arsenic is in the same period as nitrogen and phosphorus, meaning it will have five valence electrons.