Test: Electronic Structure - MCAT MCQ

A photon will be emitted each time the electron moves from shell to another.

The electron can fall to any shell between n=4 and n=1.

That means that the electron can fall through 4 paths: n=4 n=1 (1 photon) n=4 n=3 n=1 (2 photons) n=4 n=2 n=1 (2 photons) n=4 n=3 n=2 n=1 (3 photons)

So, 1, 2, or 3 photons can be emitted.

Paramagnetic materials are pulled toward magnet fields.

Paramagnetic materials have at least one unpaired electron.

All of the electrons in a noble gas configuration are paired. Rb2+’s electron configuration is [Ar]4s2 3d10 4p5, so one of the 4p electrons is unpaired.

The work function, E0, is the energy that it takes to remove an electron from the metal.

The work function can be calculated as the difference between the energy put in by the photon, E, and the kinetic energy of the ejected electron, KE. That is to say, E0=E-KE.

Since the photons all have the same wavelengths, E does not change. That means that the largest E0 will be seen in the case with the lowest KE.

Lower speeds mean lower kinetic energies. Without any calculation, we can compare cases with similar units. The electrons from Metal 1 are moving slower than those from Metal 2, and those from Metal 3 are moving more slowly than those from Metal 4.

Now, we have to compare the electrons from Metals 1 and 3. Recall that c = 2.99108m/s and note that 0.0000001c = 10-7c. This means that the electrons from Metal 3 are moving at a velocity v = 10-72.99108m/s = 2.99101m/s = 29.9m/s, which is slower than the electrons from Metal 1. With the slowest-moving electrons, Metal 3 has the largest work function.

Unless otherwise specified, atomic structures are written for neutral atoms.

Nitrogen’s atomic number is 7, so a neutral atom has 7 electrons.

The first orbital to fill is 1s, with two electrons.

The second orbital to fill is 2s, with another 2 electrons.

There are no tricks with the first two shells: the half-filled shell that you saw in “Electron configurations for the third and fourth periods” video only comes into play with higher shells and orbitals as the video showed with Cr and Cu.

The 2p shell starts to fill next, absorbing the remaining 3 electrons, so the electronic structure of nitrogen is 1s2 2s2 2p3.

The electronic structure of neutral iron is [Ar]4s2 3d6.

Fe2+ has two fewer electrons, so you might expect that the correct configuration is obtained by removing two electrons from the highest energy orbital, 3d, yielding [Ar]4s2 3d4.

However, Fe2+ would have a lower – more favorable – energy if the 3d shell is half-filled with 5 electrons, so it will take on the [Ar]4s1 3d5 configuration (like chromium).

The photon with the shortest wavelength is the one with the highest energy because Ephoton = hc means that = hc Ephoton.

Recall from the “Bohr model energy levels” video that energy for these transitions is E= -13.6eV (1/n2² -/n₁².

So, for electrons that start in the same orbit, the case with the highest ending orbit will have the highest energy and hence shortest wavelength. We can immediately rule out the n₁ = 2 and n₂ = 3 case because the n₁ = 2 and n₂ = 4 photon would have a shorter wavelength. Similarly, the n₁ = 3 and n₂ = 9 photon will have a shorter wavelength than the  n₁ = 3 to  n₂ = 6 one.

We’re only left to calculate 1/₂² -1/n₁², for these two cases, and the result with the largest absolute value yields the highest energy (and so shortest wavelength) because of the -13.6eV in E=-13.6eV

, so the photon that drives the electron from n₁ = 2, n₂ = 4, has the higher energy and shorter wavelength.

Chlorine’s electron configuration is [Ne]3s2 3p5

The highest energy electrons are in the last orbitals to fill.

The last orbitals to fill in this case are the 3p orbitals, so the principle quantum number, n, is 3 from the 3 in “3p.”

The p-orbital electrons (l=1) have ml=-1, 0, or 1 (of which 1 is the highest angular momentum number).

Spin “up” for an electron means m_s​,  = +½.

So, n = 3, l =1, ml = +1, and = m_s = +½.

This is an application of Heisenberg’s uncertainty principle.

The uncertainty of the position is determined by knowing that the electron is somewhere inside the 25 cm wide machine, so x = 0.25 m.

The minimum uncertainty is met when the inequality in the Heisenberg uncertainty inequality becomes an equation: xp > h/4pi. Solving the equation for p, we have p = kg x m/s, the most precise measurement for the electron’s momentum when it is inside the radiography machine.

The Bohr radius, r_n start subscript, n, end subscript, can be calculated in relation to the radius of the first orbit, r1 as r_n = r₁ x n²

This means that the electron that was farthest away is the one that had the highest initial energy. Since all of the electrons fell to the same ground state, the highest energy photon was given off by the electron that was most distant from the nucleus.

Recall that a shorter wavelength indicates a higher energy, and similarly photons with higher frequencies are more energetic. Comparing wavelengths, the = 12.1/h eV photon has higher energy than the one with = 10.2/hc eV Comparing frequencies, the V = 12.8/h eV photon has more energy than the = 12.1/h eV

Now compare the = 12.1/hc eV and V = 12.8/hc eV photons. Recall that E = hc/v and E = hv. Hence the photons’ energies are 12.1/hc eV x hc = 12.1eV and E = 12.8/h eV x h respectively. Therefore, the photon with frequency v = 12.8/h eV has the most energy, meaning that its corresponding electron started farthest from its nucleus.

Recall from the “Electron configurations for the third and fourth periods” video that the electron configuration for Al can be written [Ne]3s2 3p1.

All of the electrons denoted by the [Ne] noble gas configuration are in lower-energy orbitals than the n=3 electrons.

The [Ne] configuration can be written 1s2 2s2 2p6.

The lowest energy orbital is the one with the lowest quantum numbers, hence, n=1, l=0.