CHE 100 - Ch. 9

The 5p orbitals fill immediately after the 4d orbitals and immediately before the 6s because:

- observed experimental results - theoretical calculations The 5p orbitals fill immediately after the 4d orbitals and immediately before the 6s based on observed experimental results, and has been supported by theoretical calculations.

How many electrons are located in the highest energy orbital of aluminum? Al electron configuration=1s^2s^22p^63s^23p^1

1 The electron configuration for aluminum is: Al electron configuration=1s22s22p63s23p1 Looking at this electron configuration we see that 3p is the highest energy orbital. If we then look at the superscript attached to it, we see that it contains 1 electron.

How many total electrons does Cl have?

17 Yes, total # of electrons for a neutral atom is the same as it's total # of protons, which is the atomic number How many total electrons does Cl have?

Which of the following is a correct order of sublevel filling for electrons?

1s 2s 2p 3s 3p 4s 3d 4p

What is the electron configuration for the core electrons in bromine?

1s22s22p63s23p6 Bromine's full electron configuration is 1s22s22p63s23p64s23d104p5. From this configuration, we separate the valence electrons (4s24p5) from all of those that constitute full principal energy level shells; that is, the configuration of the closest previous noble gas. The closest noble gas that precedes bromine on the periodic table is argon, with an electron configuration of 1s22s22p23s23p6. These are the core electrons of bromine, which are virtually unreactive as they constitute full octets.

What is the electron configuration for K+?

1s22s22p63s23p6 To find the electron configuration of a cation, start with the electron configuration of the neutral atom and remove electrons from the outermost shell or shells first. The electron configuration of K is as shown below. 1s22s22p63s23p64s1 To find the configuration of K+, remove the single electron in the 4s subshell so that in total there is one fewer electron. This leaves a configuration of the following. 1s22s22p63s23p6

What is the ground state configuration of calcium (Ca)(Ca)?

1s22s22p63s23p64s2 The electron configuration of calcium (Ca) is 1s22s22p63s23p64s2. You can follow the steps below to figure this out: 1) Look at the periodic table to find the number of electrons that calcium has. The number of electrons is the same as the atomic number, 20. Now we know we must fill our orbitals with 20 electrons. 2) Fill the 1s and 2s orbitals first, remember the s-orbitals have 2 electrons each, giving us: 1s22s2 (4 electrons down, 16 to go) 3) After filling our s orbitals we move to the right side of our periodic table and fill our 2p orbitals. Remember that there are 6 spots for electrons in the p orbital, giving us: 1s22s22p6 (10 electrons down, 10 to go) 3) Now you should be in the 3rd row (period) of the periodic table and you can start to fill the 3s and 3p orbitals. 1s22s22p63s23p6 (18 electrons down, 2 to go) 4) For the last 2 electrons you should be on the 4th row (period) and can fill your 4s orbital which holds 2 electrons. 1s22s22p63s23p64s2

What are the core electrons of strontium?

1s22s22p63s23p64s23d104p6 Strontium's full electron configuration is 1s22s22p63s23p64s23d104p65s2. To find its core electrons, we must separate the valence electrons (in this case, 5s2) from all of the complete principal energy level shells (the configuration of the closest noble gas). Thus we isolate the electron configuration of krypton, 1s22s22p63s23p64s23d104p6, from the full electron configuration of Sr. These are its core electrons, which are virtually unreactive as they constitute full octets.

What is the electron configuration for Ru3+

1s22s22p63s23p64s23d104p64d5 The electron configuration for Ru is [Kr]5s14d7 but Ru3+ loses three electrons, so it will become [Kr]4d5. It should be noted that Ru has what is called an anomalous configuration, as there is only one electron in the 5s subshell. A number of transition metals have anomalous configurations including copper, chromium, and platinum, among many others.

How many total electrons does Vanadium have?

23

What is the frequency, in Hertz, of an electromagnetic wave with a wavelength of 625 nm? Use c=2.998×108ms for the speed of light. Report with three significant figures and in scientific notation. When reporting your answer in scientific notation, use the multiplication symbol, ×, not the letter x.

4.80 x 10^14 Hz The equation is c=λ⋅ν where λ stands for the wavelength, c is the speed of light, and ν denotes frequency. (6.25×10^−7 m)⋅(ν)=(2.998×10^8 m/s) ν=4.80×10^14 s^−1=4.80×10^14 Hz

What would the electron configuration for the transition metal, silver, end in?

4d^9

What are the valence electrons in cobalt's electron configuration?

4s23d7 Cobalt's full electron configuration is: 1s22s22p63s23p64s23d7. Argon has the electron configuration of: 1s22s22p63s23p6 which means that these electrons are stable and do not take part in bonding. The remaining electrons are the valence electrons that do take part in bonding (4s23d7).

Which are the valence electrons for tellurium?

5s2,5p4 Tellurium is in the fifth period and the fourth element in the p block, so it will have two valence electrons in the 5s orbital and four in the 5p orbitals.

What is the frequency of a particular type of radiation in Hertz given that the wavelength is 4.40×102nm? Use c=2.998×108ms for the speed of light. Report your answer in scientific notation. Your answer should have three significant figures.

6.18 x 10^14 Hz Recall that the useful wave equation is: c=λ⋅ν, where λ is the wavelength, ν is the frequency, and c is the speed of light. First, convert the wavelength from nm to m: 4.40×102 nm×1 m/^nm = 4.40×10^−7 m Next, rearrange the equation to solve for ν: v=c/λ Substitute in the known values to solve: ν = (2.998×^m/s)/(4.40×10^−7m) ν = 6.814 × 10^14 s^−1 Note that Hertz (Hz) is another name for inverse seconds (s−1). The answer should have three significant figures, so round to 6.81×1014 Hz

Consider the following electron configurations. Element A: 1s22s1 Element B: 1s22s22p63s1 Element C: 1s22s22p63s23p64s1 Which of the following is the correct order for decreasing IE1 (i.e., largest to smallest)?

A>B>C Elements A,B, and C all belong to the same group (1st group indicated by the valence s orbital). Because atomic radius increases with increasing n (A=2,B=3, and C=4) and IE decreases with increasing atomic radius, we know that ionization energy decreases from A to B to C.

According to Hund's rule, which of the following are valid orbital diagrams? (select all that apply)

An orbital diagram consists of two individual squares, labeled "1 s" and, "2 s," followed by a grouping of three connected squares labeled, "2 p." All boxes are oriented in a row. The two individual squares contain a pair of half arrows. One half arrow in each pair points up, and one points down. The remaining three connected squares each contain single upward-pointing half arrows. Five boxes are shown. The left most box contains one up arrow and one down arrow. Below the box is the label "1 s." The second box to the right contains one up arrow and one down arrow. Below the box is the label "2 s." The third, fourth, and fifth boxes are drawn in contact with one another. The third box contains one up arrow and one down arrow. The fourth box contains one up and one down arrow. The fifth box contains one up arrow. Below the fourth box is the label "2 p." Hund's rule states that every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied degenerate orbitals have the same spin. Therefore, the second and fourth options above are valid orbital diagrams. Therefore, options B and D are valid orbital diagrams.

Which of the following is the correct sequence for increasing order (i.e., smallest to largest) of first ionization energy (IE)?

B<Be<O<N Atoms with half-filled or completely filled orbitals in the outer shell are more stable and possess higher IE. As atomic radius increases, IE decreases. Although B is smaller than Be, we expect the Be atom to have a higher IE because of the stable 2s2 configuration. N's 1s22s22p3 configuration is more stable than the 1s22s22p4 configuration of O, so N has a greater IE than O. The correct answer choice is B<Be<O<N.

Which of the following is larger, Br or Br−?

Br− The larger of the two is Br− because ionic radius increases when you add an electron and decreases when you take one away. This is because electrons repel one another, so adding an electron makes an atom bigger.

Which of the following species has the smallest radius?

C4+ These species are isoelectronic, and carbon has the most protons, so it will attract the remaining two core electrons the most.

For an atom, electrostatic potential is also called:

Coulomb potential These terms are synonymous, as Coulombs deal with electric charge.

According to the Bohr model:

Electrons only emit or absorb a photon if they move to a different orbit within the atom. Bohr's model ignored the prediction that the orbiting electron in hydrogen would continuously emit light. He relied on Planck's and Einstein's work, and he used discrete values for angular momentum, energy, and orbit radius.

List S, Fr, Rb, F, and In in order of largest to smallest covalent radius.

Fr, Rb, In, S, F Figure(a) shows the increasing radii down the halogen group. Figure (b) shows the covalent radii of the elements relative to each other in the periodic table. As seen in figures (a) and (b), the general trend is that radii increase down a group and decrease across a period.

Bohr's model of the hydrogen atom, which restricts electrons to circular orbits around the nucleus parameterized by a single number, n, can best be applied to which other species?

He+ Bohr's model applies to any atom with one electron, such as He+,Li2+,Be3+, etc. Multielectron atoms like He are affected by repulsion between electrons and yield a more complicated pattern of energy levels.

Hund's rule states that the electron configuration with the lowest-energy will have the maximum possible number of unpaired electrons. Which of the elements below would require special attention to this rule to correctly depict the orbital diagram?

N The other three options only have s orbitals, so there are no degenerate orbitals. Hund's rule applies to degenerate orbitals like the 2p orbitals in a nitrogen atom, which must all contain one electron before any of them becomes completely full with two electrons.

Which of the following species has the largest radius?

O2− For any series of isoelectronic particles, the radius decreases as the atomic number (the number of protons) increases. Note that all of these species are isoelectronic with neon, meaning they all have the same number of electrons. Oxygen has the least number of protons of the species listed (smallest atomic number). Therefore, the electrons in O2− will be the least attracted to the nucleus, and so this ion has the largest radius.

Which ion is isoelectronic with krypton?

Rb+ If rubidium loses its 5s electron it will lose a shell and attain krypton electron configuration.

Which of the following can be used to show the spin state of an unpaired electron in an orbital?

Spin up Spin down Pairs of electrons in an orbital must have opposed spins, but a lone electron can have either of two possible spin states. By convention the two opposed spin states are described as spin up and spin down. Either is acceptable for a lone electron, although spin up is more commonly seen.

In which of the following element blocks does the atomic radius generally not tend to increase when moving down within a group?

The atomic radius generally increases when going down a group in all of the above element blocks. In every group, the atomic radius will increase as we move down the table, without exception. This is because each lower period features atoms with an additional shell, and that additional shell is farther from the nucleus.

Which of the following is true?

The further an electron is from the nucleus, the more energy it has. The further an electron is from the nucleus, the more energy it possesses.

Which element has an electron configuration ending in 4d2?

Zr The d block is one behind the period number in terms of energy level, so the 4d orbitals come after the 5s orbitals. Zirconium is the second element in the d block of period 5, so its electron configuration will end in 4d2.

The term nuclear charge refers to the precise charge of the nucleus, which, in units of fundamental charge, is equal to the number of protons, Z. The term effective nuclear charge is used to mean the apparent charge that determines the radius of the outermost electron(s) in an atom. Which quantity below can be used to approximate the effective nuclear charge by accounting for shielding by the inner electrons?

Z− the number of shielding electrons If Z is the number of protons, then we must subtract from it the number of shielding electrons to find the remaining positive charge that is attracting the valence electrons, which we call the effective nuclear charge.

An increase in the wavelength of electromagnetic radiation is associated with which of the following?

a decrease in frequency Wavelength and frequency are inversely proportional: As the wavelength increases, the frequency decreases. The inverse proportionality is illustrated in the figure below. This figure also shows the electromagnetic spectrum, the range of all types of electromagnetic radiation.

A larger amplitude can correspond to:

all of the above The amplitude is the magnitude of the wave's displacement, in other words one half the height between the peaks and troughs. This is how tall the wave is, which corresponds to all of these items.

Valence electrons:

all of the above: - are available to do chemistry and participate in covalent bonding - have the highest energy - are lost first These are all characteristics of valence electrons. Core electrons on the other hand do not participate in chemical reactions, have lower energy, and are lost last.

What is the magnitude of a wave's displacement known as?

amplitude Amplitude is the magnitude of a wave's displacement.

In an orbital diagram, such as the one below, each small box represents which of the following?

an individual orbital Each box represents one orbital, each of which can hold two electrons.

The electrons most responsible for shielding in an atom are the __________.

core electrons The core electrons shield the valence electrons from the nuclear charge because they are positioned between the protons and the valence electrons.

When an electron moves from a lower, more stable orbit to a higher-energy orbit:

energy is absorbed in the form of a photon When an electron moves from a lower, more stable orbit to a higher-energy orbit, energy is absorbed in the form of a photon.

The number of wave cycles that passes a specified point in space in a specified amount of time is known as:

frequency The number of wave cycles that passes a specified point in space in a specified amount of time is known as frequency.

What was a severe limitation of Bohr's model for the atom?

it only worked for one-electron systems Even when applied to the next smallest atom, a helium atom with two electrons, Bohr's model failed. It was still based on the classical mechanics notion of precise circular orbits, rather than the quantum mechanical principles that would soon follow.

That a photon of a particular energy must be emitted or absorbed when an electron in a hydrogen atom changes orbits is a ramification of the:

law of conservation of energy The electron will change its energy as it changes orbits, and thereby must emit a quantum of energy, or a photon, that is equivalent to this difference in orbital energy.

Which of the following have the last electron added in an s or p orbital?

main group elements The main group elements have the last electron added in an s or p orbital. This category includes all the nonmetallic elements, as well as many metals and the metalloids. The valence electrons for main group elements are those with the highest n level.

The more electrons an atom has the more __________________ it will need.

orbitals An orbital is a region of probability where an electron can be found. There are s, p, d, and f orbitals, and each one has a different shape. The orbitals have different shapes to show the regions in space where electrons can be found. Each orbital can hold 2 electrons, the more electrons an atom has, the more orbitals it will need to accommodate them all.

Predict what the electron configuration for Chlorine will end in (again, there is no superscript option so assume the ^ indicates superscripts)

p^5 yes, it's in group 17 (also called group VII or 7A)

For elements in the d block, the first electron removed will be a(n):

s electron Even though d electrons were the last to be added for an element in the d block, and the d orbitals are slightly higher in energy than the s orbital immediately preceding, these elements will lose their outermost s electrons first.

The electron configuration for potassium would end in what? (Note: this software does not allow for superscripts so please assume the carrot, ^, means superscript)

s^1

If we say 2p6, the 2 corresponds to:

the energy level The 2 is the n value for those orbitals, which tells us the energy level of the electrons in that subshell.

Bohr used an electron moving in a circular orbit about a nucleus to describe:

the model of a hydrogen atom Bohr used an electron moving in a circular orbit about a nucleus to describe the model of a hydrogen atom.

An especially large increase in ionization energy occurs when __________.

the next electron to be removed is a core electron If the last electron removed is the last of a valence shell, it has suddenly dropped in size significantly and has also attained noble gas electron configuration, and so the next electron to be removed is a core electron rather than a valence electron. Therefore, removing another electron will be particularly energetically unfavorable.

Which of the following results in an increase in effective nuclear charge per electron?

the removal of electrons from the valence shell The removal of electrons from the valence shell leads to an increase in effective nuclear charge per electron because there are fewer electrons to counter the positive charges of the protons.

Generally, on the periodic table, ionization energy increases __________.

up and right Atomic radius and ionization energy follow opposite trends. Atomic radius increases down and left, while ionization energy increases up and right. As we move to the right on the periodic table, we add protons, which contracts the radius and holds electrons more tightly so they are harder to remove.

According to Bohr's model, when does an electron emit electromagnetic radiation?

when it is changing from an orbit of higher energy to a lower one An electron emits electromagnetic radiation when it is changing from one orbit to another, specifically from an orbit of higher energy to a lower one.

Using the Bohr model, determine the energy of an electron with n=8n=8 in a hydrogen atom.

−3.405×10^−20 J We can use the equation below to calculate energy. E=−kZ^2/n^2 Where - k is a constant equal to 2.179×10^−18 J - Z is the number of protons in the nucleus of the atom, this is 1 because it is a hydrogen atom - n is the energy level of the electron, this is 8 because it is given in the question Now we can plug our values into the equation: E=−kZ^2/n^2 =−2.179×10^−18 × 1 / 8^2 = -3.405 x 10^-20 J