# Tag Info

25

It's a good question, and I agree that it can be quite confusing, but Chemguide is (at least partly*) correct about this. I'd phrase it this way, which I hope is clearer. First, think about what would happen if the "paired electron" factor was not a factor at all. We would then expect that, going from B to Ne, there would be a roughly constant ...

14

Of course you can take all the electrons off an atom - it is then called "fully stripped" in atomic physics. You don't need to do it to an entire mole, mind you. In accelerators one would send energetic neon ions through a background gas or a thin foil, and the interactions will result in various charge states coming out, up to and including fully stripped.

12

The easiest way to explain it is that $\ce{Al}$ has one unpaired electron in it's highest energy orbital ($\mathrm{3p}$), and $\ce{Mg}$'s highest energy orbital ($\mathrm{3s}$) the electrons are paired. It is energetically favorable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy. Here's ...

11

In terms of Bohr model ionization potential $E_\mathrm{i}$ is the work $A_\mathrm{i}$ on eliminating an electron in vacuum from its current non-excited orbital level to infinity: $$E_\mathrm{i} = \frac{A_\mathrm{i}}{e}$$ $$A_\mathrm{i} = h\nu = \frac{hc}{\lambda}$$ Unknown wavelength $\lambda$ can be determined from the Rydberg formula: $$\frac{1}{\lambda}... 10 We can use Bohr's atomic model to calculate the ionization energy of \ce{He+}. The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives IE_2=E_{\infty}-E_1 where E_1 is the energy in the first bohr-orbit (principal quantum number=1). ... 9 In fact, it is not restricted at all. You may compare the energies needed to remove one electron out of a solid; these are also meaningful, albeit in a different way, and known for a wide range of substances. But when you are talking about atoms, you want to measure atoms, and the only way to have an undisturbed lone atom is to put it in a gaseous state. And ... 9 I would love to find numbers, but alas, I have failed. If anyone can point me towards energy differences between the orbitals I am lacking, please feel free! There is no a priori physical reason why lithium, sodium, potassium and calcium give flame colours but beryllium and magnesium would not. For all these elements (and hydrogen), the principal mechanism ... 9 Oxygen has a lower first ionization energy as the electron that is removed is coming from a paired orbital. Electrons within the same orbital experience maximum repulsion as the distribution of their wavefunctions is the same, so the probability density distribution is the same and the electrons can be thought of as occupying the same space. This maximizes ... 8 Taking half the total value of both ionization energies isn't a bad initial approach. You were just one step shy of finding the answer (without calculations). Each additional ionization energy will be greater than the last because the potential will be greater; the fewer the electrons, the less electron-electron repulsive force. Thus, the first ionization ... 8 First - in chemistry there's technically no such thing as bare multivalent cation, second - as you think, there's no such energetic chemical reaction, third - ionisation energy is physical property (although important for chemistry) and ..."is usually measured in an electric discharge tube in which a fast-moving electron generated by an electric current ... 8 The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion. There are certain factors that have an impact on the ionization energy value. To understand the answer to your question, you must understand the ... 7 To reach the second ionization potential we remove a second electron from an already ionized element$$\ce{M^{+1} -> M^{+2}}Here are the electron configurations for the singly charged ions in your question, along with their second ionization potentials (kJ/mol) \begin{aligned} \ce{& Li^{+1} & 1s^2~~ &~~ 7297}\\ \ce{& Be^{+1} & 1s^2~... 7 Sure, why not. It's just the question of your bombarding electrons having the right energy. At low energies, a bombarding electron will likely be captured by one of the iodine atoms: \ce{I2 + e- -> I- + I} \, . $$But somewhere around 8.62 \pm 0.06 \, \mathrm{eV} according to this study, you'll indeed have the ion pair formation:$$ \ce{I2 + e- -> ...

7

In an infinite metal solid, there are an infinite number of electrons, so each one removed will be at the exactly identical work function with no change in energy. Intuitively, this may seem a bit strange because .. well, it should be harder to remove an electron once positive charges start to accumulate. If I have even a micron-sized piece of metal, that'...

7

Different regions of the electromagnetic spectrum correspond to different atomic and molecular processes, each with one or more associated spectroscopies. Here is a general summary, with decreasing frequency/energy going from left to right: In order to excite a valence electron, the longest wavelength or lowest energy radiation is usually in the visible ...

7

When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as: $$\psi_{1s} = \frac{2}{a_o^{3/2}} e^{-r/a_0} \tag{1}$$ the variable $r$ is the distance from the ...

7

This is an excellent question! Just to elaborate upon what @porphyrn and @M. Farooq have said, imagine that the universe is entirely empty except for one sodium atom and one photon (particle of light) that has a wavelength of $\pu{242 nm}$. If the photon bumps into the sodium atom, it might just bounce off: this is called scattering. But another big ...

6

As mentioned by Zhe, we have to look at the entire process by which an ionic compound is formed, not just the energy for a single part. For example, consider the formation reaction of $\ce{MgO}$ $$\ce{Mg(s) +1/2O2(g)->MgO}$$ To find the enthalpy of this process, we could follow a Born-Haber cycle: This demonstrates that while the ionization of $\ce{Mg}$ ...

6

Your book must be wrong. NIST gives these values (in eV) as 11.260300 for C and 12.96763 for Cl. Even not reaching that website, I would say that a halogen would have a higher ionization potential than a group 14 element of a previous period just because there are no reason for the latter to not have a higher value, as for instance N has (half filled energy ...

6

The second ionization of chlorine leaves $[\text{Ne}]3s^23p^3$ where the $3p$ subshell is half-filled with one electron per orbital and gives a favorable exchange energy. With sulfur this relatively favorable configuration is achieved instead after the first ionization, the second ionization of sulfur breaks the half-subshell and is thus less favorable. ...

6

Metal chlorides are usually much easier volatilized than oxides or sulfates. They melt or sublime at temperatures much lower than 800°C. So they are volatilized in any flame like Bunsen burners. Platinum is not attacked by hot chlorides or acids, and is not oxidized in hot air. So it can be used for a great number of consecutive tests in the flame. It will ...

6

What you have calculated is the energy required to make two separated ions out of two separated $\ce{Na}$ ad $\ce{Cl}$ atoms. When they are separated, these ions do not make a molecule or a crystal of salt. Separated ions attract one another and are able to release a huge amount of Coulomb energy when approaching each other. Anyhow your calculation should ...

5

The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} \rightarrow [Xe] 4f^{7} 5d^{1}$. Well actually, the second ionization energy being plotted in the graph is only the energy to lose the second electron so the transition is $[Xe] 4f^{7} 5d^{1} 6s^1 \rightarrow [Xe] 4f^{7} 5d^{1}$. For all the other elements Pr through Yb, there is no 5d electron, ...

5

It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects. In general, as you go across a row, the shielding effect is fairly small compared to the effect of increasing nuclear charge. As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) ...

5

Energy of electron in vacuum is zero. If it can attach (however weakly), the energy is gained and process is therefore exotermic. It is not an atom which releases energy, it is the whole system. Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined). In some ...

5

I think the most widely-used approach at the moment comes from Roi Baer and Leeor Kronik and others, e.g. Phys. Rev. Lett. (2010) 105, art. 266802 Phys. Rev. B (2011) 84, art. 075144 Basically the idea is to tune the range-separation parameter $\gamma$ between short-range and long-range electrostatic effects in a range-separated hybrid functional. You find ...

5

The ionization energy (IE) of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state. With ionization energy, an electron is not "kicked out" by other electrons, but rather it is "the energy required for the electron to 'climb out' and leave the atom." Since ...

5

The reason is the same as in the case of barium and radium, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no ...

5

I would like to clarify the matter since from OP's own answer I have a feeling he/she does not quite understand it. So, first of all, in general the ionisation energy has to be calculated as the difference in total energy and not just the electronic one. Secondly, for molecules there are two types ionization energies: adiabatic and vertical: The ...

5

It is because the shell is complete. Zinc's electronic configuration is $\rm [Ar] 3d^{10} 4s^2$, which is a complete configuration for a transition metal. The next element in the same period is $\ce{Ga}$, which starts filling $\rm 4p$ orbitals ($\rm [Ar] 3d^{10} 4s^2 4p^1$). The zinc's electronic configuration causes a sort of special stability because of ...

Only top voted, non community-wiki answers of a minimum length are eligible