Periodic Table Electron Configuration
- 1 Why is scandium 2 8 9 2?
- 2 Can a shell have 9 electrons?
- 3 Why is the S block 2 elements wide?
- 4 What is the 8 Hund’s rule?
- 5 Why does Nb have different electron configurations?
- 6 What is Kr 4d10 5s1?
- 7 Why does palladium have a different configuration?
- 8 What is Kr 5s1?
Why is NB 5s1 4d4?
Hint: The given two elements Nb and V belong to the transition metal series of the periodic table. This question can be answered based on the size of the valence orbitals, the difference in the energies between orbitals and the electron repulsions. Complete Step By Step Answer: First, let us know the energy difference between the valence orbitals in the respective elements Nb and V.
Vanadium belongs to the first transition series and Nb belongs to the second transitions metal series. The energy gap between 3d and 4s orbital in Vanadium is 2.79eV and that of 4d and 5s of Niobium is 1.64eV. Thus, we can say that the valence orbitals of Nb are very close to each other. Because of this closeness of the orbitals in Nb, electrons can easily jump from 4d to 5s.
As a result, we might expect that, based on the energy the fourth valence electron may enter into the 5s orbital instead of 4d during the filling of electrons, to reduce the electron repulsion in 4d of Nb. Based on the orbital size, we can say that the size of 4d is bigger than 3d and 5s is bigger than 4s.
- Because of the bigger orbital size, the electron repulsion can be tackled more efficiently if the electrons enter the 5s orbital than the 4d in Niobium.
- Hence we can conclude that having bigger orbitals reduces the electron repulsions and makes it more favorable to continue filling the d orbital with the fourth valence electron in Niobium.
Also, the 4d orbital being close to the 5s orbital makes it more favorable for the fourth electron to be placed in the s orbital than the d. Because of this reason niobium has a $ $ electron configuration but vanadium has a $ $ electron configuration.
What element has 1s 2 2s 2p 6 3s 2 3p 5 as its electron configuration?
Electronic Configurations of Cations and Anions – The way we designate electronic configurations for cations and anions is essentially similar to that for neutral atoms in their ground state. That is, we follow the three important rules: Aufbau Principle, Pauli-exclusion Principle, and Hund’s Rule.
The electronic configuration of cations is assigned by removing electrons first in the outermost p orbital, followed by the s orbital and finally the d orbitals (if any more electrons need to be removed). For instance, the ground state electronic configuration of calcium (Z=20) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2,
The calcium ion (Ca 2+ ), however, has two electrons less. Hence, the electron configuration for Ca 2+ is 1s 2 2s 2 2p 6 3s 2 3p 6, Since we need to take away two electrons, we first remove electrons from the outermost shell (n=4). In this case, all the 4p subshells are empty; hence, we start by removing from the s orbital, which is the 4s orbital.
The electron configuration for Ca 2+ is the same as that for Argon, which has 18 electrons. Hence, we can say that both are isoelectronic. The electronic configuration of anions is assigned by adding electrons according to Aufbau Principle. We add electrons to fill the outermost orbital that is occupied, and then add more electrons to the next higher orbital.
The neutral atom chlorine (Z=17), for instance has 17 electrons. Therefore, its ground state electronic configuration can be written as 1s 2 2s 2 2p 6 3s 2 3p 5, The chloride ion (Cl – ), on the other hand, has an additional electron for a total of 18 electrons.
- Following Aufbau Principle, the electron occupies the partially filled 3p subshell first, making the 3p orbital completely filled.
- The electronic configuration for Cl – can, therefore, be designated as 1s 2 2s 2 2p 6 3s 2 3p 6,
- Again, the electron configuration for the chloride ion is the same as that for Ca 2+ and Argon.
Hence, they are all isoelectronic to each other.
Why is scandium 2 8 9 2?
Answer: Scandium has an atomic number of 21. The outer most valence is two electrons instead of three, because the forth shell has lesser energy than the third, so the third shell add more electron become 9, so the electronic configuration of Scandium is 2,8,9,2.
What is the meaning of 1s 2s 2p 3s 3p?
1s 2s 2p 3s 3p are the electronic orbitals’ energy levels. One atom can have many electronic orbitals due to which energy levels are categorized as specific quantum numbers: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p. We can get the idea of the orbital angular momentum quantum number denoted by l.
Why is there no 4d3?
Re: Electron Configuration of Nb – Post by Andrea Sandri 3D » Fri Oct 09, 2015 10:39 am The chemical reason why the configuration of Nb is 4d4 5s1 rather than 4d3 5s2 has to do with the Aufbau principle failing to account for the effect of electron-electron and electron-nucleus interactions on the electron configurations of certain elements, in this case Nb.
Why is 4s2 filled before 3d6?
The Aufbau principle predicts that the 4s orbital is always filled before the 3d orbitals, but this is actually not true for most elements! From Sc on, the 3d orbitals are actually lower in energy than the 4s orbital, which means that electrons enter the 3d orbitals first.
What is n in 5d?
The value of the principal quantum number, n for 5d orbital is 5. The value of the azimuthal quantum number, l for 5d orbital is 2. The possible values of magnetic quantum number, are -2, -1, 0, +1, and +2. The possible values of spin quantum number, are or.
What element is Kr 5s2 4d10 5p3?
So, for an antimony atom, the noble gas configuration would be 5s2 4d10 5p3.
What is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6?
Kr (Krypton) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6. Rb (Rubidium)
Can a shell have 9 electrons?
Electron shell Principal energy levels in atomic physics This article is about the orbits of electrons. For valence shell, see, “Atomic shell” redirects here. For the weapon, see, In and, an electron shell may be thought of as an that follow around an ‘s,
The closest shell to the nucleus is called the “1 shell” (also called the “K shell”), followed by the “2 shell” (or “L shell”), then the “3 shell” (or “M shell”), and so on farther and farther from the nucleus. The shells correspond to the ( n = 1, 2, 3, 4,) or are labeled alphabetically with the letters used in (K, L, M,,).
A useful guide when understanding electron shells in atoms is to note that each row on the conventional of elements represents an electron shell. Each shell can contain only a fixed number of electrons: the first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on.
Why is the S block 2 elements wide?
The width of each orbital block is related to the maximum number of electrons that can be held by a particular orbital (i.e., s orbitals can hold up to two electrons, so the s block is two elements wide; p orbitals can hold up to six electrons, so the p block is six elements wide, etc.).
Why does fluorine need 8 electrons?
Fluorine (symbol F) is found in column 7 on the periodic table. It has 7 valence electrons. It needs to make 1 bond to get an octet. The simple logic is that 7 + 1 = 8.
Why is 5D filled before 4f?
Arun 25757 Points 4 years ago There are a number of series that rule how the electrons are configured in the orbitals. The Aufbau principle (theoretical model): 1s The Rydberg rule (from experimental/spectroscopic data): 1s The Aufbau principle, the one that you are referring to as the rule for filling orbitals is taught as a scientific law in high-school but in reality there are a lot of exceptions on this rule.
Even in high-school you might have come up to the exceptions of period 3 – the first row that includes transition metals- namely Cr (4s1 3d5) and Cu (4s1 3d10) where the 3d orbital is filled – or half-filled- prior to the 4s orbital. The logically inconsistent idea here is that in a transition metal atom, 4s is occupied before 3d but 4s is also easier to ionise.
We tell students that the (n+1)s is of lower energy, thus more stable and for that reason is filled first and then we start making excuses to explain why the ns can loose one electron to the (n+1)d orbital later due to the increased stability that a full or half-full nd orbital provides which is not exactly correct.
Here is some missing background: 1)A distinction between Kohn-Sham density functional orbitals, the ones fulfilling a simple Aufbau rule and describing the lowest experimental configuration for energy averages and the canonical Fock orbitals, describing experimental vertical ionisations and obeying a more complex Aufbau rule known for high spin/low spin complexes is necessary.2)The order of orbital filling cannot be derived directly from orbital ionisation energies, neither experimentally nor theoretically because there is no general simple relation owning to possible orbital reorganisations.
A given kind of nd and (n+1)s energies can vary from d(x-2)s(2) to d(x-1)s(1) to d(x). So we expect electrons to configure accordingly to the lowest energy Kohn-Sham orbitals, but there is always the case of them occupying the corresponding canonical Fock orbitals that are not necessarily the lowest in energy. Khimraj 3007 Points 4 years ago Orbitals fill in order of energy. So 5D fills before 4F in some cases simply because the 5D energy levels are lower than the 4F levels for some. This does not occur for all the lanthanides. Cerium, where the increase in effective nuclear charge after Lanthanum is not sufficient to stabilize the 4F2 5D0 configuration compared to 4F1 5D1.
Why is 4f before 6s?
Here, (n+l) of 6s orbital is 6 and that of a 4f orbital is 7 and hence 4f orbital is filled before 6s orbital.
Why does 4s fill before 3d?
Summary – In each of these cases we have looked at, the 3d orbitals have the lowest energy, but as we add electrons, repulsion can push some of them out into the higher energy 4s level.
- If you build up the scandium atom from scratch, the last electrons to go in are the two 4s electrons. These are the electrons in the highest energy level, and so it is logical that they will be removed first when the scandium forms ions. And that’s what happens.
- The 4s electrons are also clearly the outermost electrons, and so will largely define the radius of the atom. The lower energy 3d orbitals are inside them, and will contribute to the screening. There is no longer any conflict between these properties and the order of orbital filling.
The difficulty with this approach is that you cannot use it to predict the structures of the rest of the elements in the transition series. In fact, what you have to do is to look at the actual electronic structure of a particular element and its ions, and then work out what must be happening in terms of the energy gap between the 3d and 4s orbitals and the repulsions between the electrons.
The common way of teaching this (based on the wrong order of filling of the 3d and 4s orbitals for transition metals) gives a method which lets you predict the electronic structure of an atom correctly most of the time. The better way of looking at it from a theoretical point of view no longer lets you do that.
You can get around this, of course. If you want to work out a structure, use the old method. But remember that it is based on a false idea, and do not try to use it for anything else – like working out which electrons will be lost first from a transition element, for example.
Why is potassium electron configuration 2.8 8.1 and not 2.8 9?
According to octet rule, the outermost shell of an atom can accommodate maximum 8 electrons (except K shell which can accommodate maximum 2 electrons). Hence, the electronic configuration of potassium is 2,8,8,1 and not 2,8,9.
What is the 8 octet rule?
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Page ID 37382 The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell, When atoms have fewer than eight electrons, they tend to react and form more stable compounds. When discussing the octet rule, we do not consider d or f electrons.
What is the 8 Hund’s rule?
Example \(\PageIndex \): Carbon and Oxygen – Consider the electron configuration for carbon atoms: 1s 2 2s 2 2p 2 : The two 2s electrons will occupy the same orbital, whereas the two 2p electrons will be in different orbital (and aligned the same direction) in accordance with Hund’s rule.
Consider also the electron configuration of oxygen. Oxygen has 8 electrons. The electron configuration can be written as 1s 2 2s 2 2p 4, To draw the orbital diagram, begin with the following observations: the first two electrons will pair up in the 1s orbital; the next two electrons will pair up in the 2s orbital.
That leaves 4 electrons, which must be placed in the 2p orbitals. According to Hund’s rule, all orbitals will be singly occupied before any is doubly occupied. Therefore, two p orbital get one electron and one will have two electrons. Hund’s rule also stipulates that all of the unpaired electrons must have the same spin.
Why does Nb have different electron configurations?
2. Why does niobium have an exceptional electronic configuration? – Niobium has an exceptional electronic configuration because it follows the Aufbau principle, which states that electrons fill orbitals in a specific order. In the case of niobium, the 4d orbital is filled before the 5s orbital,
What is Kr 4d10 5s1?
The valence configuration of Ag, or silver, is 4d10 5s1 because silver is a d-block element in the periodic table.
Why does palladium have a different configuration?
CHEMISTRY COMMUNITY by » Mon Oct 27, 2014 1:30 pm Not a typo.We know that Pd has 46 electrons by looking at the periodic table, and we know that “s” orbitals can have a maximum of 2 electrons, “p” can have 6, “d” can have 10, and “f” can have 14. So the full electron configuration for Palladium would start with:
1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 Then you get to the critical point of deciding whether to fill it in as (4d 8, 5s 2 ) or (4d 10 ).1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 Hope that helps!
It comes down to stability of the sub-shell and the orbital. We know that completely filling in the 4th shell would make the atom more stable than if we were to leave the 4th shell incomplete (with only 8 electrons) while starting the 5th (and completely filling the “s” orbital).
What is Kr 5s1?
Well if it’s a neutral element, the element is rubidium.