Write the condensed electron configurations for Se,Rh,Si,Hg,Hf?
Se~[Ar] 3d10 4s2 4p4 Rh~[Kr] 4d8 5s1 Si~[Ne] 3s2 3p2 Hg~[Xe] 4f14 5d10 6s2 Hf~[Xe] 4f14 5d2 6s2
Can you tell me why the electron configuration for a lead atom is?
OK, take a look at a periodic table and find Xe. Notice that Pb is above this and so should have all the electrons in the same places that Xe has and some more. Now from Xe if we keep going up the table, we get to the 6th period which starts off with the 6s orbital where we can have 2 electrons. Going further we get into the f orbitals which are two steps back because they are "backfilling" into the principal quantum number equals 4 slots (sorry if this dosen't make sense, but the f orbitals go two steps back on the period number and the d orbitals go one step back on the period number). The 4f orbital can hold 14 electrons and is then full. We then get to the 5d orbitals (one step back from 6) where we can have 10 electrons and then finally to the 6p orbital where we can have two electrons. The configuration this way reads: [Xe](6s^2)(4f^14)(5d^10)(6p^2) Which is the same thing you have Good Luck
Write the condensed electron configurations for the following atoms.?
Ga=>[Ar]4s^23d^104p^1 with unpaired e-=1 Ca=>[Ar]4s^2 with unpaired e-=0 V=> [Ar]4s^23d^3 with unpaired e-=3 I=>[Kr]5s^24d^105p^5 with unpaired e-=1 Y=>[Kr]5s^24d^1 =1 Pt=>[Xe]6s^14f^145d^9 =2 Lu=>[Xe]6s^24f^145d^1 =1
What is the condensed electron configuration for lead?
Electron config for 82 Pb lead : [Xe] 4f14 5d10 6s2 6p2
Electron configuration of Pb?
[Xe] 4f^14 5d^10 6s^2 6p^2 (px(↑)py(↑)pz(0) Hund's Rule) Some books (and instructors) give it as [Xe] 4f^14 6s^2 5d^10 6p^2 but that is incorrect. Let's see if Wikipedia has it correct http://en.wikipedia.org/wiki/Lead Yes! RH panel P. Atkins et al, SHRIVER & ATKINS Inorganic Chemistry 4th ed. p 772
Write the condensed ground-state electron configurations of these transition metal ions?
This Site Might Help You. RE: Write the condensed ground-state electron configurations of these transition metal ions? Hf2+ and Co3+ please explain
What is the electron configuration of Zirconium?
It is [Kr] 5s2 4d2. Quora isn't the best place for electron config questions, I answer these every week please just use google.
What is the electron configuration of a cobalt 3+ ion? Is it [Ar] 4s1 3d5 or [Ar] 3d6?
It's [Ar]3d6. The Aufbau principle, which according to Wikipedia, states "...electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible."This also indicates that when electrons are stripped they will be stripped from the highest energy levels as well. Therefore, Co would be [Ar]3d74s2, while Co3+ would be [Ar]3d6. Hope this answers your question and if not, I refer you to the following page on Purdue's website on Transition Metals
What is the electronic configuration of cr2+ and O2-?
Hello I read the question, thank you for the A2A. As shown in the picture below:Chromium, Cr, has an atomic number of 24. Which indicates it’s unique number of protons. Therefore, a neutral Chromium would have 24 protons, and 24 electrons as well. When that is the case, Chromium’s electron configuration would be:[math]1s_2 2s_2 2p_6 3s_2 3p_6 3d_5 4s_1[/math]Which is a half filled d-orbital and a half filled s-orbital.Now considering Cr ion in Chromium(II) Oxide is [math]^{+2}[/math] charge ion, two of Cr’s 24 electrons will be lost. Which yields Chromium [math]^{+2}[/math], leaves Chromium with 22 electrons.NOTE: There are a few cases of some tricky metals that actually do not follow the generally rule, and electrons will jump to a higher energy orbital and complete the empty higher energy orbital, which shrinks the overall radi of the electron cloud.But in-order for the ion to have the lowest possible energy, which will in turn contract the radius of the overall electron cloud, I believe the 4s orbital will lose an electron, and the 3d orbital will lose an electron. Therefore Cr+2 will look something like this[math]1s_2 2s_2 2p_6 3s_2 3p_6 3d_4[/math]For normal rules this makes since right? s-orbital electron leaves first, followed then by another electron from 3d.But what if you saw something like this:[math]1s_2 2s_2 2p_6 3s_2 3p_0 3d_{10} [/math]HUH!!! :-/ ,,, :-(DEFINITELY NOT GONNA HAPPEN!NO WAY 6 ELECTRONS ARE GONNA JUMP p-orbital to d-orbital! REQUIRES WAY TO MUCH ENGERY FOR EXCITATION.BUT,,,, think about this for the XOR[math]1s_2 2s_2 2p_6 3s_2 3p_6 3d_2 4s_2[/math][Ar] [math]3d_2 4s_2[/math]HUH!!! Cr[math]^{+2}[/math] same electron configuration as Ti? Nice and stable looking huh? That would be 3 electrons leave the d-orbital, two leave the ion overall. :-)For Oxygen to be at it’s most stable noble gas position it would gain two electrons.O + 2e[math]^-[/math] = O[math]^{-2}[/math] . Therefore Oxygen -2 would have 10 electrons total,[math]1s_2 2s_2 2p_6[/math]I hope this helps!
What is the electron configuration of zinc?
Zinc,( chemical symbol Zn,) has an atomic number of 30, which means it has 30 protons in its nucleus. In any neutral atom, the number of electrons is equal to the number of protons. Therefore we can deduce that zinc has a total of 30 electrons. The electron configuration of any atom is represented according to its position on the periodic table, labelled by the period(ties in the periodic table) first, followed the sub atomic orbitals s, p, d, and f, and the number of electrons in. Each “s”orbital can hold a maximum of 2 electrons; each “p” There are 3 “p” orbitals with 2 electrons each, resulting in the total numbers 1–6. There are 5 “d” orbitals totalling 10, and 7 “f” orbitals totalling 14 electrons. For example:Fluorine (F), atomic # 9, is located in the “p” block, period 2, so its electron confoguration would be 1s2 2s2 2p5; if you add up the exponents (2+2+5=9), the answer will be equal to the atomic number.Zinc (Zn), atomic # 30, is located in the “d” block in period 4; it’s electron configuration would be represented as 1s2 2s2 2p6 3s2 3p6 4s2 3d10; adding up the exponents (2+2+6+2+6+2+10=30) gives us the atomic number, 30.Electron configurations can also be represented in a condensed form. In this case the valence electrons are represented with only the outer electrons in the configuration format preceded by the prior inert gas in square brackets. So, for fluorine it would be [He] 2s2 2p6, with the Helium symbol (He) representing all of the previous electron configuration.For zinc, the condensed electron configuration would be [Ar] 5s2 3d10Hope this hay this is helpful.