Write the electron configuration for the following ions. How many unpaired electrons does each contain?
D orbital has possible 5 pairs of electron. 10 electron. P 0rbital has 3 possible pairs of elctron. ( follow hund's rule to see how it is paired)./see hund's rule before reading anything below. 1) ur ion is ni+4 but u have config of ni+2. I think u meant Ni+2. Let's go to ni+4 first, When u remove 4 electron from ni, it is ni+4. It will have electronic configuration of Cr.[ar] 4s2 3d4 or more stable 4s 3d5. 5 unpaired for this. Now, Let's go to ni+2.But when u remove two electrons it has [ar]4s2 3d6, electronic config of fe(iron).. Then u'r elec config is right for ni+2. As we know d orbital can fit 5 pairs of electrons(10electron) ur d has 6 electrons so. 4 upaired electron. 2) si got two extra in ur question so it has ele config of sulfur. [Ne]3s2 3p4 . As p can hold 3 pairs of elctron(6 electron) . 2 are unpaired. 2) 2)
Electron configuration?
16 electrons: 1s2, 2s2, 2p6, 3s2, 3px2, 3py1, 3pz1 The noble gas that comes before sulfur is Neon, with 10 electrons. The first part of sulfur's configuration is identical to Neon's (1s2, 2s2, 2p6) so you can say sulfur is "everything that Neon is plus......) [Ne]3s2, 3px2, 3py1, 3pz1 (Hund's rule requires the electrons in the 3p's to be distributed evenly throughout the three oribitals).
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!
How is the ground state electron configuration of Br determined?
Just use the Aufbau principle. Bromine has an atomic number of 35, so that's 35 electrons. Start writing it out...1s2(remember the 2 is an exponent/superscript), 2s2, 2p6, so on...You'll end up with 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 or [Ar] 4s2 3d10 4p5 , whichever you choose, although notation is preferred. Hope this helps!
Electron Configuration?
The orbitals that electrons fill up in order are as follows: 1s; 2s; 2p; 3s; 3p; 4s; 3d; 4p; 5s; 4d; 5p; 6s; 4f; 5d; 6p; 7s; 5f; 6d; 7p S-orbitals hold up to 2 electrons. P-orbitals hold up to 6 electrons. D-orbitals hold up to 10 electrons. F-orbitals hold up to 14 electrons. First lets start with N(Nitrogen) - the first thing you have to do is find out how many electrons the element has. to do that, you simply look at the Periodic Table and use its Atomic #. Since all elements have the same # of Protons as electrons, Nitrogen has 7 electrons in its orbitals. To calculate the electron configuration, start filling up the orbitals in order: [Things in parenthesis () are superscripts] N = 1s(2) 2s(2) 2p(3) Fluorine has 9 electrons: F = 1s(2) 2s(2) 2p(5) Oxygen has 8 electrons: O = 1s(2) 2s(2) 2p(4) Neon has 10 electrons: Ne = 1s(2) 2s(2) 2p(6) When you get to the large elements, instead of writing out very long configurations, you can just put the previous Nobel Gas in brackets and use its electron configuration For Example: Magnesium has 12 electrons: Mg = [Ne] 3s(2) As for the charge after the N, F, and O, those are just the oxidation numbers and are not used to write the electron configuration. If you were to write the electron configuration for an element in its oxidation state, it would be the same as the next Nobel Gas in the period(row) if it has a negative oxidation #.If an element has a positive oxidation #, the electron configuration is the same as the Nobel Gas one period up. To find the electron configuartion for the transition metals requires a slightly different method, but I think I've answered this question.
Electron configurations?
Think of it as the first two columns are the "s" configurations, columns 3A-8A are p, and the metals are d's. The number infront of the letter is the row, and the number after the letter is the column. Sometimes to write the configurations shorter, you can use the final column element and then follow it with the rest of the comfiguration. Depending on if the element has a + or - charge, you add or subtract them from the orbital (second number) so it moves you forward or backward elements Example) for Al, instead of putting 1s2 2s2 2p6 3s2 3p you can put [Ne] 3s2 3p but since it is a 3+ charge, then there are 3 electrons missing so you go back 3 elements and end up on the Ne element so I believe the configuration for Al3+ is just [Ne] Same for the others Mg = [Ne] 3s2 but Mg2+ = [Ne] K = [Ar] 4s but K+ = [Ar] Al = [Ne] 3s2 3p but Al3+ = [Ne] P = [Ne] 3s2 3p3 but P3- = [Ar] S = [Ne] 3s2 3p4 but S2- = [Ar] N= [He] 2s2 2p3 but N3- = [Ne] Cl = [Ne] 3s2 3p5 but Cl- = [Ar] I hope that helps!
Write the electron configuration for the element sulfur?
When you do an electron config., the first thing to do is look up the element's atomic number on the periodic table, then just fill out the energy level formula until you reach that number of electrons. The formula is like this: 1s 2s 2p 3s 3p 4s...and so on Each shell can hold a maximum # of electrons: s can hold 2, p can hold 6, for example. So sulfur (atomic # =16) would look like this: S(16)=1s2 2s2 2p6 3s2 3p4 (The electrons 2+2+6+2+4= 16)
How do you read an electron configuration table?
How do you read an electron configuration table?1The strict aufbau filling order from Pauli and others is:That is the first step.You get:1s2, 2s2, 2p6, 3s2, 3p4 (or I often write 3p4of6) - where the critical issue is the last subshell is not completea) that 4 of 6 leaves two (2) slots for valance, receiving bonding.b) That also shifts where and how bonding (double vs single) occursThis covers 90% of electron configurations. Figure out the last subshells, and if not complete.2People shortcut starting from the full noble gas to avoid repeating all the inner electron subshells. So, the below is the same as the above full configuration.[Ne] 3s2, 3p43The ions tend towards the full shells nearest.4However, in the last few years, there are found significant variations from that aufbau filling order, particular in the transition metals (probably the start of the greatest challenge and opportunity in chemistry today is explaining these variances). Those changes impact a) spectrum, electrical resistance, bonding, magnetism, and other properties. So, quite important.So, you can gethttps://web.chem.ucsb.edu/~devri...Copper per strict aufbau - [Ar] 4s2 3d9 of 10Copper per current model - [Ar] 4s1 3d10 of 10[Note that my model actually has a intermediate, transitional subshell that is even more different, 4s2, 4p6, 3eq3, but that is still not reached peer-reviewed approval.]5You get filling in Pauli ‘pairs’ where you fill the up or down pair before any 2nd pair fills.Again, combinations like Carbon are up, up, up, down so it can bond triple, but not quadruple. [I think of the 3rd in the 2nd hemisphere Pauli ‘spin’ pair which in 3D cannot bond because it is on the other side.]
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