how many electrons could fit in the n = 3 level?
145018
It does not just have to do with Bohr's quantum number (n). It is also to do with the angular momentum quantum number(L) and electron spin (+1/2 or -1/2). Therefore, if n=1, L can only = 0 and the result is 2 electons. One with a spin of +1/2 and the other with a spin of -1/2. \nIf n=3, L can be either 0,1, or 2. \nMeaning, \nif n=3 and L=0 this results in 2 electrons on the sub-shell. \nIf n=3 and L=1 this results in 6 electrons on the sub-shell. \nIf n=3 and L=2 this results in 10 electrons on the sub-shell.\n\nThis results in a maximum of 10+6+2 = 18 electrons in the shell.\n\nThis is because if L=2, the maximum number of possible electron states on a sub-shell of n=3 are as follows:\n\nGiven, m = -L, -(L-1)..0..(L-1), L\n\nm(magnetic quantum number) can have a value of -2, -1, 0, 1, 2 with the electrons having a spin of +1/2 or -1/2 resulting in a maximum of 10 electons in that sub-shell.\n\nThis is all based on the Pauli Exclusion Principle which states no two fermions (electrons) can exist on the same sub-shell with the same spin and magnetic quantum number(m).
656882
a) No unpaired electrons - [Ar] 4s2 3d10\n\nb) Three unpaired electrons - [He] 2s2 2p3, by Aufbau principle the three 2p electrons are in three different orbitals.\n\nc) One unpaired electron - [Ne] 3s2 3p5\n\nd) Two unpaired electrons - [Ar] 4s2 3d8
454087
Use the formula:\n\nmagnetic moment = sqrt(n(n+2)\nwhere n = number of unpaired electrons.\n\nEg Mn has 5 d electrons, therefore, Mn2+ could have a maximum of 3 unpaired electrons. Plugging 3 into the above formula gives:\n sqrt(3(3+2) = sqrt(15) = 3.87 BM.
493143
Considering the conflicting answers, perhaps I should add something:\n\na sodium *atom* carries 11 electrons and has a neutral charge, therefore having a configuration of 2:8:1 as one person wrote -- 2 electrons in the first (ground) level, 8 electrons at the second level, and 1 (valence electron) in the third level.  As a configuration written 1s2 2s2 2p6 3s1 -- 2 electrons in the first (ground) level, 8 electrons total at the second level (2 in s, 6 in p), and 1 electron in the third level.  These are the two possible ways of writing the electron configuration -- remembering of course the superscript that we don't have here!  \n\nThe other answers that I saw were describing sodium as having 10 electrons.  These with 10 electrons are not sodium atoms but sodium *ions.*  If sodium becomes ionized it loses an electron and becomes positively charged -- in that case losing the valence electron in the third level.
756384
Indeed:\nGroup 13 (3 valence electrons)\nPeriod 3 (3 energy levels)\n\nGet an "A"
917414
You apparently know what the diagonal rule looks like, but the link below gives a quite nice picture of it.\n\nBasically it is a summary of the order in which electron shells will be filled up. Follow the arrows round to discover where the electrons will go, starting at the 1s level.\n\nSo for 15:\n2 electrons will go into 1s, filling that up\n1s2\nthe next electrons will go into 2s2, 2p6, 3s2 and then the remaining 3 electrons into the 3p orbital, giving us a structure of:\n1s2 2s2 2p6 3s2 3p3\nRemeber, on your diagram, the first two characters (2p, 4d etc) are the names of the orbital, and the final digit is the number of electrons in that orbital. On your diagram the final digit shows the maximum number of electrons on the orbital; when you write out the structure the first few orbitals will be full (1s2 2s2 2p6 3s2) and the last one will be part full of however many electrons you have left (3p3.)
949019
The number of electrons in any shell/orbit is 2n^2 (2 )(n square), where "n" is the number of the orbit.\nSo in the second in the second shell, it is 2(2)(2)= 8 electrons.\n2 electrons in the "s" orbital, and 6 in the "p" orbital
275083
Write down what you know about the complex:\n\n1) it's tetrahedral (sq planar)  --> ∆t  (t2 and e levels)\n2) CN   --> low spin\n3) d electrons of Ni(2+)  --> [Ar],d8\n\nBefore drawing the CFSE, state that ∆ is large for a square planar complex.  Since ∆ is large, electrons will always pair up, rather than occupying the dx²-y² level.  Since there will be no unpaired electrons, the complex will always be diamagnetic.  I cant draw the diagram here, but show the z²,x²-y² and xz,yz,xy levels, fill the diagram with 8electrons and pair them.
774337
An electron has the greatest (ionisation) energy in the energy level n = 1.\nAn electron cannot occur between energy levels, unless it is moving from one to the other.
722082
s,p,d and f are designations for the four different types of orbitals.\n\ns is spherical in shape and has only one orientation in three dimensional space.\n\np has a dumbbell shape and has three orientations in 3-d space:\nPx,Py and Pz.\n\nd has a double dumbbell shape and has 5 orientations in 3-d space.\n\nf has a complex shape and has 7 orientations.\n\nEach orbital can hold 2 electrons. s having only 1 orientation can hold only 2 electrons. p having 3 orientations, has 3 orbitals, each having the capacity to hold 2 electrons, so it can totally hold (3*2=) 6 electrons.\nSimilarly d can hold (5*2=) 10 electrons and f can hold (7*2=) 14 electrons.
191934
Two in the first shell. 10 in the second shell. 10 in the third shell. 5 in the 4th shell for neutral Manganese. Mn 2+ is missing 2 electrons, so it would only have 3 in the outer shell. The rest will stay the same.\n\nHere's a cool interactive periodic table that may help you with your homework. \n\nhttp://www.webelements.com/\n\nYou can click on the the element you are interested in, then in the column on the left you can go to electronic properties and it will show you how the electrons are distributed. You can add or remove electrons to configure the electron arrangement of various ions of the substance you are interested in.
970362
Yes.  All p shells will have a total of 3 atomic orbitals.\n\nhttp://en.wikipedia.org/wiki/Electron_configuration#Worked_example
832259
N cannot expand its outer level to more than 8 electrons, because it is in Period 2.  NF5 would require there to be 10 electrons in the outer level of N.
714831
the atomic number is 5 b/c atomic number equals the number of elctrons and protons. You have 5 electrons so it should be 5.
937724
Mg has 12 electrons\nwe will fill them from lower orbitals to higher\none orbital needs 2 electrons to fill in\n1s : one orbital, 12 electrons: 2e\n2s : one orbital, 10 electrons : 2e\n2p : three orbitals, 8 electrons : 6e\n3s : one orbitals, 2 electrons : 2e\n\nSeveral rules govern the placement of electrons in orbitals (electron configuration). The first dictates that no two electrons in an atom may have the same set of values of quantum numbers (this is the Pauli exclusion principle). These quantum numbers include the three that define orbitals, as well as (the hitherto unmentioned) s. Thus, two electrons may occupy a single orbital, so long as they have different values of s. However, only two electrons, because of their spin, can be associated with each orbital.\n\nAdditionally, an electron always tries to occupy the lowest possible energy state. It is possible for it to occupy any orbital so long as it does not violate the Pauli exclusion principle, but if lower-energy orbitals are available, this condition is unstable. The electron will eventually lose energy (by releasing a photon) and drop into the lower orbital. Thus, electrons fill orbitals in the order specified by the energy sequence given above.\n\nSo you will have to fill the electrons that have the same spin\nevery two electrons of an orbital have different spin but two electrons of two orbital can have the same spin\n\nyou can use a rectangle to represent for an orbital, an arrow to stand for an electron,( up or down is the spin of that electron )\ngood luck
850007
The electron configuration of Sc3+ is [Ne] 3s2 3p6\nThe value of "n" is the energy level, which is 3 here.\nThe value of "l" is the orbital type. For a "p" orbital, l = 1. \nSo, the statement is true.\n\nChromium has an unusual electron configuration.\nIt is actually 1s2 2s2 2p6 3s2 3p6 4s1 3d5.\nIt seems that the atom is more stable when it has all of the orbitals in a group filled the same way, ie. half-full.\n\nGround state means that the electrons are in the lowest energy levels available. In an excited state, energy has been added to the atom, and the electrons are (temporarily) at higher levels. Which ones depends on how much energy you put into the atom.  Too much, and you remove the electron entirely.\nSo, it's not really predictable without the energy values.
802906
they fill the s orbitals with 2 electrons