Explain why the second ionization energy is higher than the first, the third higher than the second, and so on
822048
in the first ionization only need enough energy to seperate the electron from the nuclear charge (Z*). For second ionazation energy   you need enough energy to overcome Z* plus the positive charge on the cation. third ionazation you have to overcome Z* plus the even higher charge on the cation and this continues on.
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.
479233
The energy of the incoming photons is given by Planck's relationship:  energy = Planck's constant * frequency.\n\nYou are given the wavelength, not the frequency, so you need to know that for any wave, the wavespeed = frequency * wavelength, so frequency = wavespeed/wavelength.  For photons, wavespeed = c.\n\nh = 6.626068 × 10^-34 34 m^2 kg / s (Planck's constant)\nc = 2.998 × 10^8 m / s (speed of light)\nwavelength = 48.2 x 10^-9 m (given in problem)\n\nPutting this together, the energy of each incoming photon is:\n\nE_photon = h*c/(48.2 x 10^-9 m) = 4.121 x 10^-18 J\n\nThe outgoing electron has kinetic energy = 1/2 * m * v^2.  You are given both m and v:\n\nE_electron = 1/2 * 9.109 x 10^-31 kg * (2.371 x 10^6 m/s)^2 \nE_electron = 2.56 x 10^-18 J\n\nThe ionization energy is the difference in energy between the incoming photon and the kinetic energy of the outgoing electron.  Because the question asks for the energy in joules per mole, we must multiply the energy difference for a single photon/electron pair by Avagadro's number:\n\nE_ionization = (E_photon - E_electron)* N_A\n\nE_ionization = (4.121 x 10^-18 J -2.56 x 10^-18 J)*6.022 x 10^23 mol^-1\n\nE_ionization = 9.4 x 10^5 J/mol = 940 kJ/mol
895120
The correct answer is 3.\nThe 1st spectral line is caused by the electron falling from state 2 to state1.  The 2nd line is caused by a shift from 3 to 2, and the third, a shift from 3 to 1.
817827
energy levels/orbitals
430403
i presume you have the energy values for the different levels?\n\nif so you use the equation E=hc/lamda to find the wavelength for each energy transition in the series.\n\nThe red line will be about 575nm and the purple line will be about 375nm.
627895
When an electron goes from lower energy orbital to higher energy orbital it absorbs energy and when an electron drops from a higher energy orbital to a lower energy orbitaltehn it emits energy.\n\n(a) Emission (read the reason above)\n(b) Absorption (Read reason above)\n(c) Emission (read reason above)\n(d) Absorption (read reason above)
590559
(a) Across the row (called the Period), the magnitude of the 1st IE increases. At the stage where we label an element as a non-metal, this value is too high for the outermost electrons to be lost, so instead of Oxygen with a charge of 6+, it actually gains electrons to obtain the stable structure; i.e. oxygen with a charge of 2-.\n\n(b) Down the group, the size of the atom increases due to an increase in the number of shells. Due to this the outermost electrons (-ve) are further away from the nucleus (+ve) and so the 1st IE decreases.
175956
C, remember that the electron loses and gets energy several times through the chain.
926360
Four.\n\nThe first one is usually called "n" and it gives what people think of when they say "energy level".  It's also called the "principal quantum number", I think.\n\nThe next one is usually called "l" (el) and for a given energy level, it can be anything from 0 up to "n-1".  These correspond to what people sometimes call "shells", such as the s, p, d, f, etc. shell.\n\nThe next one is called "m" and is sometimes called the "sub-shell".  Each shell can have up to "2*l+1" sub-shells.  So an "s shell" (l=0) can have 1 sub-shell.  A "p shell" (l=1) can have 2*1+1=3 sub-shells, etc.\n\nThe final quantum number is called "s" or "spin".  For each sub-shell, this can take on one of two values, +1/2 or -1/2  in some books.  This means that only two electrons are allowed in any given sub-shell.\n\nSo for n=1, the only option is l = 0, which means further that m must be 1, with 2 choices for s.  This all means that the first energy level can only hold 2 electrons.\n\nFor n=2, l can be either 0 or 1.  For l =0, m must again be 1, allowing 2 electrons.  For l = 1, m can be as much as 3, each allowed to have 2 electrons.  So adding the 2 electrons for the l=0 subshell to the 6 allowed in the 3 possible subshells for l=1 (6 electrons we get a total of 8 electrons allowed in the second energy level.\n\nYou can continue on to get totals of 18 (2 + 6 + 10) for the 3rd energy level (n=3; l =0, 1, or 2;  and m = 1, 3 and 5, each with 2 electrons allowed).\n\nThis is where the numbers that correspond to "filled energy levels" come from in the periodic table.
901447
Note that we're not taking energy differences into account.  In cases where the difference in orbital energy is greater than 15eV you get negligible overlap.  Now onto the symmetry-allowed overlaps.\n\n1s and 2s are both spherical and overlap to form a sigma bond.\n\n1s and 2p(x) form a sigma bond.  You can tell because if you draw the 2p(x) lobe you'll see it overlap with the corresponding 1s.\n\nThe 2p(x) and the 2p(y) do not overlap.  There is no possible bond.  Draw the 2p(x) extending toward the other atom as positive or shaded (remember: positive does not denote charge).  The other atom has its 2p(y) pointing up and down (perpendicular) and there is equal positive and negative overlap, thus canceling both of them out.  No net overlap.\n\n3p(y)-3p(y) form a pi bond.  Once again, draw these out and see it for yourself.  Draw the 3p(y) orbitals as rather fat so you can see how they overlap.  Since they are not cylindrically symmetric they are not a sigma bond.  Pretend you were to rotate one of the atoms about the x-axis... you would change orientation and there'd be less overlap.  For the sigma bonds above, rotation does not affect the overlap.\n\n2p(x)-2p(x) have positive overlap.  Draw them and see why.  They point toward one another.  Moreover, if you rotate one of the 2p(x) orbitals about the x-axis you keep the same structure/orientation since the 2p(x) is along the x-axis.  Thus, you can tell that this gives a sigma bond.
562475
The electron in 4p orbital is farther than 3p orbital because the electron in 4p belongs to the 4th energy level while the electron in 3p orbital belongs to 3rd energy level. Use Aufbau's principle to answer questions similar to this.
682608
The hydrogen atom attracts more electrons than the oxygen atoms do. It's uneven.
919041
We're basically looking for a ratio of ionization energy E(n), where n =4 and n = 1 (ground state). Electron energy levels of a hydrogen atom is given by:\n\nE(n) = 13.6/n^2 eV\n\nTherefore,\n\nE(4)/E(1) =  1/16\n\nIt takes only 1/16 of 13.6 eV , or 0.85 eV, to ionize the ground state electron from a hydrogen atom on this very hot planet.
317931
There are two transitions in the energy levels of mercury that are close to each other in energy difference, therefore the wavelength of the light from these two transitions are very close. See also "fine structure".
967786
I'm not going to tell you but look use Saha's equation to find out the energy in ev of each level. Then you can work out the energy value of the jump. It's just a simple bit of maths. But you are wrong so have a think
745401
There are two rules to use:\n1) The lattice energy between ions of opposite charge is strongest when the ions are small.\n2) The lattice energies  increase rapidly as the charge on the ion becomes larger.\n\nThe first rule implies that  LE(PbF2) > LE(PbCl2), in absolute value.\nThe second implies that   LE(PbF4) > LE(PbF2) and  LE(Pb2O3) > LE (PbF4)\nTherefore we have  LE(Pb2O3) > LE (PbF4) > LE(PbF2) > LE(PbCl2)  still in absolute value.\n\nConclusion:  3c 2d 1a 2b