Which of oxygen and Beryllium has the largest electron affinity?
610804
Oxygen has the largest electron affinity. Oxygen has a valency of 2- because the outer shell is deficient of 2 electrons, thus it would accept two electrons to become stable - to complete its octet. If the number of electrons on the outer shell is greater than the number required to fill the octet (8 electrons to do so), then there will be greater electronegativity and less electron shielding between the electrons of the outer shell and the electrons on the subsequent shells, thus the electrons are attracted to the nucleus. You will then find it will then need more electrons to fill the octet. Beryllium however, as small as the atom is, has a smaller electron affinity. With 2 electrons on the outer shell, there is repulsion between those and the electrons from the inner shells, thus less attraction between the valence electrons and the nucleus. It will have a tendency to lose the electrons, rather than to keep them and gain 6 more to fill the octet.
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
537794
Okay, to begin with; the molecular orbital theory is based on the "octet rule."  This means that, with exceptions of hydrogen and helium of course, an atom is most stable with 8 electrons (the octet) in its outer shell.  \n\nA diatomic molecule is a pair of identical atoms forming a small molecule.  The most common diatomic molecules are in the halogens; Cl, F, I, Br.  These atoms each lack one electron from a full octet, and will covalently bond in pairs to each gain a full octet.  Oxygen gas (O2) will bond in a double pair, sharing four electrons to gain an octet on each atom.  \n\nBeryllium contains 4 electrons in total.  The electronic configuration is 2:2 or if you use the other method is 1s2 2s2.  A full octet would be 2:8 or 2:0 to gain the most stability, and the Be2+ is actually in use in labs (2:0).  \n\nBecause it would need to gain electrons from or donate them to some other element with a greater number or greater need, beryllium can't form a diatomic molecule or a pair with another of itself.  \n\nIt is difficult to explain without pictures, but there are a few shell pictures available at Chemical Elements which will let you compare these elements.
763730
If you're in the US, 1 is true, otherwise not, eg in the UK.\nTransition metals are the most dense.\nFluorine, 1s2, 2s2, 2p5 is the most reactive halogen.\nFluoride ion is 1s2, 2s2, 2p6.
754664
Los Alamos national lab has a good periodic table/ elemental factoids.\n\nLanthanum is La.  It is the beginning of the "Lanthanides".
352406
The predicted adiabatic electron affinities, obtained at the DZP++ BHLYP level of theory, are 0.91 eV (AsH), 1.09 eV (AsH2), -1.03 eV (AsH3), 0.91 eV (AsH4) and -0.033 eV (AsH5). The adiabatic electron affinity at this level of theory for AsH5 is much smaller (0.37 eV) than the one obtained using other functionals. BHLYP vertical detachment energy for AsH2 (1.08 eV) is in good agreement with experiment, while other values range from -1.03 eV (AsH3) to 1.30 eV (AsH4).
813548
The answer is a) Krypton.\n\nThe bromide anion exists as Br-.\nWith the single negative charge, you know that it has an extra electron, therefore it's electron config is like the noble gas nearest to it on the periodic table: krypton.
752142
The element with the largest atomic radius is cesium.  Its electronic configuration is [Xe] 6s1.\n\nFrancium could be larger but I do not believe there is data on its atomic radius yet.
570696
It´s 10. Neon. \nIt has two energy levels, filled with 2 and 8 electrons respectively.
452173
the short answer is valence shell electrons.
788408
Oxidizing agents remove electrons from other atoms to complete a stable outer octet.  Therefore, oxygen and fluorine are oxidzers.  (Hydrogen and lithium, because they can lose electrons more easily than accept them, are reducing agents.)\nSo, of the two oxidizers which is better?  It will be the one that is more electronegative.  Electronegativity increases left to right on a row in the periodic table*.  F is therefore more electronegative than O, so the winner is F.\n'Chemistry is full of exceptions and this rule does not hold for the noble family (VIIIa) because their electron structures are fully satisfied.
887107
B.the electron subshell
712728
choose the noble gas that comes before the element.   in Mn(2+)'s case, that's Ar\n\nwrite the noble gas down in brackets, and continue on with the electron config. as normal\n\n[Ar}4s23d3\n\nFor O(2-), it turns into Neon so it would just be [Ne]
714217
The metal which is the furtherest to the right-hand top corner of the periodic table will be the most electronegative. Those who said fluorine didn't read the question properly because you are asked for a metal and fluorine, although the most electronegative 'element', is not a metal. In answering this question, I'm going to assume that the metalloids aren't included and not a subset of the cagetory of metals. The most electronegative metal would be aluminium (13). Have a look at the periodic table, it's the element which is closest to the top-right hand side but is still classified as a metal.
986606
O has a Pauling electronegativity of 3.5, while Se has one of 2.4.  Thus, the oxygen atom is more electronegative, and so the shared electron pair is "more closely drawn" to O.\n\nIn regard to valence electrons for neutral Se, there are 6, because Se is group 6A.
928933
The electron affinity is a measure of the energy change when an electron is added to a neutral atom to form a negative ion. For example, when a neutral chlorine atom in the gaseous form picks up an electron to form a Cl- ion, it releases an energy of 349 kJ/mol or 3.6 eV/atom. It is said to have an electron affinity of -349 kJ/mol and this large number indicates that it forms a stable negative ion. Small numbers indicate that a less stable negative ion is formed. Groups VIA and VIIA in the periodic table have the largest electron affinities.\n\n* Alkali earth elements (Group IIA) and noble gases (Group VIIIA) do not form stable negative ions. \n\nElectron Affinities in kJ/mole\nPeriod IA  IIIA  IVA  VA  VIA  VIIA  \nPeriod 1 H  ...  ...  ...  ...  ...  \n-73  ...  ...  ...  ...  ...  \nPeriod 2 Li  B  C  N  O  F  \n-60  -27  -122  0  -141  -328  \nPeriod 3 Na  Al  Si  P  S  Cl  \n-53  -44  -134  -72  -200  -349  \nPeriod 4 K  Ga  Ge  As  Se  Br  \n-48  -30  -120  -77  -195  -325  \nPeriod 5 Rb  In  Sn  Sb  Te  I  \n-47  -30  -121  -101  -190  -295  \nPeriod 6 Cs  Tl  Pb  Bi  Po  At  \n-45  -30  -110  -110  -180  -270  \n\n\n\nElectron Affinities in Electron Volts\n1 kJ/mol = .010364 eV/atom\nPeriod IA  IIIA  IVA  VA  VIA  VIIA  \nPeriod 1 H  ...  ...  ...  ...  ...  \n-0.76  ...  ...  ...  ...  ...  \nPeriod 2 Li  B  C  N  O  F  \n-0.62  -0.28  -1.27  0  -1.43  -3.41  \nPeriod 3 Na  Al  Si  P  S  Cl  \n-0.55  -0.46  -1.39  -0.75  -2.08  -3.62  \nPeriod 4 K  Ga  Ge  As  Se  Br  \n-0.50  -0.31  -1.25  -0.80  -2.02  -3.37  \nPeriod 5 Rb  In  Sn  Sb  Te  I  \n-0.49  -0.31  -1.26  -1.05  -1.97  -3.06  \nPeriod 6 Cs  Tl  Pb  Bi  Po  At  \n-0.47  -0.31  -1.14  -1.14  -1.87  -2.80  \n\nPeriodic Table  Chemical bonds
45574
a) Sr, Mg, Be \nRationale: they are all in the same group (II), but because Sr has the most energy levels, it has greater shielding and lower effective nuclear force, so it is easiest for it to give up an electron.  Be has the lowest number of energy levels, so it has less shielding and greater effective nuclear force, so it clings to its electrons more, requiring greater force to take away an electron.  \nb) Cs, Ba, Bi\nRationale: They are all in the same period, so shielding and effective nuclear force are not variables.  Instead, we look at stability.  Cs is in group one, meaning that it only needs to lose one electron to reach a stable octet.  The electron configuration of Bi gives exactly 1 electron per p orbital (try writing it out), so it's somewhat stable already; taking away an electron would mess with that stability.  Ba is somewhere in between.  \nc) Na, Al, S\nRationale: group I elements always have lower ionization energies in the same period because they only have to lose one electron, so sodium goes first.  Next is aluminum, which only has to lose one electron to be totally rid of electrons in its p orbitals.  Last is sulfur, which will lose one electron to have one electron in each of its p orbitals, which is less stable than the other two.