What is the electron affinity of arsine? (Arsine is AsH3.)?
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).
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
763345
Let's say you want to know what is the formula of Aluminum oxide.\n\nYou know that oxides always have a negative 2 charge and Aluminum is always a positive 3 charge.  So if you have 2 positive 3 charges that would be a positive 6, and if you had 3 negative 2 charges that would be a negative 6 and that would allow the charges to cancel out to a zero.\n\nSo the formula would be Al2O3 for aluminum oxide.\n\nIn other words, the criss cross method of determining the formula uses the charge of the opposite ion to determine how many atoms to use.  It criss crosses.
843792
Be very careful:\nFirst, look at the electronegativity differences, the presence of groups like OH- or cations (like Na+ of soap)\nThen, you must know molecular geometries, it is highly important:\nFor BF3 this is a  trigonal planar compound, all differences of electronegativity B<F are canceled due to the position of the F atoms (every 120º on the same plane). It is not polar.\n[SnCl3]- this is a trigonal pyramid compound.\nCl\n I\n:Sn-Cl\n I\nCl\nAS you can see, Sn has a lone pair of electrons, making Cl atoms accommodate in the base corners of an imaginary pyramid. Even if Cl cancel each others dipolar momentum on the sides, they all point ↓. This is a polar compound.
927706
Al2(SO4)3 ==> Al2O3 + 3SO3
165799
You want us to calculate acitivites?\n\nYou need to determine the ionic strength and atomic radius in order to do that.
855087
Ca2+=[Ar]\nMg2+=[Ne]\nI-=[Xe]\nF-=[Ne]
775039
For the groups on the table, the ion charges are\n\n          1       +1\n          2       +2\n         13      +3\n         14    +4/-4\n         15       -3\n         16       -2\n         17       -1\n\nThe ion charge equals the number of electrons gained (-) or \nlost (+) by the atom to form the ion.
739878
I am not sure what you are asking but the answer could be a monoatomic ion. The positive being a cation and the negative being an anion.\n\nYou might ask over at http://forums.mychemistrytutor.com though.
729388
effective nuclear charge:\nZ* = Z - S\nZ is the number of protons in the nucleus and S is the average number of electrons between the nucleus and the electron in question.\n\nIf we are looking at the 3s Obital, the "S" will be the number of electrons in the orbitals in the 1s,2s,2p orbitals. *this number will remain constant as the elements mentioned are all below Neon.\n\nS = 2+2+6 = 10\n\nZ*\nNa = 11-10 = +1\nCa = 20 - 10 = +10\nS = 16-10 = +6\nNi = 26 - 10 = +16\nMo = 42-10 = +32\n\nso ranking from lowest > highest\nNa, S, Ca, Ni, Mo.\n\nthis is based on the 3s orbital, not on the valence electrons.
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.
873575
Reactivity...Nitrogen requires 3 electron to fill its outer shell and oxygen only two.  10 points please.
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
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.
948794
protons is the atomic number\nvalence electrons corresponds to this:\nGroup 1 (I) (alkali metals) :1\nGroup 2 (II) (alkaline earth metals) :2\nGroups 3-12 (transition metals) :1 or 2\nGroup 13 (III) (boron group) :3\nGroup 14 (IV) (carbon group) :4\nGroup 15 (V) (nitrogen group) :5\nGroup 17 (VII) (halogens) :7\nGroup 18 (VIII or 0) (noble gases) :8\n\n\nEvery element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. It goes down the periodic table like that. At this time, the maximum number of electron orbitals or electron shells for any element is seven.
362157
Atomic radii decrease as one moves from left to right across a period and while moving up a group. The atomic radius is the distance between the nucleus and the valence electrons.\n\nElectron affinity, ionization energy, and electronegativity increase as one moves from left to right across a period and while moving up a group. \n\nElectron affinity is the ease with which an atom can accept an electron. The halogens (F, Cl, I, etc.) have the highest electron affinity, with F having the highest of all elements. The alkaline metals (Li, Na, K, etc.) have the lowest electron affinity, with Fr having the lowest of all elements.\n\nIonization energy is the energy needed to remove an electron from an atom. All atoms want to form a stable octet. Since alkaline metals only have one valence electron, it takes very little energy to remove an electron from them. In contrast, it takes a lot of energy to remove an electron from a halogen.\n\nElectronegativity is the strength of the attraction an atom has for an electron in a chemical bond. Halogens are the most electronegative, while the alkaline earths are the least. The electronegativity of two elements can inform you of what type of chemical bond they will form. If Na (low EN) and Cl (high EN) form a bond, Cl will rip the lone valence electron away from Na, forming an ionic bond. If C and O form a bond, the atoms will share electrons because their EN's are very close. Since O is more EN, it will have a slight negative charge, and the bond will be a polar covalent bond. If C and C form a bond, both have the same EN, and they will form a non polar covalent bond.
933205
electron affinity\n\nas you are changing Cl2 ->2Cl-