Which atom will have higher ionization energy ?\n\nLi or O\n\nis it like down and right of the periodic table have higher I.E ?
132077
Oxygen will hv higher IE. IE increases on moving left to right in a period and decreases on moving top to bottom in a group. So elements with higher IE are in the top and right corner of the periodic table.
425294
No concrete way to work it out. Groups one and two take positive charges equal to their group number (Mg2+, Li1+ etc). Groups 3, 4, 5, 6 and 7 ionise to a negative charge of 8 minus their group number (Cl1-, O2- etc), though 3, 4, and 5 tend to resist ionisation and are much more commonly found bonded covalently. The Transition elements ion charges are virtually impossible to predict without an extensive knowledge of d orbitals etc..
538785
Huge order. Let's see how far it can be served.\n\n1. Monoatomic negative ions: take the element name (sometimes shortened) and add "-ide" (as in hydride, nitride, oxide, sulphide, phosphide, chloride...)\nPositive ions: element-name, charge (if several charges are possible), ion (sodium ion, iron II ion...)\n\n2. Polyatomic ions:\nComplexes with various ligands list the ligands and the central ion/atom.\nOxygen as ligands has its own nomenclature, depending on oxidation state. E.g. Sulphate, sulphite, perchlorate, chlorate, chlorite, hypochlorite...\n\n3. Binary molecular compounds: Halogen, oxygen and sulphur compounds usually have the correspondent "-ide" as if you had ionic compounds. Usually the more electropositive element is used as center.\n\n4. Acids: Oxygen acids have a similar nomenclature to oxygen ions.
112166
Well, the elements in column 1 when ionized have a +1 charge. The elements in column 2 have a +2 charge when ionized. The elements in column 7 when ionized have a -1 charge, and the column 6 elements form -2 charged ions.\n\nThe column 8 elements are called noble gases, because they have a full valence shell of electrons and don't bond readily with anything. They don't normally ionize.
17958
Atoms have what are called "orbitals" which are sorts of energy states that electrons can acheive. A certain set of orbitals are considered the ground state, which is to say that a neutral atom (meaning the atom has as many electrons as protons) will be at its lowest energy state when all electrons occupy the ground state orbitals. However, an electron can acquire extra energy which will raise the electron to a higher orbital. This can be done by heat or light. When an electron is elevated to a higher orbital, it is unstable, and it will eventually drop. When it drops back down to groundd state, it will release a photon which has a specific amount of energy--the difference in enrgy between the higher and lower orital.\n\nEnergy in a photon relates to its wavelength, and wavelength relates to color. Hence, for a given type of atom, the transition between two orbitals (which is set by the nucleus involving physical chemistry matters that are way too complex to speel out here) will always require a certain amount of energy, and the relaxation of an electron between those states will always release a photon of a specific color.\n\nThe continuous spectrum will result only from what is called "black body radiation," which results from a substance being thermally heated to the point where it throws off light.  This is generally seen in solids, like the filament of a lightbulb (incandescent light).  Conjugation of orbitals in a conductive metal like that make essentially infinite combinations capable of producing any number of wavelengths of light.  By contrast, spectra result usually from atoms in a gas phase, where there are limits to the sorts of energy they can absorb and emit.
131318
A two-color three-photon ionization technique, using two pulsed dye lasers, was employed to explore the high-lying energy levels of U I. A number of new U I levels in the 34 290--35 120-cm/sup -1/ region was observed.^Uncertainty in the assignment of J values to some levels known from earlier studies was also reduced or removed.\n\nHUH? Is that what you want?\n\nEDIT\nAbstract. Two-colour resonance ionization spectroscopy was used for identification of 240 new levels of 238U in the energy range of 38 500 to 41 200 cm-1. A precision of +or-0.03 cm-1 (1 SD) for the energies was obtained by using laser-induced fluorescence signals of I2 for spectral calibration. The angular momentum quantum number, J, was uniquely assigned for 134 of these levels by using angular momentum selection rules. \n\nThats a little more specific\nAnd some more\n\nAbstract. Visible and infrared absorption measurements on the  ion in tetragonal zircon-type matrix  are reported and analysed in terms of the standard parametrization scheme. The observed 17 main peaks and a number of less intense lines have been assigned and fitted to most of the 32 allowed electric dipole transitions with the root mean square error equal to . The free-ion parameters obtained for the model Hamiltonian, , ,  and , as well as the corresponding crystal-field parameters, , , ,  and , agree fairly well with the initial theoretical estimations. The results are discussed in relation to the previous spectroscopic study on the scheelite-type matrix .
499586
Which energy?\n\nCheck out the Periodic Table below.  For Hydrogen, the Ionization Potential is listed as 13.598 volts.  The electron configuration (energy level of the outer most electron) is 1s^1 (one electron in the 1s orbital).\n\nThe rows are the Levels, the columns are the individual orbitals:\nI & II are the "s" orbitals (He is "out of place").\nIIIB to IIB (under Sc to Zn) are the "d" orbitals.\nIII to 0 (under B to Ne) are the "p" orbotals.\nThe Lanthanides & Actinides are the "f" orbitals.
15670
Oxygen atoms like to form two bonds with other atoms. They will be more stable that way. It is related to the outer shell of electrons(valence electrons). Individual oxygen atoms have 6 valence atoms and would prefer to have 8(it's called the octet rule!)
144252
Here is the answer.. Just remember periodi table...\nElements on left most side are electro positive and elements on right most are electro negative. (The last row is nobel gases..so leave it)\nNow Cl- belongs to right most side... its more electro-ve\nRb, Ca are on left side.... they are electro positive(They will give  electron with out much force)\nSi, S are in middle-- their properties are in between...\n\nThats it!
475956
Sulphur would be the biggest atom because it has the most protons, electrons, & neutrons.\n\nThose with a tendancy to gain electrons are further to the right of the periodic table. This inlcudes O, S and P (to a lesser extent).\nAl and B are further to the left on the periodic table and would have a tencancy to lose electrons.
342426
B. electron\n\nis the right answer.
446589
Iodine is larger than bromine which is larger than chlorine. Larger atoms have their valence electrons farther from the nucleus so those electrons are easier to move around. This is called polarizability. Because iodine is the most polarizable, it's molecules have stronger London dispersion forces holding the molecules together so it is a solid. Bromine is less polarizable and it's a liquid. Chlorine is least polarizable and it's a gas.
483740
You could work out that in I2O7, the iodine is in a +7 oxidation state, so it can't go any higher.\nSimilarly, in Cl- it is -1, and it can't go any lower.
378101
s p d f blocks tell which set of orbitals are the last to fill for those elements.\n\n+ or - depending on whether it will lose or gain an e-\n\ncations are smaller than atoms b/c they lose e-\nanions are larger than atoms b/c they gain e-\n\neach successive ionization energy gets higher becasue the #protons is more and more unbalanced by # e-\n\nhalogens have 7 valence e- and noble gases have 8.  by gaining one e- the halogens now have the e- configuration of a noble gas and are more stable.  gain 1 e- = 1- charge
179124
Atoms/elements that are higher atomic number on the periodic table are lower in entropy because they contain more bonded protons, neutrons, and electrons.  \n\nThus F2 has more entropy than Cl2, and so on.
279213
For this you should see the electronic configuration of those elements.\nThe metals with atomic number 21-30 are transition metals; they have outer 4s shell complete and the inner 3d orbital is 'in the process' of being filled up. They all have 2 outermost electrons in the 4s2 shell. \nIonization energy is the energy required in removing an outer most electron from a neutral atom. The transition metals and calcium have their outermost 2 electrons in 4s which means that their electrons experience the same force of attraction from the nucleus (being at the same distance from the nucleus) ; hence almost the same amount of energy will be required to overcome that attractive force to remove the electrons.
4533
Hi again,\n\n    Well you see, as you know electrons are added up in the 3d orbital in galium, right? So with the increase in number of electrons, the no of protons i.e atomic number also increases. So now the outermost p electron in Al is attracted by 13 protons. But that in galuim is attracted by 31 electrons. So the force of attraction is much more. Now in other atoms, these outer electron attraction is balanced by counter electron electron repulsion. But the added d electrons  cannot shield the outer p electrons much from the force of attraction of the highly charged nucleus. This is similar to the reason why the transition elements contract on increase in at no. Other examples are lanthanide and actinide contractions. If you still dont get it,\nmeet me on Yahoo Messenger. My id is jchemisch. I will talk to you about it. All the best. Hope i can help you if you need!!