How do you find the electrn configuration of an Ion?
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for positive ions take the electron configuration for the neutral atom and subtract the charge from the end.\nfor anions take the electron configuration for the neutral atom and add that many electrons to the end.
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WL = h/(mv)\n\nh is the planck constant\nm is the mass of the elecron\nLook for them in you book.  Plug the numbers.
48721
The periodic table elements atomic number will tell the number of protons found in the atom.  The number electrons will be same as the number of protons.\n\nHowever, if the atom is an "ion", it could more or fewer electrons than there are protons.  These are refered to as anions and cations (cat-eye-ons) respectively.\n\nThe net electric charge of N+ or N- written as a superscript to the atom will reflect the number of additional or fewer electrons for the ion.
851721
complex ions - which then form ionic molecules
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The answer is A\n\nCoBr2 + Pb(NO3)2 ==> PbBr2  + Co(NO3)2\nPb(C2H3O2)2 + 2KBr ===> PbBr2 + 2KC2H3O2\n\nIn both cases, the net ionic equation is:\n\nPb+2  + 2 Br-  ====> PbBr2(s)
787571
When the charges are different, you give the electron configuration for the element that has that number of electrons. For example, Na+ is like Na, but it is missing an electron, so the configuration is simply [Ne]\n\nfor O2-, it has gained 2 electrons, so it is also [Ne]\n\nCo2+ has lost 2 electrons so it is [Ar] 4s2 3d5
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Forget the potassium ions - they don't do anything. The aluminium metal dissolves in the potassium hydroxide solution, forming aluminate ions and hydrogen gas.\n\nThe ionic equation is:\n\n2Al + 2OH- + 6H2O ---> 2[Al(OH)4]- + 3H2\n\nTo explain how to get this equation, you need the ideas of half equations, and oxidation and reduction.\n\nFor example, Al + 4OH- ---> [Al(OH)4]- + 3e-\n\nand 4H2O + 4e- ----> 2H2 + 4OH-\n\nNow combine them, getting the number of electrons equal, and cancelling down some hydroxide ions.\n\nIf it's any consolation, this exercise would be beyond most A-level pupils in the UK.
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Layer, orbital (azimuthal), magnetic and spin.\n\nNo two electons in the same atom can all the 4 numbers the same: be in the same layer, in the same orbital, sharing the same magnetic and spin value.
706819
this is double replacement, group a cation wit ha different anion. Then, using solubility rules, find the precipitate, this is your solid. Write the rest as ions.
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Ruthenium has the configuration [Kr] 4d7 5s1. Since the ion is a 3+, that means it loses three electrons. When losing electrons, the electrons are lost from the shell that is furthest from the nucleus first. It'll therefore lose electrons from the 5s shell first. The final configuration means that it loses three electrons, one from the 5s orbital, and the other two from the 4d orbital. The configuration will come up to be [Kr] 4d5
138200
This is based on the electons in the outer shell of the atom you are drawing. From the periodic table you can see how many electrons are in the outer shell e.g. Na has 1, Cl has 7, to create a complete shell (8) they combine to form NaCl. \n\nIf you wanted to draw all of the electrons then you would draw the first shell with 2, which completes the first row of the periodic table, then the second with 8, then the third with 1 (for Na). For Cl you would draw the first shell with 2, which completes the first row of the periodic table, then the second with 8, then the third with 7. \n\nIf you wanted to show the bond between them then you would draw the shells with the final circles combining and two electrons touching the circles.\n\nI did a search and found the website below, there are probably better ones out there but this gives a pictorial representation of some of what I say
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You mean the electron configuration?\n\n1s2\n2s2 2p6\n3s2 3p4
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when an atom loses an electon from its outermost shell then it forms an electropositive ion while when an atom gains an electron it forms an electronegative elctron.\n\nfor e.g. during the formation of nacl na atom loses an elcron to the cl atom to form a +ve ionwhile the cl atom which gains that electon forms a -ve ion.
1049588
I think that you are trying to find the net ionic equation. \nWhich in this case would be \nAg+ (aq) + Br- arrow AgBr (solid)\nif you are trying to find a single replacement reaction?\nthen you need to put Br- instead of NO3- and cancel out the NO3 - on the product side? \nTried my Best.
69353
d)Argon: 1s2 2s2 2p6 3s2 3p6\ne)Could be carbon, silicon, germanium, tin, or lead.  It's not specific enough.  And there is no such thing as an "elemen." Please treat chemistry with enough respect just type a "t" at the end of a word.  \nCarbon: 1s2 2s2 2p2\nSilicon: 1s2 2s2 2p6 3s2 3p2\nGermanium:  1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2\nTin: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4d6 5s2 4d10 5p2 \nLead: \n1s2 2s2 2p6 3s2 3p6 4s2 3d10 4d6 5s2 4d10 5p6 6s2 4f14 5d10 6p2
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Positive ions (Cation) are formed by a neutral atom losing electron(s), and Negative ions (Anion) are formed by atom gaining electron(s).  \n\nThe exact number of electrons an atom gains or loses are predictable.  Atoms gain or lose electrons so that they turn isoelectric with the closest neighboring Noble Gas.
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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