I understand that the electron configuration for GOLD (Au) is [Xe] 4f14 5d10 6s1, can someone explain please?
365879
see [Xe] means the element xenon . before 4f14 5d10 6s1 the electronic configuration is same as that of xenon.it is just for the sake of conveniance that it is written like this.\nnow if u know the EC of xenon just add 4f14 5d10 6s1(i hope that u understand what that means.)after the EC of xenon and u get the EC of gold.\n4f14 5d10 6s1 maens that in gold,the 4th f orbital has 14 electrons 5th  d orbital has 10 electrons in it and the 6ths orbital has 1 electron.well i tried my best to tell u \nfor further queries e mail me at koolguyrik@yahoo.co.in
706975
5.56 x 10^(21) = 5,560,000,000,000,000,000,000 conduction electrons.\n\nYou can calculate this two ways if you use the Wikipedia references of 5.90 x 10^(22) free electrons per cm^3, 19.3 grams per cm^3 for the density of solid gold or 196.966569 grams per mol, and 6.0221415 x 10^(23) for Avogadro's number.\n\nIn any case, the first step is to calculate the volume = (pi) x R^(2) x L, where pi = 3.1416, R = 1 mm = 0.1 cm, and L = 30.0 cm. Do the math: V = 9.425 x 10^(-2) cm^3.\n\nIf you multiply the calculated volume by 5.90 x 10^(22) free electrons per cm^3 you arrive at 5.56 x 10^(21) free (conduction) electrons.\n\nOr, as a check, calculate the mass of this volume: 9.425 x 10^(-2) cm^3 x 19.3 grams per cm^3 = 1.819 gm. Divide this mass by 196.97 grams per mol to obtain 0.009235 mols of gold and then multiply by Avogadro's number to obtain 5.56 x 10^(21) gold atoms. Each gold atom has just one valence electron to act as a conduction electron, so the number of gold atoms is also equal to the number of conduction electrons.
766178
The electron configuration of Si is:\n\n[Ne] 3s2 3p2\n\nSi2+ is Si with 2 less electrons so the electron configuration is:\n\n[Ne] 3s2\n\nThe 2 electrons are taken from the p orbital as it requires less energy to remove 2 electrons.
758277
the amount of electrons in an atom.\n\nif you're going to use noble gas configuration, you would use the noble gas and then continue the configurations.\n\nex. i want the configuration of... flourine (F)\nfor noble gas configuration i would say: (He) 2s 2 2p 6\n\nthe parenthasese is actually a bracket, and the numbers that are spaced should be in the upper righthand corner of the letter before it.
778586
It's in the 6th row, so n= 6.\nIt's in the first column of the "p" block of the table, so it's highest energy electron is 6p1\n\nStarting from Xe, and following the atomic numbers, these are the orbitals that are filled :\n\n[Xe] 6s2 4f14 5d10 6p1\n\nThe valence electrons are the ones with the highest quantum numbers: 6s2 6p1.
443303
It only has 7 valence electrons :)\n\nRival
869755
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
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Au - 6s2 4f14 5d9\nNd - 6s2 4f3 5d1\nCl - 3s2 3p5\nununillium - 7s2 5f14 6d9\nZr - 5s2 4d2
16898
In fact the gold has two main valences 1 and 3 but due to his noble character it doenst react with other atoms so probably the concept of valence is not so clear as in more reactive elements.
245727
Xe, element # 54 I believe
939139
Li would expect to have the following electron configuration:\n\n1s2 2s1\n\nIn this case, the free (valance) electron is in an "excited" state.  Some energy had caused the electron to jump up to the 3p orbital.  Typical excited states are sigma to sigma* transitions for these electrons, this one has transitioned in a "sigma to pi*" manner.
224420
how far it is from the edge of the periodic table. i.e., Francium has a charge of -1 because to get it to the noble gases it has to gain an electron. on the other hand, lithium would need to lose an electron to make it happy, giving it a charge of +1. Same concept for all elements.exclude scandium's period thru zinc's, as they are very different. Carbon's group is able to go either way, as it is equidistant from the noble gases. email me if you have any more questions.
981192
http://periodic.lanl.gov/default.htm\n\nClick on the element, and it's in the top section of information. To write out the electron configuration for every element would take quite a while, so if you need it for a specific element or a few elements, I'd reccomend asking just for the few.
263276
You mean the electron configuration?\n\n1s2\n2s2 2p6\n3s2 3p4
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THE CORRECT BOILING POINT OF AU OR GOLD IS 2807.0 DEGREE CELCIUS ( 3080.15K,5084.6 DEGREE F)
1021964
usually the valence in an ion leaves a noble gas config (ngc) so if the it is a cation, it looses electrons to have a ngc and for an anion, it gains electrons to have a ngc.
148525
Do you mean what where are the d sublevels? If you do they are the transition metals.  But i do not think that is your question. Do you mean the fourth row? That would be the latest last noble gas configuration then s2 of next level up then d2 of same level of last noble gas. I hope it helped. Maybe you should clarify your question.