Which of the following elements has the highest fourth ionization energy?
814112
For Al .. \n\nElectron configuration : [Ne] 3s2 3p1 ... so after losing 3 electrons its stable and need high energy to destabilize it ..\n\nwhile for As [Ar] 3d10 4s2 4p3 .. \nfor Ga [Ar] 3d10 4s2 4p1\nfor Se [Ar] 3d10 4s2 4p4\nfor S [Ne] 3s2 3p4\n\n4th ionization energy in kJ/mol :\n\nAl  11577 \nGa 6180 \nAs 4837 \nS 4556\nSe 4144
935698
nonmetals have much higher ionization energies than metals do. this is because non-metals have more protons in their nucleuses to attract their electrons with. metals do not have as many, as compared to the rows they are in, so the electrons will move away from the nucleus of the metal if given a little push.
811889
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..
955362
Violet12.. is right on both questions. Now, about the second question:\n\nFor the 4s orbital: n + l = 4 + 0 = 4.\nFor the 3d orbital: n + l = 3 + 2 = 5\n\n4 < 5, so the 4s orbital has less energy than 3d, so we put electrons first in 4s orbital and then in 3d.
637738
You must be after the points because it is so easy to Google this, or visit Wikipedia.\n\nBasically, the n-th ionization energy is the energy required to remove the n-th electron (starting from the outside working in) to a distance of infinity from the nucleus after the previous n-1 electron have been removed.  For example, using sodium Na, it has 1 electron in the outer ring, or valence electron.  The first energy will be needed to bump that guy to infinity.  The next electron will take more energy because it is close to the nucleus.\n\nHope it helps.
848192
Ti(4+) = 18 electrons.\nSo the atom, which forms a stable compound without forming an ion, is first of all an inert gas and has 18 protons and electrons. that's argon.\n\nHope this helps=)
43823
Carbon, with 4 electrons in its outer shell is the most versatile element. It can form single, double, and triple bonds with a wide variety of elements, creating just about every living thing we see. Carbon is known to form at least 10 million different compounds. Elements, such as Nitrogen, with 5 electrons in the outer shell, and Boron with 3, are much more limited.
756384
Indeed:\nGroup 13 (3 valence electrons)\nPeriod 3 (3 energy levels)\n\nGet an "A"
328894
A potential answer would be the Positron (the electron's antiparticle)\n\nFor a "normal" 4H->1He fusion cycle it really depends, there are several paths that it can take, but for the sake of argument let's look at the one most common in the sun.  The excess mass that 4 hydrogen nuclei have over 1 helium nuclei is converted into energy.  Some of that energy is carried off in gamma rays - ready to use light energy, that will then interact and heat the surroundings.\n\nSome of that mass-energy is given off in positrons.  These are the anti-particle to the electron.  These particles will quickly meet up with an electron, and mutually anhilate relesing energy equal to their combined mass.  Some of the energy is carried of by neutrinos.  They don't anhilate and thus leave the system quite rapidly.
562475
The electron in 4p orbital is farther than 3p orbital because the electron in 4p belongs to the 4th energy level while the electron in 3p orbital belongs to 3rd energy level. Use Aufbau's principle to answer questions similar to this.
926360
Four.\n\nThe first one is usually called "n" and it gives what people think of when they say "energy level".  It's also called the "principal quantum number", I think.\n\nThe next one is usually called "l" (el) and for a given energy level, it can be anything from 0 up to "n-1".  These correspond to what people sometimes call "shells", such as the s, p, d, f, etc. shell.\n\nThe next one is called "m" and is sometimes called the "sub-shell".  Each shell can have up to "2*l+1" sub-shells.  So an "s shell" (l=0) can have 1 sub-shell.  A "p shell" (l=1) can have 2*1+1=3 sub-shells, etc.\n\nThe final quantum number is called "s" or "spin".  For each sub-shell, this can take on one of two values, +1/2 or -1/2  in some books.  This means that only two electrons are allowed in any given sub-shell.\n\nSo for n=1, the only option is l = 0, which means further that m must be 1, with 2 choices for s.  This all means that the first energy level can only hold 2 electrons.\n\nFor n=2, l can be either 0 or 1.  For l =0, m must again be 1, allowing 2 electrons.  For l = 1, m can be as much as 3, each allowed to have 2 electrons.  So adding the 2 electrons for the l=0 subshell to the 6 allowed in the 3 possible subshells for l=1 (6 electrons we get a total of 8 electrons allowed in the second energy level.\n\nYou can continue on to get totals of 18 (2 + 6 + 10) for the 3rd energy level (n=3; l =0, 1, or 2;  and m = 1, 3 and 5, each with 2 electrons allowed).\n\nThis is where the numbers that correspond to "filled energy levels" come from in the periodic table.
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.
176528
http://www.chem1.com/CQ/ionbunk.html\n\nskim this article.
36855
This element is Barium, Ba.  It has 56 electrons.\n1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2
977750
Mn(+3) has the electronic configuration of 1s(2) 2s(2) 2p(6) 3s(2) 3o(6) 3d(2)\n\nthe one before me missed the 4s2 so his is wrong
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.
452173
the short answer is valence shell electrons.