How can you tell which element is the largest by looking at their electron configuration?
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You need to know more than just the electron configuration.  There is contraction from left to right on the periodic table, plus additional periodic trends that must be considered.   \n\nEven more importantly, you need to define whether you are looking for the largest element IN ITS NATURAL  STATE (i.e. Cl2, Hg metal, etc) or AS THE ION.   Atomic and ionic radii differ.\n\nMore clarity is needed here.
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The position of the elements in the periodic table gives you clues about the valences of each element. Also, if you remember the formulae for familiar substances (like carbon dioxide, water, sulfur dioxide, hydrochloric acid, etc) that will also give you clues about possible or likely valences. \n\nThe final answer is:\n\nmemorize.
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determine how many valence electrons each atom has. locate the central atom and spread out its valence electrons(each electron is represented by a dot). then pair up one from the atoms attached to the central atom. finally put all the other electrons for each outer atom around it.\nyou should see all electrons paired up, many times each atom will end up with eight electrons around it but many times there will be more or less than eight.
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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.
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The atomic radii for Cr, Mn, Fe, Co, Ni, and Cu are\n\nCr: 125 pm\nMn: 137 pm\nFe: 124 pm\nCo: 125 pm\nNi: 125 pm\nCu: 128 pm\n\nSo, you can see, there is a bit difference between their sizes. Although among them, Cr, Co, Ni, and Cu do have very similar atomic radius. \n\nThe reason that these elements have similar radius is due to their electron configurations. So, let's look at their corresponding electron configuration first (hope you know what I am referring to) :\n\nCr: [Ar] 3d5 4s1\nMn: [Ar] 3d5 4s2\nFe: [Ar] 3d6 4s2\nCo: [Ar] 3d7 4s2\nNi: [Ar] 3d8 4s2\nCu: [Ar] 3d10 4s1\n\nAs you may realize, the major difference between their electron configuration is mainly in the "3d" orbital, which has a quantum number (n) of 3; this 3d orbital has a smaller radius than the next orbital, 4s. The reason why 4s orbital is filled BEFORE 3d is due to the lower energy level of 4s compared to 3d.\n\nIn short, the major reason why these elements have similar radii is because the newly added electrons are not added to the "outermost electron shell", instead, they are added to the "inner electron shell". Therefore, no SIGNIFICANT changes to their radius.
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A major research institution has recently announced the discovery of the heaviest element yet known to science. This new element has been tentatively named "Administratium." Administratium has 1 neutron, 12 assistant neutrons, 75 deputy neutrons, and 111 assistant deputy neutrons, giving it an atomic mass of 312.\n\nLightest would be Hydrogen.\n\nelements consist of atoms which have protons, neutrons and electrons.
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the s is the orbital in which 2 electrons can occupy\nthe orbitals are the s which can hold 2 electrons, the p which can hold 6 electrons, the d which can hold up to 10 electrons and so on \nhydrogen has an electron configuration of 1s1\nhelium has its elec. conf. of 1s2\nLithium has its elec. conf. of 1s2 2s1\nkeep adding \nNeon has its elec. conf of 1s2 2s2 2p6 \n\nTHOSe two 2s u add the elctrons which are 2 & 6 and u get get which makes the octect of NEON (noble conf.)
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As you go down, the electrons occupy energy levels whose energy is greater. That is, they are farther away from the nucleus. For example an element in period 6 has its outtermost electrons in the n=6 energy level. These electrons have a greater potential energy than those in the n=2 level because they are farther away from the nucleus. This is similar to an object that is higher up: It has a greater gravitational potential energy. The size of the atom increases because as more electrons are present in an atom, the more energy levels they fill. The outtermost occupied energy level of an atom is what determines size.
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Atomic radius (or size) increases as you go down and to the left on the periodic table.  Therefore, out of these, the largest would be K.
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You need to look at the radius of the atom, and the number of protons in the nucleus.\n\nGroups I and 2 in the Periodic Table have large atoms, with good screening and low IE values.  The transition metal atoms are smaller, and the outer electrons are therefore harder to remove.
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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.
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B. electron\n\nis the right answer.
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K, Na, Li, Be ... its already arranged from largest to smallest atomic radius. \n\nFor Potassium, Sodium and Lithium, they belong to the same group, so size increases from top to bottom. Lithium being on the top of the other is the smallest. \n\nBeryllium is on the right of Lithium, so it has one proton extra than Lithium but the number of shell is the same so its nucleus pulls the outer electronic shell with more force, hence its atomic radius is even smaller than Lithium.
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The more at the top of the table, the more electronegative the element is.  So it's I Br Cl F
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Take radium and subtract two from its atomic number. Then find the element with that result as its atomic number. I'm too lazy to see what that is right now.
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you need a periodic table... just look for the symbol NA... and the atomic number 15, also look for it on the table.. it shud be under the 2-3 letter symbol of the element.
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There are various means of determining this, and without any specific data on what elements are involved in the answer you are looking for, the link below is the best reference I can find that can answer this question in a mostly generalized way.