It increases.
It becomes a negatively charged ion.
it S2-, because when you add electrons, you are making the atomic radius bigger. In S2-, you add 2 electron, making it bigger. hope that helped.
NaCl Add water. Na + Cl - Electrons have flowed and that is all electricity is, electron flow.
The element, oxygen, is in group 16, period 2 of the periodic table. Thus its electron configuration is 1s2 2s2 2p4. Oxide's ionic state is O2-, so to get its electron configuration we just need to add two electrons to the old one. That yields 1s2 2s2 2p6.
A-transfer an electron from the beryllium atom to the chlorine atom B-tranfer an electron from the chlorine atom to the beryllium atom C-add another beryllium atom D-add another chlorine atom The answer is D. :) Good luck!!
In this case the atom become an anion with the charge -1.
It becomes a negatively charged ion.
you cant add proton to an atom...u can only add electron....
it S2-, because when you add electrons, you are making the atomic radius bigger. In S2-, you add 2 electron, making it bigger. hope that helped.
The atomic radius increases down a column. More electron shells are added and the nucleus had less pulling power on those electrons. decreasesdLo
NaCl Add water. Na + Cl - Electrons have flowed and that is all electricity is, electron flow.
The element, oxygen, is in group 16, period 2 of the periodic table. Thus its electron configuration is 1s2 2s2 2p4. Oxide's ionic state is O2-, so to get its electron configuration we just need to add two electrons to the old one. That yields 1s2 2s2 2p6.
A-transfer an electron from the beryllium atom to the chlorine atom B-tranfer an electron from the chlorine atom to the beryllium atom C-add another beryllium atom D-add another chlorine atom The answer is D. :) Good luck!!
well pi=3.14 then you have to times it by the radius number given in your question then add 999!!
remove either a proton or electron OR add a proton or electron...
The ionic charge of potassium iodide is determined by the groups of the elements. Potassium is in the 1st column of the periodic chart, so it is group 1, and forms a +1 cation. Iodine is a halogen (group 7) and will add an electron to fill its shell with 8 electrons, so its charge is -1.
because they are the 1st discoverers.