The electron configuration of phosphorus is [Ne]3s2.3p3.
The valence electrons are the outermost (highest energy) s and p sublevels. There are 5 valence electrons in a phosphorus atom, and it is in period 3, so its valence electron configuration is 3s23p3.
For phosphorus [Ne]3s23p3
It would be 3 electrons!Why?Antimony: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3 Nitrogen: 1s2,,2s2,2p3Phosphorus: 1s2 2s2 2p6 3s2 3p3Arsenic: 3d10 4s2 4p3Bismuth: 4f14 5d10 6s2 6p3so on..
1s22s22p63s23p3 is for Phosphorus and the most likely ion is to be a 3- because it wants to have a full outer shell therefore the answer is 1s22s22p63s23p6
2Added explanation:In the outer orbitals of Pt (Period VI, 5d-blocktransition elements, after the 4f-block-lanthanides) there are two unpaired electrons:the first is the odd one in 4f17 (or maybe by exchange with 5d10>9)and the other one is the odd one in 6s1
The valence electrons are the outermost (highest energy) s and p sublevels. There are 5 valence electrons in a phosphorus atom, and it is in period 3, so its valence electron configuration is 3s23p3.
The valence electrons are the outermost (highest energy) s and p sublevels. There are 5 valence electrons in a phosphorus atom, and it is in period 3, so its valence electron configuration is 3s23p3.
the correct electronic confrigration of phosperous is 2,8,5
[Ne]3s23p3
Selenium: [Ar]3d104s24p4 Phosphorus: [Ne]3s23p3
[Ne]3s23p3
PD (Palladium) element 46 has an electron configuration281818
[Ne]3s23p3
2, 8, 5 [Ne]3s23p3
"Noble gas configuration" means that in writing out an electron configuration for an atom, rather than writing out the occupation of each and every orbital specifically, you instead lump all of the core electrons together and designate it with the symbol of the corresponding noble gas on the periodic table (in brackets). For example, the noble gas configuration of phosphorus will be [Ne]3s23p3
For phosphorus [Ne]3s23p3
For phosphorus [Ne]3s23p3