The dots represent the electrons in valence shell. Phosphorus= P it has five electrons on the outer shell.
There are three dots around unpaired while the dots on the right of the P are paired.
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The orbital diagram of phosphorus (P) includes 15 electrons distributed in the following manner: 1s^2 2s^2 2p^6 3s^2 3p^3. This distribution represents the arrangement of electrons in the energy levels and orbitals of a phosphorus atom.
like this:
+15p + -15e = 0 net charge
and that's your bohr model for phosphorus!!!!
1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1
Im in A.P. Molecular Physics as a 8th grader and after 2 hours of work i found this as an answer.
it could be kinda a dicfficult to answaer this questin but just ask your teacher how to answer it ok
A phosphorus atom has 3s, 3p, and 3d orbitals for a total of 5 orbitals.
For a neutral magnesium atom, the orbital diagram would show two electrons in the 1s orbital, two electrons in the 2s orbital, and six electrons in the 2p orbital, following the Aufbau principle and Hund's rule. This configuration can be represented as 1s^2 2s^2 2p^6 in the electron configuration notation.
An electron in a phosphorus atom would experience the greatest shielding in the 3s orbital. This is because electrons in inner shells provide greater shielding than those in outer shells, and the 3s orbital is closer to the nucleus compared to the higher energy orbitals.
Yes, phosphorous (and sulfur) have access to a d orbital. It's a bit weird (as is most chemistry), in the ground state phosphorous does not have any d orbital electrons, however, d orbital hybridization is used to explain why phosphorous can form more than the "octet" number of bonds, such as PCl5. This d orbital is also used when describing phosphorous as a pi-acceptor ligand, and the reason it can be considered a pi-acceptor ligand is because it does have access to that d orbital, which can accept the metal's e- density. Hope that helped.
ml = 0