P and f only
An element with 5 electrons in the third energy level has the electron configuration of 3s² 3p³. This indicates that there are 2 electrons in the 3s subshell and 3 electrons in the 3p subshell. The atomic number of this element is 15, which corresponds to phosphorus (P).
The electron configurations of chromium (Cr) and copper (Cu) are exceptions to the expected order due to the stability gained from half-filled and fully filled d subshells. For chromium, having a half-filled 3d subshell (3d^5) provides extra stability, so one electron from the 4s subshell is promoted to the 3d subshell, resulting in 3d^5 4s^1. Similarly, for copper, a fully filled 3d subshell (3d^10) is more stable than having one more electron in the 4s subshell (3d^9 4s^2), leading to the configuration of 3d^10 4s^1.
Two electrons are in the external shell of vanadium.
The number of electrons in a subshell of a copper atom depends on which subshell you are referring to. Copper has 29 electrons, so its electron configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d9. The 3d subshell in copper contains 9 electrons.
1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 4 d 10 4 f 14 5 s 2 5 p 6 5 d 4 6 s 2 The link below disagrees with the above, group 6 elements are special. By the Aufbau principle I believe. It is is more stable to have one electron in each subshell than to have 2 electrons in the s subshell and 4 in the d subshell.
3s and 3p, remember Cl has 7 valance electron. 3s^2 and 3p^5
Bromine is paramagnetic. This is because the 4p subshell has a value of 4p5. 5 electrons in a p subshell (which has space for 6), means the last electron is unpaired. The unpaired electron gives it the property of paramagnetism. Alternatively, Selenium is diamagnetic, as it has the outer subshell 4p4, meaning no electrons are unpaired.
Halogens have 5 electrons in their outermost p shell in their electrically balanced state. The p shell has 3 orbitals in each energy level. Halogens have 2 filled orbitals each with 2 electrons in them and one orbital with only one electron in it. In order to obtain the stable noble gas electron configuration, halogens gain one electron to completely fill the p shell on the outermost energy level. This gives halogens a charge of -1.
There are 5 electrons in the outermost electron shell of a phosphorus atom. Phosphorus has an electron configuration of 2, 8, 5.
There are 4 electron sub-shells: s, p, d, and f. These letters stand for sharp, principal, diffuse, and fundamental, but the names are not important. s subshells have 2 electons, while p subshells have 6, d subshells have 10, and f subshells have 14. There can be higher subshells, but these subshells require too much energy to fill and no element with a g subshell (the next subshell after f) has ever been synthesized. The first shell (i.e. the first period of the periodic table) has only s. Thus, the first shell has 2 electrons. The second shell has s and p subshells, so it has 2+6 or 8 electrons. The third shell has s, p, and d subshells. It ultimately has 18 electons. This can be misleading, however. The d subshell requires more energy to fill than the higher-shell s subshell. This is why the third period of the periodic table does not have a d section: the d electron subshell of the third Bohr shell does not fill until after the s subshell of the fourth Bohr shell has filled. Looking at the periodic table, you can see that the third period only has 8 electrons, while the 4th period has 18. The 18 electrons in the fourth period are the s subshell of the fourth shell, the d subshell of the 3rd shell, and the p subshell of the 4th shell. The fourth shell is similar to the third shell, but more extreme. The fourth shell has s, p, d, and f subshells, but the f subshell is not filled until two higher s shells have been filled. It does, however, fill out to 32 electrons in the 6th period of the periodic table. In the 6th period, the first period to have 32 electrons, there are 32 electrons, filling these subshells: s subshell of the 6th shell, f subshell of the 4th shell, d subshell of the 5th shell, and then the p subshell of the 6th shell. The fifth shell would ultimately fill out to a full 50 electrons and would do so in the 8th period of the periodic table. However, as previously noted, no substance has ever been found or generated with that many electrons. It would fill the s subshell of three shells above (i.e. shell 8) before it filled the g subshell of shell 5. No element in the 8th period has ever been synthesized, so a filled fifth Bohr shell has never been found. A good example for a Bohr diagram would be Astatine, which is in the 6th period. In the first shell of the Bohr diagram, you have 2 electrons (s subshell only). It is filled completely. In the second, you have 8 electrons (s and p subshells) and in the third you have 18 electrons (s, p, and d), and both shells are filled completely. In the fourth shell, you have 32 electrons (s, p, d, and f), and it is filled completely. In the fifth shell, you have 18 electrons. This is because only the s, p, and d subshells are filled. It would require too much energy to fill the f subshell of the 5th shell, so the electrons just go to the s, p, and d subshell of higher shells. The 6th shell has 7 electrons. The 2 electrons of the s subshell are filled first, and then 5 electrons go into the p shell.
There are 5 electrons in the outermost electron shell of a phosphorus atom. Phosphorus has the electron configuration 2-8-5, so it has 5 electrons in its outermost shell.
An element with 5 electrons in the third energy level has the electron configuration of 3s² 3p³. This indicates that there are 2 electrons in the 3s subshell and 3 electrons in the 3p subshell. The atomic number of this element is 15, which corresponds to phosphorus (P).
Nitrogen has 5 electrons in its outer shell, so there are 3 electron pairs in the outer shell of nitrogen.
9. The number of orbitals in a given shell fit the equation 2(L)+1, where L=the angular quantum number. L=0 corresponds with the s orbital, L=1 with p orbital, L=2 with d orbital, L=3 with f orbital, L=4 with g orbital, and L=5 with h orbital.
An element with the electron configuration Ne3s^23p^5 is in period 3 of the periodic table. The electron configuration indicates that the element has 3 energy levels, with the last electron being in the 3p subshell.
There are 6 electrons in the third principal level (n = 3) of a chromium atom. The electron configuration of chromium is [Ar] 3d5 4s1, so there are 5 electrons in the 3d subshell and 1 electron in the 4s subshell.
The electron configurations of chromium (Cr) and copper (Cu) are exceptions to the expected order due to the stability gained from half-filled and fully filled d subshells. For chromium, having a half-filled 3d subshell (3d^5) provides extra stability, so one electron from the 4s subshell is promoted to the 3d subshell, resulting in 3d^5 4s^1. Similarly, for copper, a fully filled 3d subshell (3d^10) is more stable than having one more electron in the 4s subshell (3d^9 4s^2), leading to the configuration of 3d^10 4s^1.