nope :)
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.
Carbon is the group 14, period 2 chemical element. Its electron configuration is 1s2 2s2 2p2. Thus, carbon has 2 electrons in its 2p subshell.
The period.
Selenium. As you go across the groups, not counting the transition metals, a valence electron is added.
non-metals have between 3 and 8 electrons in the outer shell. You can tell how many they have by looking at which period they're in on the periodic table, period 13 elements have 3, period 15 elements have 5 ect.
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.
It depends which n since n is the row (period) number. 1st n = 1-s subshell, 1 orbital, and 2 electrons. 2nd n = 2-s subshell with 1 orbital and 2 electrons + 2-p subshell with 3 orbitals and 6 electrons.
Carbon is the group 14, period 2 chemical element. Its electron configuration is 1s2 2s2 2p2. Thus, carbon has 2 electrons in its 2p subshell.
Fluorine is the group 17, period 2, halogen. Thus, its electron configuration is 1s2 2s2 2p5. So, as you can see, there are 5 electrons is fluorine's 2p subshell.
Aluminum is the group 12, period three chemical element. It has an electron configuration of 1s2 2s2 2p6 3s2 3p1. That means that its 2p subshell is full, having 6 electrons.
The period.
There are no metals in the first period. Only hydrogen and helium.
6 electrons can ocupy the 2p, 3p, 4p, and so on. each p subshell has 3 orbitals, and each orbital can hold up to 2 electrons, so each p subshell can hold up to 6 electrons total.
Selenium. As you go across the groups, not counting the transition metals, a valence electron is added.
The valence electrons are added to d orbitals in the case of transition metals (or d block elements).
because its a metalloid and the rest are all metals
It is not an abnormal configuration. For example let us consider the first series of transition elements in the fourth period. They are filling the 3d subshell. According to Aufbau Theorem, the electrons fill in such a way in the order of the increasing energy, (not in the order of the principal energy levels). 4s has a lesser energy than 3d, therefore the latter fills later.