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Q: Does the s or d shell lose electrons first?
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What a maximum of eight electrons can occupy the first shell?

the first shell (the K shell) can only contain a maximum of two electrons, as it consists of just an s orbitalthe second shell (the L shell) can contain a maximum of eight electrons, as it consists of an s orbital and three p orbitalsthe third shell (the M shell) can contain a maximum of eighteen electrons, as it consists of an s orbital, three p orbitals, and five d orbitalsthe fourth shell (the N shell) can contain a maximum of thirty two electrons, as it consists of an s orbital, three p orbitals, five d orbitals, and seven f orbitalsthe fifth shell (the O shell) can contain a maximum of fifty electrons, as it consists of an s orbital, three p orbitals, five d orbitals, seven f orbitals, and nine g orbitalsetc.


Which orbital is rhodium's last electron in?

If you are filling in the electrons it will be in the 4d orbital. If you are removing electrons the first to come out is in the 5s electrons since transition metals lose 's' electrons before 'd' electrons


How many places are there for electrons in the third shell of an atom?

The third shell of an atom can hold a maximum of 18 electrons. This shell consists of three subshells - s, p, and d - with each subshell being able to accommodate a certain number of electrons. The s subshell can hold up to 2 electrons, the p subshell can hold up to 6 electrons, and the d subshell can hold up to 10 electrons, totaling 18 electrons in the third shell.


Why do 'd' block elements exhibit variable valencies?

The "d" block elements have electrons in a "d" in a lower shell than the valence electrons in the "s" orbital. These "d" electrons can be moved to the outer shell to give a metal a higher valence then the value of 2 that might otherwise be expected. Osmium can have a valence as high as 8. Additionally, the outer shell electrons can be moved from the "s" orbital down to the "d" orbital to give a metal a lower valence, perhaps the most notable case of this is silver which almost always forms a 1+ ion rather than 2+.


How many electrons can 3rd shell hold?

In (3s2 3p6 3d10) = 2 + 6 + 10 = 18 electrons in the 3rd shell (3(s+p+d))

Related questions

How many electrons would make the d shell complete?

The d shell needs 10 electrons to be complete.


How many electrons does Magnesium need to complete last shell?

Magnesium, in the second group, can either gain six electrons or lose two electrons to achieve a full outer shell.


How many electrons in 2d xy?

The 2d sub energy level does not exist. The first shell to contain a d sub-shell is the third shell: the 3d sub-shell contains a maximum of 10 electrons, with two electrons in each of five different d orbitals.


What a maximum of eight electrons can occupy the first shell?

the first shell (the K shell) can only contain a maximum of two electrons, as it consists of just an s orbitalthe second shell (the L shell) can contain a maximum of eight electrons, as it consists of an s orbital and three p orbitalsthe third shell (the M shell) can contain a maximum of eighteen electrons, as it consists of an s orbital, three p orbitals, and five d orbitalsthe fourth shell (the N shell) can contain a maximum of thirty two electrons, as it consists of an s orbital, three p orbitals, five d orbitals, and seven f orbitalsthe fifth shell (the O shell) can contain a maximum of fifty electrons, as it consists of an s orbital, three p orbitals, five d orbitals, seven f orbitals, and nine g orbitalsetc.


Each carbon atom can share 4 pairs of electrons because a. It only has 6 protons. b. It only has 2 electrons in its first shell. c. It only has 4 electrons in its last shell. d.?

because it has 4 valance e- or 4 e- in it's outermost shell


Which orbital is rhodium's last electron in?

If you are filling in the electrons it will be in the 4d orbital. If you are removing electrons the first to come out is in the 5s electrons since transition metals lose 's' electrons before 'd' electrons


An atom of chlorine has several valence electrons in its what A nucleus Bfirst shell C second shell D third shell?

Third Shell


In a Bohr diagram how many electrons are in each shell?

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.


Do main group metals from period 5 onward tend to lose electrons from the d subshell first?

nope :)


How many places are there for electrons in the third shell of an atom?

The third shell of an atom can hold a maximum of 18 electrons. This shell consists of three subshells - s, p, and d - with each subshell being able to accommodate a certain number of electrons. The s subshell can hold up to 2 electrons, the p subshell can hold up to 6 electrons, and the d subshell can hold up to 10 electrons, totaling 18 electrons in the third shell.


Why do transition metals have variable oxidation states?

Transition metals can move electrons between the outer shell and the d d orbital. For example, copper can either lose its 2 s electrons or move one of those s electrons into d orbital, which is one electron short of being full.


How many electrons fit in the 3rd shell?

To find this the equation is 2n2, while n being the shell number, so in this case 2*32 = 18