Please much more detailed on how to get the answer. I"m really puzzled on this chapter of valance electrons and etc.

Thank you so much!!!


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They encompass the #1s# electrons in its electron configuration, so, there space two the them.

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#overbrace(1s^2)^"core" " "overbrace(2s^2 2p^6)^("valence")#

I expect I"ll do a comprehensive review from particle properties every the way through quantum numbers and also electron configurations.

FLUORINE ELECTRONS

#"F"# as-written as the fluorine atom, through atomic number #9#, providing it #9# protons by definition.

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As a neutral atom, the number of protons it has must be equal come the number of electrons, together they room particles of the contrary charge and also a neutral atom has a full charge of zero.

Therefore, #"F"# atom has actually #9# electron total and #"F"^(-)# has #10# electrons total. Note that these include both the core and valence electrons, the last being the ones used most regularly to react.

FLUORINE ORBITALS AS concerned QUANTUM NUMBERS

An atomic orbital is offered as the symbol #nl#, which we will define later.

As #"F"# is on the 2nd heat of the regular table, it has access up to (and including) the atomic orbitals belonging come principal quantum number #n = 2#, whereby we can potentially have

#n = 1, 2, 3, . . . #

defining each power level. These room the coefficients of atom orbital symbols.

For each power level #n#, the orbitals the exist have certain shapes given by #l#, the angular momentum quantum number, where

#l = 0, 1, 2, . . . , n-1#.

Since the best #l# is #n - 1#, fluorine offers orbitals up to and also including #l = 1#, which are uncovered in the subshells the has accessibility to. Together it transforms out,

#l = 0 harr s# subshell#l = 1 harr p# subshell#l = 2 harr d# subshell#l = 3 harr f# subshell

and therefore on. This completes the orbital symbol given by #nl#.

Therefore, fluorine atom includes electrons in ~ #1s#, #2s#, and also #2p# subshells.

Furthermore, each subshell (#s,p,d,f, . . . #) consists of a certain variety of orbitals, every assigned a worth of #m_l#, wherein #m_l# is the magnetic quantum number:

#l = 0 harr m_l = 0# #-># #1 xx s# orbital for any type of #n#

#l = 1 harr m_l = -1,0,+1# #-># #3 xx p# orbitals for any type of #n#

#l = 2 harr m_l = -2,-1,0,+1,+2# #-># #5 xx d# orbitals for any kind of #n#

etc.

This method we have one #1s#, one #2s#, and also three #2p# orbitals.

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"PUTTING" ELECTRONS into ORBITALS

Each of this orbitals (of a certain #m_l#) will certainly contain particular numbers that those #10# electrons we say #"F"^(-)# has.

By the Pauli exclusion Principle, no 2 electrons have the right to be in the exact same orbital and have the very same spin worth #m_s = pm1/2# at the exact same time.

Since 2 electrons in the exact same orbital have actually the same #n#, #l#, and #m_l#, these 2 electrons have to then have actually a spin quantum number that #m_s = pm1/2#.

As a result, each orbital deserve to only save two electrons, maximum.

See more: What Is 1.32 As A Percentage

This then leads come the electron configuration of #"F"^(-)#:

#"F"^(-): " "1s^2 2s^2 2p^6#

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where the superscripts denote how numerous electrons room in the subshell characterized by #nl#.

CORE VS VALENCE ELECTRONS

Since helium has actually two electrons, that is practically to specify the noble gas shorthand # = 1s^2#. Similarly, we could say # = 1s^2 2s^2 2p^6#, and also so on.

We likewise call this the noble gas core for a reason! It consists of the core electrons (most that the time)!

As such, it should at this allude be straightforward that there room #bb2# core electrons in #"F"^(-)#, and also #bb8# valence electrons: