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!!!

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*.

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.

**"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.*

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

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

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 #

*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**: