Practice Problem A What are the possible values for m e when the principal quantum number n is 2 and the angular momentum quantum number C is 07 Practice Problem B What are the possib
Trang 1214 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms
Think About It Consult Table
6.2 to make sure your answer is
correct Table 6.2 confirms that it
is the value of C, not the value of n,
that determines the possible values
e = 2 -1-2 ~ - llQ]EJEJ -BBCDI +1 ~ +2 ~ -e = 21 dsubShelll
e = 1 BI 0 ~ +1 ~ -BQ]EJ -B[DB - e = 11 p subshelll
Strategy Use the rules governing the allowed values of m e Recall that the possible values of me
depend on the value of C, not on the value of n
Setup The possible values of m e are - C, ,0, , +C
Solution The possible values of m e are -1, 0, and + 1
Practice Problem A What are the possible values for m e when the principal quantum number (n) is
2 and the angular momentum quantum number (C) is 07
Practice Problem B What are the possible values for m e when the principal quantum number (n) is 3 and the angular momentum quantum number (C) is 27
Whereas three quantum numbers are sufficient to describe an atomic orbital, an additional tum number becomes necessary to describe an electron that occupies the orbital
quan-Experiments on the emission spectra of hydrogen and sodium atoms indicated that each line in the emission spectra could be split into two lines by the application of an external mag- netic field The only way physicists could explain these results was to assume that electrons act like tiny magnets If electrons are thought of as spinning on their own axes, as Earth does, their magnetic properties can be accounted for According to electromagnetic theory, a spinning charge
Trang 2SECTION 6.6 Quantum Numbers 215
generates a magnetic field, and it i s thi s motion that causes an electron to behave like a magnet
Figure 6.16 s hows the two possible spinning motion s of an electron, one clockwise and the other
counterclockwise To s pecify the electron's s pin , we use the electron spin quantum number (ms)
Because there are two po ss ible direction s of spin, opposite each other : m ; has ' two possible vaiiies : "
+~ and -~
Conclusive proof of electron s pin was established by Otto Stern 13 and Walther Gerlach 14 in
1924 Figure 6.17 shows the basic experimental arrangement A beam of gaseous atoms generated
in a hot furnace passes through a nonuniform magnetic field The interaction between an electron
and the magnetic field causes the atom to be deflected from it s straight-line path Becau se the
direction of spin is random, the electrons in half of the atoms will be spinning in one direction
Those atoms will be deflected in one way The electrons in the other half of the atoms will be s pin-
ning in the opposite direction Tho se atoms will be deflected in the other direction Thus, two spots
of equal intensity are observed on the detecting sc reen
To summarize, we can designate an orbital in an atom with a set of three quantum number s
These three quantum numbers indicate the size (n), shape (€), and orientation (m e ) of the orbital A
fourth quantum number (m s ) i s necessary to designate the spin of an electron in the orbital
Checkpoint 6.6 Quantum Numbers
6 6 1 Which of the following is a legitimate 6 6.3
set of three quantum numbers: n , e, and
m e 7 (Select all that apply.)
a) 1, 0, ° b) 2,0,0
c) 1, 0, + 1 d) 2, 1, + 1
e) 2,2, - 1
6 6 4
6 6 2 How many orbitals are there in a
subshell designated by the quantum numbers n = 3, e = 27
a) 2
b) 3
c) 5 d) 7 e) 10
Atom beam
Detecting screen Magnet
How many subshells are there in the
What is the total number of orbitals in the shell designated by n = 37
a) 1 b) 2
c) 3 d) 6 e) 9
Slit screen Oven
13 Otto Stern ( 1 888-1969) German physicist H e made important co ntr ibutions to the s tudy of the magnetic prop e rtie s of
at o ms and the kinetic th eory of gases Stern was awarded the Nobel Pri ze in Physic s in 1943
l ·t Walther Gerlach (1889-1979) G erm an physicist Gerlach's main area of research was in quantum theor y
Two electrons in the same orbital with opposite spins are referred to a s "paired."
Figure 6.16 (a) Clockwise and
(b) counterclockwise spins of an electron The magnetic fields generated by these two spinning motions are analogous to those from the two magnets The upward and downward arrows are used to denote the direction of spin
Figure 6.17 Experimental arrangement for demonstrating the
spinning motion of electrons A beam
of atoms is directed through a magnetic field When a hydrogen atom, with
a single electron, passes through the field, it is deflected in one direction or the other, depending on the direction
of the electron's spin In a stream consisting of many atoms, there will be equal distributions of the two kinds of
spins, so two spots of equal intensity are detected on the screen
•
Trang 321 6 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms
The radial probability d i stribution can be
tho ug ht of as a map of "where an electron
spends most of i ts time."
.-.0
'"
.0
0 A:
~
'" ~
- '"
Strictly speaking, an atomic orbital does not have a well-defined shape because the wave function
characterizing the orbital extends from the nucleus to infinity In that sense, it is difficult to say what an orbital looks like On the other hand, it is certainly useful to think of orbitals as having
specific shapes Being able to visualize atomic orbitals is essential to understanding the formation
of chemical bonds and molecular geometry, which are discussed in Chapters 8 and 9 In this
sec-tion, we will look at each type of orbital separately
s Orbitals
For any value of the principal quantum number (n), the value 0 is possible for the angular tum quantum number (f), corresponding to an s subshell Furthermore, when f = 0, the magnetic quantum number (m e ) has only one possible value, 0, corresponding to an s orbital Therefore, there is an s subshell in every shell, and each s subshell contains just one orbital, an s orbital
momen-Figure 6.18 illu s trates three ways to represent the distribution of electrons: the probability
d e n SIty,' ' di e ' spheriC ' <il ' di ' stn 6ut!o ' n ' of erectioii density; ' arid : the ' radia l probability distributio n (t h e
~ '" ~
~
>-~ - '" ~
(c)
Figure 6.18 From top to bottom, the probability density and corresponding relief map, the distribution of electron density represented spherically
with shading corresponding to the,relief map above, and the radial probability di st ribution for (a) the Is, (b) the 2s, and (c) the 3s orbitals of hydrogen
Trang 4SECTION 6.7 Atomic Orbitals 217
probability of finding the electron as a function of distance from the nucleus) for the Is , 2s, and 3s
orbitals of hydrogen The boundary surface (the outermost surface of the spherica l representation)
is a common way to repre se nt atomic or bital s, incorporating th e vo lume in w hich there is about a
90 percent probability of finding the electron at any given time
All s orbitals are spherical in shape but differ in size, which increases as the principal
quan-tum number increa ses The radial probabilit y distribution for the I s orbita l exhi bit s a maximum at
o
52.9 pm (0.529 A) from the nucleu s Intere s tingly , this di sta nce is equal to the radiu s of the n = 1
orbit in the Bohr model of the h y drog e n atom The radial probability distribution plots for th e 2s
and 3s orbitals exhibit two and three maxima , re spect i vely, w ith the grea te st probability occ urring
at a greater di s tance from the nucleu s as n in c rea ses Bet ween the two maxima for th e 2s orbital
there is a point on the plot where the probability dr o p s t o zero Thi s corresponds to a node in the
electron den s ity , where the sta nding wave h as zero amplitude There a re two suc h nodes in the
radial probability di s tribution plot of the 3s orbital
Although the detail s of electron d e n sity variation wit hin each boundary s u rface are lost, the most important features of atomic orbitals are their overall s hape s and relati ve s i zes, w hich are
adequately repre se nted by boundary s urface diagram s
p Orbitals
When the principal quantum number (n) i s 2 or greater, the value 1 is po ss ible for th e angular mom e
n-tum quann-tum number (C), corresponding to a p s ub s hell And, when C = 1, the magnetic quantum
num-ber (me) has three po ss ible values: -1,0, and + 1, each corresponding to a differentp orbital Therefore,
there is a p s ub s hell in every s hell for which n :> 2, and eac h p sub s hell co ntain s three p orbital s The se
three p orbitals are labeled 2p x , 2p y , and 2p z (F i g ur e 6.19) , w ith the subsc ript ed lett ers indicating the
axis along which each orbital is oriented The se t hre e p orbital s are identical in size, shape, and energy;
they differ from one another only in orientation Note, ho weve r , that there is no sim ple relation betwe e n
the values of m e and th e x, y, and z directions For our purpose, yo u ne ed only remember th at because
there are three possible values of m e , there are three p o rbital s w ith different orientations
The boundary s urface diagram s of p orbitals in Figure 6.19 s h ow that eac h p orbital can be
thought of as two lobe s on opposite sides of the nucleu s Like s o rbital s, p orbitals incr ease in s i ze
f rom 2p to 3p to 4p o rbital and so o n
When the principal quantum number (n) i s 3 or greater, the va lu e 2 i s p oss ibl e fo r the angular
momentum quantum number (C), corresponding to a d s ub s hell When C = 2, the magnetic
quan-tum number (m e ) ha s five po ss ible values, - 2, - 1 , 0, + I , and +2 , each correspo ndin g to a differ e nt
d orbital Again there i s no direct correspondence between a given o ri en tati on and a particular m e
va lue All the 3d orbitals in an atom ar e identical in e nerg y and are lab e led w ith su b sc ript s den ot
-ing their orientation with r es pect to th e x, y, and .G axes and t o t h e pl anes defined by them Th e d
o rbital s that have higher principal quantum numb e r s (4d, 5d, etc.) ha ve s hape s s imilar to th ose
s hown for the 3d orbital s in Figure 6.20
s urfac es for the P x ' P l" and p z orbitals
Figure 6 20 B o undar y s urfa ces for
the d orb it al s
Trang 5218 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms
Think About It Consult Figure
6.15 to verify your answers
Figure 6 21 Orbital energy level s
in the hydrogen atom Each box
represents one orbital Orbitals with the
sa me principal quantum number (n) all
have the same energy
The! orbitals are important when accounting for the behavior of elements with atomic bers greater than 57, but their shapes are difficult to represent In general chemistry we will not concern ourselves with the shapes of orbitals having e values greater than 2
num-Sample Problem 6.8 shows how to label orbital s with quantum numbers
Sample Problem 6.8
Li s t the va lues of n, e, and m e for each of the orbitals in a 4d subshell
the va lue s of nand e There are multiple po ss ible values for m e , which will have to be deduced from the value of e
The letter in an orbital de s ignation gives the value of the angular momentum quantum number (e)
The magnetic quantum number (m s ) can have integral values of -e, , 0, , + e
+1, and +2
Trang 6-~ -SECTION 6.8 Electron Configuration 219
Checkpoint 6.7 Atomic Orbitals
Electron Configuration
6.7.3 In a hydrogen atom, which orbitals
are higher in energy than a 3s orbital?
(Select all that apply.)
6.7.4 Which of the following sets of quantum
numbers, n, f, and me , corresponds to a
The hydrogen atom is a particularly simple system becau se it contains only one electron The
o rbital (an excited s tate) With many-electron systems, we need to know the ground-state electron
configuration that is, how the electrons are distributed in the various atomic orbitals To do this,
we need to know the relative energies of atomic orbitals in a many-electron system, which differ
fro m those in a one - electron syste m such as hydrogen
Energies of Atomic Orbitals in Many-Electron Systems
Co nsider the two emission s pectra shown in Figure 6.22 The spectrum of helium contains more
correspond-ing to emission in the visible range, in a helium atom than in a hydrogen atom Thi s i s due to the
s plitting ' o(eiiergy 'ieveis ' c ' aus ecl ' by ' eiec ' tro s ' tatlc ' inter actIons ' betwe ' en heilum ; s ' two · eiectron · s.' ·
to the hydrogen atom, in which the energy of an orbital depends only on the va lue of n ( Figure
6 21), the energy of an orbital in a many-electron sys tem depends on both the value of n and the
the 3s orbital and lower in energy than the 3d orbitals In a many-electron atom, for a given value
o f n, the energy of an orbital increases with increa s ing value of e One important consequence of
m e splitting of energy levels is the relative energies of d orbital s in one s hell and the s orbital in the
n ex t higher shell As Figure 6.23 shows, the 4s orbital is lower in energy than the 3d orbitals
Like-w ise, the 5s orbital is lower in energy than the 4d orbital, and so on This fact bec o mes important
w hen we determine how the electrons in an atom populate the atomic orbitals
" Splitting" of energy levels refers to the splitting
of a shell i nto su bshells of different energies, as
shown in Figure 6.23
emission spectra of H and He
Trang 7220 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms
in many-electron atoms For a given
value of n, orbital energy increases with
the value of e
15 ' i s read as " one 5 o ne."
T he groun d st ate for a m a ny-e l e ctron atom
is the o n e in whi ch a l l th e e lect ron s o cc u p y
o rb i ta ls of t he low e st pos s ibl e ene r gy
-~ 4d ~ 4d r~~B
~-
- ~~~- - -
-n _ _ _ _ _ _ _ _ _ _ _ _ _ _ ~ 3d ~ 3d r~~~ -
-~ -
-~~~ - ~-
-
-~~~ -
-~ -The Pauli Exclusion Principle According to the Pauli1 5 exclusion principle, no two electrons in an atom can have the same four quantum numbers If two electrons in an atom have the same n, e, and m e values (meaning that they occupy the same orbital), then they must have different value s of m s ; that is, one must have m s = +~ and the other must have m s = - ~ Because there are only two possible values for m s , and no two electrons in the same orbital may have the same value for ms , a maximum of two electrons may occupy an atomic orbital, and these two electrons must have opposite spins We can indicate the arrangement of electrons in atomic orbitals with labels that identify each orbital (or subshell) and the number of electron s in it Thus, we could describe a hydrogen atom in I the ground s tate using Is ~ denote s the number of electrons lsI in the orbital or subshell denotes the principal / ~ denotes the angular momentum quantum number n quantum number e We can also represent the arrangement of electrons in an atom using orbital diagrams , in which each orbital is represented by a labeled box The orbital diagram for a hydrogen atom in the ground state i s The upward arrow denotes one of the two possible spins (one of the two possible m s values) of the electron in the hydrogen atom (the other possible spin is indicated with a downward arrow) Under certain circumstances, as we will see shortly, it is useful to indicate the explicit locations of electrons
The orbital diagram for a helium atom in the ground state is
15 Wolf g ang Pauli (1900-195 8) Au s trian ph ys ici s t One o f the founder s of quantum mechanic s , Pauli was awarded the
N o bel Pri z e in Ph ys i cs in 1945
Trang 8SECTION 6.8 Electron Configuration 221
The label I i ·i~di~~t~~· th~;~ ~~ t;;~ ~l~~tr~~~ ·i~· the i ; ·~~bit~l·.· N~te· ~so that · th~ ~~~~·i ·i~· the ·bo~ ·
dia-gram includes an orbital with a single electron, we represent it with an upward arrow alth ough
effect on the energy of the electron
The Aufbau Principle
orbital energies and the Pauli exclusion principle Thi s proce ss i s based on the Aujbau principle,
which makes it pos sible to "build" the periodic table of the elements and determine their electron
configurations by step s Each step involve s adding one proton to the nucleus and one electron to
the appropriate atomic orbital Through thi s proce ss we gain a detailed knowledge of the electron
c onfigurations of the elements As we will see in later chapters, knowledge of electron
configura-t io ns help s us understand and predict the propertie s of the elements It also explains why th e
ele-me nts fit into the periodic table the way they do
After helium , the next element in the periodic table is lithium, which ha s three electrons
Because of the restriction s imposed by the Pauli exclusion principle , an orbital can accommodate
n o more than two electrons Thu s, the third electron cannot res ide in the Is orbital Instead, it mu st
r es ide in the next available orbital with the lowest poss ible energy According to Figure 6.23 , this is
m e 2s orbital Therefore , the electron configuration of lithium i s l i2s1, and the orbital diagram i s
irnilarly, we can write the electron configuration of beryllium as li2i and represe nt it with the
o rbital diagram
ith both the Is and the 2s orbitals filled to capacity, the nex t electron, w hich i s needed for the
el ectron configuration of boron, must re side in the 2p sub shell Becau se all three 2p orbital s are of
,h ow the fir st electron to occupy the p s ub shell in the first empty bo x in the orbital diagram
1 1 1
Hund's Rule
• T ill the sixth electron, which is needed to represent the electron configuration of carbon, reside in
iIe 2p orbital that is already half occupied, or will it re side in o ne of the other, empty 2p orbitals?
-\c c ording to Hund'sl 6 rule, the most s table arrangement of electrons in orbitals of e qual energy
l.S me one in which the number of electrons with the same spin is maximized As we have seen , no
0':0 electron s in any orbital may have the sa me spin, so maximizing the number of electrons with
iIe same spin requires putting the electrons in separate orbital s Accordingly, in any subshell, an
:: l e ctron will occupy an empty orbital rather than one that already contains an electron
The electron configuration of carbon is, therefore , l i2i2p 2 , and its orbital diagram is
' F rederick Hund (1896-1997) German physici s t H u nd ' s work was ma inl y in quantum mec hani c s H e also helped to
1 S2 i s r ead a s " o ne s tw o " , not as " one s
s quared "
Trang 9222 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms
fill with electrons
Remember that in this context, de g enerate
means "of equal energy " Orbitals in the same
s ubshell are degenerate
• • •
".,., ~==~ Multimedia
Atomic Structure electron configurations
6.23 again to make s ure you have
filled the orbitals in the right order
and that the s um of electrons i s 20
before the 3d orbitals
Similarly, the electron configuration of nitrogen is l i2i2p3, and its orbital diagram is
General Rules for Writing Electron Configurations
Ba se d on the preceding examples we can formulate the following general rules for determining the
electron configuration of an element in the ground sta te:
1 Electron s will re s ide in the available orbitals of the lowest possible energy
2 Each orbital can accommodate a maximum of two electrons
" "' :' ( Eiectrons ' wiii ' not ' paIr ' i"n ' degenerate orbitals if an empty orbital is available
4 Orbitals will fill in the order indicated in Figure 6.23 Figure 6.24 provides a simple way for
you to remember the proper order
Sample Problem 6 9 illustrate s the procedure for determining the gro und· state e lect ron con· figuration of an atom
Sample Problem 6.9
Strategy Use the general rules given and the Aufbau principle to "build" the electron configuration
Setup Becau se Z = 20, we know that a Ca atom ha s 20 electrons They will fill orbitals in the order
the following o rder: I s, 2s, 2p, 3s, 3p, 4s Each s s ubshell can contain a maximum of two electrons,
Trang 10SECTION 6.9 Electron Configurations and the Periodic Table 223
6.8 1 Which of the following electron configurations correctly 6.8.3 Which orbital diagram is correct for the ground state S atom?
represents the Ti atom?
The electron configurations of all elements except hydrogen and helium can be represented u si ng
m ost recently precedes the element in que s tion , followed by the electron co nfiguration in the
out-er most occupied subshells Figure 6.25 gives the ground - state electron configurations of elements
fro m H (Z = 1) through Rg (Z = 111) Not i ce the s imilar pattern of electron configurations i n
the element s lithiu m (Z = 3) through neon ( Z = 10) and tho se of sodium (Z = 11) through argon
( Z = 18 ) Both Li and Na, for example, have the configuration ns1 in their outermost occupied
s ubshells For Li, n = 2; for Na, n = 3 Both F and CI have electron configuration nin/, where
n = 2 for F a nd n = 3 for CI, and so on
As mentioned in Section 6.8, the 4s s ub s hell i s filled before the 3d su b s hell in a many
-e lectron atom (see Figure 6.23) Thu s, the electron configuration of potassium (Z = 19) is
l i 2i2p63s23p 64s1 Because 1 i 2 i 2p 6 3s23p6 is the e l ec t ron configuration of argon, we can s
im-pl ify the electron configuration of pota ss ium by writing [Ar]4 s1, where [Ar] den otes the "a rgon
co re."
The placement of the outermost electron in the 4s orbital (ra ther than in the 3d orbital) of
po tassium is strongly supported by experimental evidence The physical and chemical
In both lithium and s odium , the outermost electron is in an s orbital (there i s no doubt that
t he ir outermost electrons occupy s orbitals becau se there i s no 1d or 2d subshell) Ba sed on
its s imil arities to the other alkali metal s, we expect pota ss ium to have an analogous electron
o nfiguration; th at is, we expect the last e l ectron in pota ss ium to occupy the 4s rather than the
3d orbital
The elements from Group 3B through Group 1B are transition metals [ ~~ Section 2.4]
Tra nsition metals either ha ve incompletely filled d s ubshell s or readily give ri se to cations that
h ave incompletely filled d subshells In the first transition meta l series, from scandium (Z = 21)
IHI H I HI [IT] IHIHI
Although zinc and the other elements in Group
"t rans ition metals," they neither have nor
thro u gh copper (Z = 29), additional electrons are placed in the 3d orbitals according to Hund's
ru le However, there are two anomalies The electron configuration of chromium ( Z = 24) is
I
Trang 11224 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms
1
Core
2 [He]
3
[Ne]
4 [Ar]
5 [Kr]
6 [Xe ]
7 [Rn]
11
Na 3s1
55
Cs 6s1
53
I 4d lOSs2
Figure 6.25 Ground-state electron co nfi g uration s for the known e l ement s
E lectron configurations such as these may also
be written with the d subshell first For example,
[Arl4s'3d' o can also be w ri tten as [ArI3d104s'
Either way i s acceptab l e
cop p er: w h ose ' eiectron ' cc )Ji ' flguratlon ' ' {i ' [Ar]4S13dlO rather than [Ar]4s 2 3Jl T h e reason for these
anomalies is t h at a slightly greater s tability is associated with the half - fi ll ed (3d5 ) and completely filled (3dlO) subs h e ll s
For elements Zn (Z = 30) through Kr ( Z = 36), the 3d, 4s , and 4p s ubshell s fill in a s
through s ilver (Z = 47)] are al so irregular, b u t t h e detail s of ma n y of these irreg ul arities are beyond
gap in the period i c table where the lanthanide (rare earth) series be l ongs T h e lanthanide s are a
Trang 12SECTION 6.9 Electron Configurations and the Periodic Table 225
series of 14 elements that have incompletely filled 4f s ub s hell s or that readily give rise to cations
that have incompletely filled 4fsubshells The lanthanides (a nd the actinides, to be discussed next )
are shown at the bottom of the periodic tabl e to keep the table from being too wide
In theory, the lanthanide s arise from the filling of the sev en degenerate 4f orbitals In reality,
30 wever, the energie s of the 5d and 4f orbitals are very close and the electron configurations of
:h ese elements s ometimes involve 5d e l ectrons For example, in lanthanum it se lf (Z = 57) the 4f
or b ital is s li ghtly higher in energy than the 5d orbita l Thus, lanthanum' s electron configuration i s
;:X e J6s25d1 rather than [Xe J6s24l
After the 4f subs hell i s completely filled, the next e l ectron enters the 5d subshell of lutetium
Z = 71) This series of elements, including lutetium and hafnium (Z = 72) and extending through
;n erc ury (Z = 80), i s characterized by the filling of the 5d su bshell The 6p s ub s hell s are filled
je xt, which takes u s to radon (Z = 86)
The la s t row of elements begin s with francium (Z = 87; electron configuration [RnJ7s1)
:::n d radium (Z = 88; electron configuration [RnJ7 i), and then continues with the actinide series,
, ,
· hic h starts at ac tinium (Z = 89) and ends with nobelium ( Z = 102) The actinide series ha s
;>art ially filled 5fand/or 6d s ubshell s The elements lawrencium (Z = 103 ) through darmstadtium
Z = 110) have a filled Sf subshell and are characterized by the filling of the 6d s ub s hell
Wit h few exceptions, you should be able to write the electron configuration of any ele
-=e nt, using Figure 6.23 (or Figure 6.24) as a guide Elements that require particular care are the
=<lIls ition metals , the lanthanides, and the actinides You may notice from looking at the electron
:n nfig urations of gadolinium (Z = 64) and curium (Z = 96) that half- filled f s ub s hell s also appear
:0 exh ibit slightly enhanced stability As we noted earlier, at larger values of the principal quantum
- ':J mber n, the order of s ubshell filling may be irregular due to the closeness of the energy level s
Fig ure 6.26 groups the elements according to the type of subshell in which the outermost
=~ec trons are placed Elements who se outermost electrons are in an s subshell are referred to as
- al ock elements, tho se whose o ut ermost electrons are in a p subshell are refened to as p-block
: ' ements, and so on
an f subshell Th ere are seven possible valu es for
m e when e = 3: - 3, -2 , -1 ,0, +1, +2, and +3,
Trang 13226 CHAPTER 6 Quantum Theory and t he Electronic Structure of Atoms
groups of elements in the periodic table
according to the type of subshell being
filled with electrons
Think About It Arsenic is a
p-block element; therefore, we
should expect its outermost
electrons to reside in a p subshell
Setup The noble gas core for As is [Ar], where Z = 18 for AI' The order of filling beyond the noble
gas core is 4s, 3d, and 4p Fifteen electrons must go into these subshells because there are 33 - 18 =
15 electrons in As beyond its noble gas core
Practice Problem A Without referring to Figure 6.25, write the electron configuration for a radium
atom (Z = 88) in the ground state
Practice Problem B Without referring to Figure 6.25, determine the identity of the element with the
following electron configuration:
Checkpoint 6.9 Electron Configurations and the Periodic Table
6.9.1 Which of the following electron 6.9.3 Which of the following is ad-block
Ag atom?
a) Sb
a) [Kr]S i 4cf
b) Au b) [Kr ] Si4i o
c) Ca c) [Kr]Ss14io
d) Zn d) [Xe]Si4cf
e) U e) [Xe]Ss14dlO
6.9.4 Which of the following is a p -block
6.9.2 What element is represented by the element? (Select all that apply.)
following electron configuration:
a) Pb
[Kr]Ss 24dlO Sp 5?
b) C a) Tc
c) Sr b) Br