Identify whether the electron configurations are correct or incorrect. If they are incorrect, explain why and write the corrected configuration. · Fe: 1s?2s2p?3s²3p²4s²3d? . In: 1s252p°3s?3p 3d104s²4p64d105s²5p1 • K: 1s22s2p°3s²3p64s1

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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We need to find an element that is able to
have a lot of un-paired electrons. The more
degenerate (same energy) orbitals a
sublevel/subshell has, the more unpaired
electrons the atom can contain. The periodic
table below has highlighted all of the
elements that are filling electrons into
sublevels/subshells with degenerate orbitals
(multiple orbitals of equal energy, such as the
three p orbitals in a p subshell).
PERIODIC TABLE OF THE ELEMENTS
H
Не
17
VIA
14
15
VA
VA
VIA
Li Be
В
CN
F
Ne
694
14
" 12 Al Si PS CI Ar
Na Mg
22.9 2430
19 20
K Ca Sc Tiv Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
* ve ve
ve
VI
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
Cs Ba 'La Hf Ta w Re Os Ir Pt Au Hg TI Pb Bi Po At Rn
87
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg
*Lanthanide Series Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
94 95 % 97 8 9 100 101 102
"Actinide Series Th Pa U x Pu Am Cm Bk Cf | Es Fm Md No Lr
90 91 92
93
214 204
Explain which regions of the periodic table
would be best for holding multiple unpaired
electrons. Make sure to discuss what
subshell is being filled with electrons in each
of those regions of the periodic table
Transcribed Image Text:We need to find an element that is able to have a lot of un-paired electrons. The more degenerate (same energy) orbitals a sublevel/subshell has, the more unpaired electrons the atom can contain. The periodic table below has highlighted all of the elements that are filling electrons into sublevels/subshells with degenerate orbitals (multiple orbitals of equal energy, such as the three p orbitals in a p subshell). PERIODIC TABLE OF THE ELEMENTS H Не 17 VIA 14 15 VA VA VIA Li Be В CN F Ne 694 14 " 12 Al Si PS CI Ar Na Mg 22.9 2430 19 20 K Ca Sc Tiv Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr * ve ve ve VI Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba 'La Hf Ta w Re Os Ir Pt Au Hg TI Pb Bi Po At Rn 87 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg *Lanthanide Series Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 94 95 % 97 8 9 100 101 102 "Actinide Series Th Pa U x Pu Am Cm Bk Cf | Es Fm Md No Lr 90 91 92 93 214 204 Explain which regions of the periodic table would be best for holding multiple unpaired electrons. Make sure to discuss what subshell is being filled with electrons in each of those regions of the periodic table
D
Magnets are magnetic because the electrons
are unpaired and all the spins point in the
same direction. If you can accomplish this
over a solid material, then you can have a
magnet! The more unpaired spins, the
stronger the magnet!
D Question 1
We'll get to choosing a magnet material, but
fırst we need to have a solid understanding
of where the electrons reside in each
element. Let's look at the following three
electronic configurations, generated by a
summer intern the day before they finished.
Identify whether the electron configurations
are correct or incorrect. If they are incorrect,
explain why and write the corrected
configuration.
· Fe: 1s22s?2p?3s²3p²4s?3d?
1 In:
1s22522p°3s?3p°3d104s²4pº4d105s²5p1
• K: 1s22s22p°3s²3pº4s1
Transcribed Image Text:D Magnets are magnetic because the electrons are unpaired and all the spins point in the same direction. If you can accomplish this over a solid material, then you can have a magnet! The more unpaired spins, the stronger the magnet! D Question 1 We'll get to choosing a magnet material, but fırst we need to have a solid understanding of where the electrons reside in each element. Let's look at the following three electronic configurations, generated by a summer intern the day before they finished. Identify whether the electron configurations are correct or incorrect. If they are incorrect, explain why and write the corrected configuration. · Fe: 1s22s?2p?3s²3p²4s?3d? 1 In: 1s22522p°3s?3p°3d104s²4pº4d105s²5p1 • K: 1s22s22p°3s²3pº4s1
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