5. The elements Si, Cu, O, Y, Ba, Tl, and Bi are all found in high-temperature c
ID: 571283 • Letter: 5
Question
5. The elements Si, Cu, O, Y, Ba, Tl, and Bi are all found in high-temperature ceramic superconductors. For these atoms, (a) write the expected ground state electron configuration (b) write the ground state noble gas electron configuration (c) write the complete set of quantum numbers for the ground state valence electrons for each atom (d) state whether the ground-state atom is diamagnetic or aramagnetic and if it is paramagnetic, state the number of unpaired electrons in the ground state atom.Explanation / Answer
a)
Si -14 - 1s2 2s2 2p6 3s2 2p2
Cu -29 - 1s2 2s2 2p6 3s2 2p6 3s2 3p6 4s1 3d10
O - 8 - 1s2 2s2 2p4
Y - 39 - 1s2 2s2 2p6 3s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1
Ba - 56 - 1s2 2s2 2p6 3s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5s2 5p6 6s2
Tl - 81 - 1s2 2s2 2p6 3s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 4f14 5s2 5p6 5d10 6s2 6p1
Bi - 83 - 1s2 2s2 2p6 3s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 4f14 5s2 5p6 5d10 6s2 6p3
b]
Si - [Ne] 3s2 3p2
Cu - [Ar] 3d10 4s1
O - [He] 2s2 2p4
Y - [Kr] 4d1 5s2
Ba - [Xe] 6s2
Tl - [Xe] 4f14 5d10 6s2 6p1
Bi - [Xe] 4f14 5d10 6s2 6p3
c]
Quantum numbers for valence electrons
n = represents outer shell number
l = orbital quantum number
m = magnetic quantum number
S = spin qunatum number = (+/-) 1/2
l = 0 for s ; 1 for p and 2 for d and so on
m = - l to +l values
for s ,
m = 0
for p ,
m = -1 , 0 , 1
for d ,
m = -2 , -1 , 0 , 1 , 2
I have given all details ..please write yourself using the information
d]
Unpaired electrons rises to magnetism
Among the given elements
Cu , Tl and Bi have unpaired electrons in their outershells
Y has unpaired electron in 4th shell but it is not outer shell (so no magentism)
Si --> Diamgnetic
Cu - > Paramagnetic - one unpaired electron
O - diamagnetic
Y -> Diamagnetic
Ba -> diamagnetic
Tl -> paramagnetic - one unpaired electron
Bi -> paramagnetic - 3 unpaired electrons
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