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1_ Assume that you travel through spy tO a gtrangc univteEr2. where the allowable values for the quantum numbers are quite different from those in our own universe_ In this new universe„ s orbitals...

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1_ Assume that you travel through spy tO a gtrangc univteEr2. where the allowable values for the quantum numbers are quite different from those in our own universe_ In this new universe„ s orbitals may anti in a toth.I Of 2 electrons,. orbitals may conthin a total of 4 electrons,. and d orbitals. may contain a tab.' of electrons (as opposed to our 2„15„ and 10.. tespectively). The other rules of electron configuration in this new univerSe are the Si.rne as in Our Own urtiverSe_ AriSwtr the- fallOveing about elements in this new universe.
a. How many eirbritalS weluld these be. in each of the s, and d level? b. Write the rule. for the allowable values of in e What is the ground s.ilite electronic configuration of element number 10 in this new universe? d. ghat would be the ateirniC number of the first tranition metal hi this nCw univurse? You can assume the s.airrie. order of energy levels as in our univere.
2. a. Cak-ulatc the wavelength of a 5001:1 kg elephant travelling at 1_50 m s b. Comment ork the wave nature of this elephant.
3. a.What is the wavelength (in nanometres) of blue light which has a fropericy of 6_62 X 'OIL" -]?
b. Cakulatt the energy, in joules., of a photon with this frequency_
4. For each of the following specks: Cu}', Tc Sn, V •1-.. SC2- a. Give the. electron. configuration_ b. Give the. numbee of unpaired eled-rons. r_ Indicate which would be diarnagn.ai.c.
Answered Same Day Dec 29, 2021

Solution

David answered on Dec 29 2021
111 Votes
1)
a) Every o
ital occupied by maximum of 2 electrons, hence there should be 1 o
ital for “s”(2e
electrons) o
ital, 2 o
itals for “p” (4 electrons), 3 o
itals (6 electrons) for “d”
) Since , ml = l+1 to l-1
for s o
ital ml = 0
for p o
ital ml = +1 and -1
for “d” o
ital , ml = +2,+1,-1 and -2
c) 2 2 4 21 2 2 3s s p s
d) 2 2 4 2 4 2 1Atomicnumbershould be17 1 2 2 3 3 4 3s s p s p s d

2)
a)
  
34
386.626 10 J.s 8.834 10
5000kg 1.50m/s
h
m
m v


   


) The velocity of the elephant is very less because of its greater weight, the elephant’s wave nature
is thus very less and wave length obtained is thus less.
3) a)
8
14
3.0 10
453
6.62 10
c m s
nm


  


)
  
 
34 8
19
7
6.626 10 . 3 10
4.38 10 J
4.53 10
J s m sh c
E




 
   


4) a)
 
 
 
 
2 9
4 2
1 1 2
2 10 2 6
10 2 2
5
2
3
3
[Xe]4 5 6
[Kr]4 5 5
[Kr]4 5 5
Cu Ar d
Mo Kr d
Ce f d s
Te d s p
Sn d s p
V A
Sc A













) Number of unpaired electrons in the ions are –
2+
4+
2
5+
2+
Cu 1
Mo 2
Ce 2
Te 0
Sn 2
V 0
Sc 0
e
e
e
e
e
e
e



...
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