What is the correct config?
3d 4s or 4s 3d
And pls also explain why its filled before. And also who will lose electron first?
Answers
Answered by
1
4s 3d is correct
This is because electrons are filled in subshell from lower state to higher
And 4s has less energy as compared to 3d
You can compare the energy by comparing the (n+l) value of 4s and 3d
n is the principal quantum number and l is the azimuthal quantum number
For 4s
n=4 and l=0
so (n+l)=4
For 3d
n=3 and l=2
so (n+l)= 5
Thus the energy of 4s is less than 3d and hence it is filled first.
But 4s will lose electron first as the n value is more, it is the outermost orbit and thus will lose electron first.
HOPE IT HELPS..... :-)
This is because electrons are filled in subshell from lower state to higher
And 4s has less energy as compared to 3d
You can compare the energy by comparing the (n+l) value of 4s and 3d
n is the principal quantum number and l is the azimuthal quantum number
For 4s
n=4 and l=0
so (n+l)=4
For 3d
n=3 and l=2
so (n+l)= 5
Thus the energy of 4s is less than 3d and hence it is filled first.
But 4s will lose electron first as the n value is more, it is the outermost orbit and thus will lose electron first.
HOPE IT HELPS..... :-)
AR17:
elements that have exceptional electronic configuration are chromium, Copper, Niobium, Molybdenum, Technitium, Ruthenium, Rhodium, Palladium, Silver, Tungsten, Platinum and gold....
for Cu the electronic configuration according to general rule must be 1s2 2s2 2p6 3s2 3p6 4s2 3d9
But one electron from the 4th shell jumps to 3rd shell's d orbital to attain stability
It would be 1s2 2s2 2p6 3s2 3p6 4s1 3d10
hope you undeestood....
Okay so to attain stability copper fills its d subshell and hence its an exception
yup...
Thanks now its clear!
fills the 3d orbital and 4s orbital is half filled too
:-)
Answered by
0
4s 3d is the correct answer
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