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falgunichapekar:
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6. C)
8.D) n=3 to n=5
10.A)
11.D)
13.C)
8.D) n=3 to n=5
10.A)
11.D)
13.C)
8. If we calculate using the energy formula then the max energy req will be from n=3 to n= 5
Cu= 29
Electronic configuration
[Ar] 3d10 4s1
Because 3d9 4s2 would be partially filled which would be unstable but 3d10 4s2 is stable
Electronic configuration
[Ar] 3d10 4s1
Because 3d9 4s2 would be partially filled which would be unstable but 3d10 4s2 is stable
Thanks
Sry 3d10 4s1 is stable
Welcome
13 th one
13. Cl= 17
Electronic configuration
= 1s2 2s2 2p6 3s2 3p5
So unpaired electrons present in 3p5
A) n=2 l=1 so 2p orbital
B) n=2 l=0 so 2s orbital
C) n=3 l=1 so 3p orbital
D) n=3 l=0 so 3s orbital
Thus C) is the correct answer
Electronic configuration
= 1s2 2s2 2p6 3s2 3p5
So unpaired electrons present in 3p5
A) n=2 l=1 so 2p orbital
B) n=2 l=0 so 2s orbital
C) n=3 l=1 so 3p orbital
D) n=3 l=0 so 3s orbital
Thus C) is the correct answer
I think u understood now??
Ya
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