Question

Total no. of electrons present in an orbital having n+l = 4, its atomic number(Z) = 55.

Answer 1

Answer:

The electronic configuration of element with at.no. 105 is

1s 2,2s 4, 4s 8, ...

Also we know than angular quantum no. of sub-orbits are

for s, l=0

p, l=1

d, l=2

f, l=3

in this configuration orbitals with n+l=8 is

5f,(n+l)=5+3=8, and no of electron present in 5f is 14

6d,(n+l)=6+2=8 and no of electron present in 6d is 3

so total no of electron present in n+l=8 is 14+3 = 17

answer is 17

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