Question

The number of electrons in an atom with atomic number $105$ having $(n+l)=8$ are:

• A. $30$
• B. $17$
• C. $15$
• D. $21$

Answer 1

The correct option is B $17$

The electronic configuration of element with at.no. 105 is
$1s^2,2s^2{p}^6,3s^2{p}^6{d}^{10},4s^2{p}^6{d}^{10}4f^{14},5s^2{p}^6{d}^{10}{f}^{14},6s^2{p}^6{d}^{3}7s^2$
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 $14+3=17$