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How many electrons in an atom with atom number 107 can have \mathrm{(n+\ell)=8}

Option: 1

17


Option: 2

19


Option: 3

30


Option: 4

Unpredictable


Answers (1)

best_answer

Atomic No \rightarrow 107

Outer electronic configuration \mathrm{=5 f^{14} 6 d^{5} 1s^2}

                   \mathrm{n+\ell=8}

Possible combination \mathrm{=5+3\;}\& \; 6+2

i.e 5f & 6d 

\left.\begin{array}{c} 3+5 \rightarrow 3 h \\ 4+4 \rightarrow 4 g \end{array}\right] \rightarrow \text{does not exist}

\mathrm{No. of\; e_{s}^{-} in\; 5 f=14}

\mathrm{No. of\; e_{s}^{-} in\; 6 d=5.}

Total electrons = 14+5=19

Option (2) is the right answer.

Posted by

Devendra Khairwa

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