The electronic configuration of an element in ultimate and penultimate orbitals is $(n\ \ 1)s^{2}(n\ \ 1)p^{6}(n\ \ 1)d^{x}\ ns^{2}$. If $n\ \ 4\$and$\ x\ \ 5$ then number of protons in the nucleus is

Question

Dr. Abubacker · Accepted Answer

(c)
When $n = 4\ and\ x = 5$ then electronic configuration can be written as
$(4 - 1)s^{2}(4 - 1)p^{6}(4 - 1)d^{5}4s^{2}$
This electronic configuration represents Mn and its atomic number is 25. Hence, number of protons are 25 in its nucleus.

Q. The electronic configuration of an element in ultimate and penultimate orbitals is $(n\ \ 1)s^{2}(n\ \ 1)p^{6}(n\ \ 1)d^{x}\ ns^{2}$. If $n\ \ 4\$and$\ x\ \ 5$ then number of protons in the nucleus is

Explanation

Reviewed by Dr. Abubacker

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