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The electronic configuration of an eleme...

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

A

15

B

724

C

25

D

30

Text Solution

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The correct Answer is:
C
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General configuration of outermost and penultimate shell is (n - 1) s^(2) (n - 1) p^(6) (n - 1)d^(x) ns^(2) . If n = 4 and x = 5 then no. of protons in the nucleus will be