Assign the position of the element having outer electronic configuration

(i) nsnp4 for n=3

(ii) (n-1)dns2 for n=4, and

(iii) (n-2)f7 (n-1)dns2 for n=6, in the periodic table.

Asked by Pragya Singh | 1 year ago |  59

##### Solution :-

(i) Here, n = 6 so the element is in the sixth period. The element is an ‘f-block element’ because the last electron enters in the f-orbital. As the f-block elements are in the third group. They are having electronic configuration.$$[Xe]4f^{7}5d^{1}6s^{2}$$ So, the atomic number can be calculated as 54 + 7 + 2 + 1 = 64. Thus, the required element is Gadolinium.

(ii) Here, n = 3 so the element is in the third period. The element is in ‘p-block element’ because the last electron enters in the p-orbital. It contains 4 electrons in p-orbital.

For a group of the element

= No. of s – block groups + No. of d – block groups + No. of p – block groups

= 2 + 10 + 4 = 16

Thus, the given element is in the third period and sixteenth the group in the periodic table. Thus, the required element is Sulphur.

(iii) Here, n = 4 so the element is in the fourth period. The element is in ‘d-block element’ because the last electron enters in the d-orbital but this orbital is incompletely filled. It contains 2 electrons in d-orbital.

For a group of the element

= No. of s – block groups + No. of d – block groups

= 2 + 2 = 4

Thus, the given element is in the fourth period and fourth the group in the periodic table. Thus, the required element is Titanium.

Answered by Abhisek | 1 year ago

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