Identify the period and group of the element having outer electronic configuration as(i) ns2np4for n = 3(ii) (n – 1) d2ns2for n = 4 and(iii) (n – 2) f 7(n  1) d1ns2for n = 6 in the periodic table.

(i) For ns2np4 where n = 3, the element is in the p-block. The period is determined by the value of n, so the period is 3. The group is determined by the number of electrons in the outermost shell, which is 2 (from ns2) + 4 (from np4) = 6. Therefore, the group is 16 (10 + the number of electrons in the p orbital).

(ii) For (n – 1) d2ns2 where n = 4, the element is in the d-block. The period is determined by the value of n, so the period is 4. The group is determined by the number of electrons in the outermost shell, which is 2 (from ns2) + 2 (from d2) = 4. Therefore, the group is 4 (the number of electrons in the d orbital + 2).

(iii) For (n – 2) f 7(n – 1) d1ns2 where n = 6, the element is in the f-block. The period is determined by the value of n, so the period is 6. The group in the f-block is not determined by the number of electrons in the outermost shell, but rather by the specific electron configuration. In this case, the element is in the lanthanide series.