Assign the position of the element having outer electronic configuration (i) ns2np4 for n = 3 (ii) (n - 1)d2ns2 for n = 4, and (iii) (n - 2) f7 (n - 1)d1ns2 for n = 6, in the periodic table.

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(i) Since n = 3, the element belongs to the 3rd period. It is a p–block element since the last electron occupies the p–orbital. There are four electrons in the p–orbital. Thus, the corresponding group of the element = Number of s–block groups + number of d–block groups +...
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(i) Since n = 3, the element belongs to the 3rd period. It is a p–block element since the last electron occupies the p–orbital. There are four electrons in the p–orbital. Thus, the corresponding group of the element = Number of s–block groups + number of d–block groups + number of p–electrons = 2 + 10 + 4 = 16 Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur. (ii) Since n = 4, the element belongs to the 4th period. It is a d–block element as d–orbitals are incompletely filled. There are 2 electrons in the d–orbital. Thus, the corresponding group of the element = Number of s–block groups + number of d–block groups = 2 + 2 = 4 Therefore, it is a 4th period and 4th group element. Hence, the element is Titanium. (iii) Since n = 6, the element is present in the 6th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] 4f7 5d1 6s2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium. read less
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