(i) Since n = 3${}^{rd}$ 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 3${}^{rd}$ period and 16${}^{th}$ group of the periodic table.

Hence, the element is Sulphur.

(ii)Since n = 4, the element belongs to the 4${}^{th}$ 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  4${}^{th}$ 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 $\left[Xe\right] 4 f^7 5d^{1}6s^2$

Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.