Question

Assign the position of the element having outer electronic configuration
(i) n$s^2$ n$p^4$ for n = 3

(ii) (n - 1)$d^2$ n$s^2$ for n = 4,

(iii) (n - 2) $f^7$ (n - 1)$d^1$ n$s^2$ for n = 6, in the periodic table.

Answer 1

(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

It means $3s^{2}3p^4$.
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

It means (4 - 1)$d^2$ 4$s^2$ or $3d^{2}4s^2$ .
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] 4$f^7$ 5$d^1$ 6$s^2$.

It means (6 - 2) $f^7$ (6 - 1)$d^1$ 6$s^2$ or $4f^{7}5d^{1}6s^2$

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