Question

Assign the position of the element having outer eletronic configuration
i) $n{s}^{2}n{p}^4$ for $n=3$
ii) $(n-1)d^{2}n{s}^2$ for $n=4$ and
iii) $(n-1)d^{1}n{s}^2$ for $n=6$, in the periodic table.

Answer 1

(i) Since n = 3, the element belongs to the $3^{rd}$ 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$\left(3s^2\right)$ + number of d–block groups($[Ne{]}^{10}$ + number of p–electrons$\left(3p^4\right)$

= 2 + 10 + 4

= 16

Therefore, the element belongs to the 3rd period and 16th group of the periodic table. Hence, the element is Sulphur$\left(\right[Ne{]}^{10}3s^{2}3p^4)$

(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$\left(4s^2\right)$ + number of d–block groups$\left(3d^2\right)$

= 2 + 2

= 4

Therefore, it is a 4th period and 4th group element. Hence, the element is Titanium$\left(\right[Ar{]}^{18}3d^{2}4s^2.$

(3)$(n-1)d^{1}n{s}^2$

For n = 6

= $(6-1)d^{1}6s^2$

= $5d^{1}6s^2$

Now if we write it in the order of filling the sub shells, we know that first 6s fill then 6d fill so

$6s^{2}5d^1$

Now

period = Highest principal quantum number = 6th period

Group = 1 + 2 = 3 group

And the atomic number is 57