**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 $n=4$, and

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

**Solution:**

(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] 4*f*7 5*d*1 6*s*2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.

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