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Question -

Assign the position of the element having outer electronicconfiguration

(i) ns2 np4 for = 3 (ii) (–1)d2 ns2 for = 4, and (iii) (n –2) f7 (n –1)d1 ns2 for = 6, in the periodictable.

 



Answer -

(i) Since n =3, the element belongs to the 3rd period.It is a p–blockelement since the last electron occupies the p–orbital.

There are four electrons in the p–orbital. Thus, thecorresponding group of the element

= Number of s–block groups + number of d–blockgroups + number of p–electrons

= 2 + 10 +4

= 16

Therefore, the element belongs to the 3rd period and 16th groupof the periodic table. Hence, the element is Sulphur.

(ii) Since n =4, the element belongs to the 4th period.It is a d–blockelement as d–orbitalsare incompletely filled.

There are 2 electrons in the d–orbital.

Thus, thecorresponding group of the element

= Number of s–block groups + number of d–blockgroups

= 2 + 2

= 4

Therefore, it is a 4th periodand 4th group element. Hence, the element is Titanium.

(iii) Since n =6, the element is present in the 6th period.It is an f –blockelement as the last electron occupies the f–orbital. It belongs to group 3of the periodic table since all f-block elements belong to group3. Its electronic configuration is [Xe] 4f7 5d1 6s2. Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence,the element is Gadolinium.

 

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