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.

Answer 1

Hint: To find the position of element across the period to check the number of n and to find the position of element along the group use the equation:
 Group= Number of s-block group + number of d –block group + number of p electrons + number of f –block group

Complete step by step answer:
Periods are the horizontal rows of the periodic table and the number of periods in a periodic table is 7.
In a period all the elements have the same number of shells.
Groups are vertical rows of the periodic table and the number of groups in a periodic table is 18.
In the question we are asked to find the position of the element in a periodic table.
To find the position of an element in a period check the highest energy level of the shell or the principal quantum number (n).
To find the group of an element check the number of valence electron of an element or the numbers of electron present in the highest energy level of an element,
Group= Number of s-block group + number of d –block group + number of p electrons
 Let us take each case one by one,

(i) $n{{s}^{2}}n{{p}^{4}}$ for n = 3,
 In this case the value of n is given as 3 , so the elements will belong to the 3rd period. The elements belong to the p- block because the last electrons are present in the p-orbital. The number of electrons in the p-orbital are 4. Thus the group of the element will be,
 Group= Number of s-block group + number of d –block group + number of p electrons
\[=3{{s}^{2}}+{{[Ne]}^{10}}+3{{p}^{4}}\]
\[=2+10+4=16\]

Thus the element belongs to 3rd period and 6th group in a periodic table and therefore the element is sulphur $({{[Ne]}^{10}}3{{s}^{2}}3{{p}^{4}})$.

(ii) $(n-1){{d}^{2}}n{{s}^{2}}$ for n = 4,
In this case the value of n is 4, so the element belongs to the 4th period. This element belongs to d-orbital because the number of electrons in d-orbital is incomplete; it has only 2 electrons in it.
Hence the group of the element will be,
 Group= Number of s-block group + number of d –block group
\[=4{{s}^{2}}+3{{d}^{2}}\]
\[=2+2=4\]

Thus the element belongs to the 4th period and 4th group of the periodic table. Therefore the element is Titanium $({{[Ar]}^{18}}3{{d}^{2}}4{{s}^{2}})$.

(iii) $(n-2){{f}^{7}}(n-1){{d}^{1}}n{{s}^{2}}$ for n = 6
In this case the value of n is 6 , so the element belongs to the 6th period of the periodic table. The element s belongs to the f block because the last electron is filled in the f-orbital. In a periodic table all the f-block elements belong to group 3.

Thus the elements belong to the 6th period and 3rd group of the periodic table. Therefore the element will be Gadolinium $({{[Xe]}^{54}}4{{f}^{7}}5{{d}^{1}}6{{s}^{2}})$.

Note: To write the electronic configuration of elements having atomic number more than 11 we usually use noble gas shorthand rule. Because the noble gases are already stable and there is no need of losing or gaining electrons. In this way we can write the electronic configuration of any element easily and it will be shorter.