Question

Consider the following electronic configuration of an element (P) :
$\text{ }\left[ \text{Xe} \right]\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{1}}}\text{6}{{\text{s}}^{\text{2}}}\text{ }$
The correct statement about the element ‘P’ is?
A) It belongs to $\text{ }{{\text{6}}^{\text{th}}}\text{ }$ a period and $\text{ }{{\text{1}}^{\text{st}}}\text{ }$ group
B) It belongs to $\text{ }{{\text{6}}^{\text{th}}}\text{ }$ period and group
C) It belongs to $\text{ }{{\text{6}}^{\text{th}}}\text{ }$ period and $\text{ }{{\text{3}}^{\text{rd}}}\text{ }$ group
D) None of these

Answer 1

Hint: The periodic table is a regular arrangement of elements. The period is a horizontal arrangement of elements. The electronic configuration of an element is used to determine its position in the periodic table. The principal quantum number ‘n’ gives an idea about the highest occupied energy level or the period of the element.

Complete step by step answer:
The periodic table is a systematic arrangement of the elements according to their atomic number.
The periodic table is composed of:
1) Vertical column OR Groups
2) Horizontal rows OR Period
In groups, the elements are arranged in increasing atomic numbers, such that each atom has the same number of valence electrons. As we move along the group, the general valence shell electron configuration remains unaffected. For example, for group II, the general valence shell configuration is given as$\text{ n}{{\text{s}}^{\text{2}}}\text{ }$ , where n is the number of shells. The period is the horizontal arrangement of elements in the periodic table. The elements are arranged based on the principal quantum number. The principal quantum number ‘n’ defines the highest energy electron in an element. Along the period, the electrons enter the same shell. We are given the electronic configuration of an element P. It is given as follows,
$\text{ P = }\left[ \text{Xe} \right]\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{1}}}\text{6}{{\text{s}}^{\text{2}}}\text{ }$
The last electrons for the element P enters the $\text{ 6s }$ or the sixth energy shell. Here, the principal quantum for the element is 6. Thus, it is a sixth-period element. The $\text{ 6s }$which is the last shell contains the two electrons in it. We know that group II elements have an electronic configuration$\text{ n}{{\text{s}}^{\text{2}}}\text{ }$. Thus given elements should be a group II element. It is an f-block element .Its name is lutetium and its atomic number is 71.$\text{ Lu = }\left[ \text{Xe} \right]\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{1}}}\text{6}{{\text{s}}^{\text{2}}}\text{ }$
It belongs to the f-block and a 6-period element. None of the given statements correctly defines the position of elements.

Hence, (D) is the correct option.

Note: Note that f block or the inner transition elements are situated in the sixth and seventh period. This is in group II, such that the 14 elements are given a single place in the periodic table. Here, the last shell electrons enter the f orbitals. The general electronic configuration of the f block elements is $\text{ n}{{\text{f}}^{\text{1-14}}}\text{ (n+1)}{{\text{d}}^{\text{1}}}\text{(n+2)}{{\text{s}}^{\text{2}}}\text{ }$ .