Question

Match list I (atomic number of the elements) with list II (position in the periodic table) and select the correct answer using the codes given below the lists.

List I	List II
A. 52	1. s-block
B. 56	2. p-block
C. 57	3. d-block
D. 60	4. f-block

A. A-2, B-1, C-3, D-4
B. A-2, B-1, C-4, D-3
C. A-1, B-2, C-3, D-4
D. A-1, B-2, C-4, D-3

Answer 1

To solve this we must know the elements having the given atomic numbers. From the atomic number of the elements we can determine the electronic configurations of the elements. And thus, we can then determine the block of the periodic table to which the element belongs.

Complete answer:
We know that the atomic number is defined as the number protons in the nucleus of an atom. We are given four atomic numbers 52, 56, 57 and 60.

The element having atomic number 52 is tellurium. The chemical symbol for tellurium is ${\text{Te}}$. The electronic configuration of tellurium is as follows:
$1{s^2}\,2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^4}$
As the valence shell of tellurium is 5p, tellurium belongs to the p-block of the periodic table.
Thus, the element having atomic number 52 belongs to p-block. Thus, A-2.

The element having atomic number 56 is barium. The chemical symbol for barium is ${\text{Ba}}$. The electronic configuration of barium is as follows:
$1{s^2}\,2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^2}$
As the valence shell of barium is 6s, barium belongs to the s-block of the periodic table.
Thus, the element having atomic number 56 belongs to s-block. Thus, B-1.
The element having atomic number 57 is lanthanum. The chemical symbol for lanthanum is ${\text{La}}$. The electronic configuration of lanthanum is as follows:
$1{s^2}\,2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^2}5{d^1}$
As the valence shell of lanthanum is 5d, lanthanum belongs to the d-block of the periodic table.
Thus, the element having atomic number 57 belongs to d-block. Thus, C-3.

The element having atomic number 60 is neodymium. The chemical symbol for neodymium is ${\text{Nd}}$. The electronic configuration of neodymium is as follows:
$1{s^2}\,2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^2}4{f^4}$
As the valence shell of neodymium is 4f, neodymium belongs to the f-block of the periodic table.
Thus, the element having atomic number 60 belongs to f-block. Thus, D-4.

**Thus, the correct answer is A-2, B-1, C-3, D-4.

Note:**
The electrons fill the orbitals according to Aufbau's principle. The Aufbau’s principle states that in the ground state of the atoms, the orbitals are filled with electrons in order of the increasing energies. The order of energy of different orbitals in an atom is as follows:
$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d$ and so on.
The maximum number of electrons that can be accommodated is s-orbital are 2, p-orbital are 6, d-orbital are 10 and f-orbital is 14.