Question

What is the electronic configuration for tin?
A. $[Kr]5{{s}^{2}}5{{p}^{2}}$
B. $[Kr]4{{d}^{10}}5{{s}^{2}}5{{p}^{2}}$
C. $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{2}}$
D. $[Kr]4{{d}^{11}}5{{s}^{2}}5{{p}^{1}}$

Answer 1

Hint: The periodic table is defined as a tabular array of the chemical elements organized by atomic numbers, from the element starting with the lowest atomic number, hydrogen, to the element with the highest atomic number, oganesson.

Complete Solution :
A periodic table consists of seven horizontal rows and 18 vertical groups.
We know that an electron in an atom is characterised by a set of four quantum numbers (n, l, m and s) where the principal quantum number (n) defines the main energy level known as the shell. We have also studied about the filling of electrons into different subshells (s, p, d and f ) in an atom. This distribution of electrons into various subshells of an atom is called its electronic configuration. The location of any element in the periodic table is reflected by the quantum numbers of the last orbital filled.

a) First period : It correspond to the filling of electrons in the first energy shell i.e., (K shell), n = 1, since this energy shell has only one orbital, i.e., 1s which can accommodate only two electrons, therefore the first period contains only two elements. Hydrogen and Helium.

(b) Second period : It corresponds to the filling of electrons in the 2nd energy shell (L shell) i.e. n = 2. Since this shell has four orbitals (one 2s and three 2p) can accommodate eight electrons therefore the 2nd period contains eight elements. It starts with Li (Z = 3) which has three electrons where two electrons are in the 1s orbital and the third electron enters the 2s orbital. The next element Be(Z = 4) has four electrons with configuration $[He]2{{s}^{2}}$. From next element B (Z = 5) the 2p orbitals are filled with electrons and the L shell completes at Ne (Z = 10) with electronic configuration $[He]2{{s}^{2}}2{{p}^{6}}$.

(c)Third period : This period corresponds to the filling of electrons in the third shell i.e., n = 3. This shell has nine orbitals (one 3s, three 3p and five 3d). However according to the energy level diagram of multielectron atoms, 3d-orbitals have high energy so filled after 4s-orbital. Consequently the third period contains only four orbitals (3s and 3p) accommodating only eight electrons. Thus it has only eight elements. It begins at sodium (Z = 11) where the added electron enters the 3s orbital. Successive filling of electrons completing the 3s and 3p orbitals ends at Argon (Z = 18) with electronic configuration $[Ne]3{{s}^{2}}3{{p}^{6}}$.

(d) Fourth period : It corresponds to the filling of electrons in the fourth energy level i.e. n = 4. It starts with potassium (Z = 19) and the added electron goes to 4s-orbital which completes at calcium (Z = 20). After filling 4s orbital filling of five 3d-orbital starts that can accommodate ten electrons. Before the 4p-orbital is filled 3d orbitals are filled first being energetically more favourable and we come across the so called 3d transition series of elements. This series starts at scandium (Z = 21) with electronic configuration $[Ar]3{{d}^{1}}4{{s}^{2}}$. The 3d orbitals get completely filled at zinc (Z = 30) with electronic configuration $[Ar]3{{d}^{10}}4{{s}^{2}}$. Therefore, the filling of 4p orbitals starts at gallium (Z = 31) with electronic configuration $[Ar]3{{d}^{10}}4{{s}^{2}}4{{p}^{1}}$ and ends at krypton (Z = 36) with electronic configuration $[Ar]3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}$. Therefore in the fourth period there are nine orbitals in total which can accommodate eighteen electrons i.e., eighteen elements.
So, the correct answer is “Option B”.

Note: Tin is a chemical element with the symbol Sn and it has an atomic number 50. It is a silvery metal that characteristically has a faint yellow hue.