Question

The element with atomic number 35 belongs to:
A) s-block
B) p-block
C) d-block
D) f-block

Answer 1

Let us understand what decides the place of an atom in the periodic table. The elements are arranged in periodic tables, their location is reflected by the quantum numbers filled in its last orbitals. The distribution of electrons into orbitals of an atom is called its electronic configuration. The sum of all its electrons in a neutral atom gives its atomic number.

Complete answer:
There are certain rules that the elements of the periodic table follow, therefore the elements with similar properties are placed together. The aufbau principle and the electronic configuration of atoms provide a theoretical foundation for the periodic classification. The elements in the particular group or family have similar chemical behavior due to the same number and electrons distribution in the outermost orbitals.
The elements are classified into four blocks s,p,d,f block. Therefore let us write electronic configuration to the atomic number 35 and check which orbital contains the outermost electrons.
The electronic configuration of the element with atomic number 35 is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^5}$ the last valence electrons enter into p orbital. Hence the option A is incorrect. The last valence electrons enter into the p-orbital hence they are classified under p-block elements. From the periodic table we know that Atomic number 35 belongs to Bromine.
We know that electrons enter into the orbitals in a fixed trend like 1s,2s,2p,3s,3p,3d,4s,4p… the electronic configuration of bromine is $4s^2,4p^5$ ie, $ns^2np^5$ belongs to the configuration of the halogen family which belongs to 17th group, which is p-block.
The p-block comprise group 13 to 18 and these together with the s-block elements are called representative elements or main group elements. Its outermost electronic configuration varies from $n{s^2}n{p^1} - n{s^2}n{p^6}$. the outermost electronic configuration for s block,d block and f block are:
$n{s^1} - n{s^2},(n - 1){d^{1 - 10}}n{s^{0 - 2}},(n - 2){f^{1 - 14}}(n - 1){d^{0 - 1}}n{s^2}$ respectively.

Hence we can conclude that the correct option is B.

Note: We must know that group 17 and group 16, halogens and chalcogens respectively have highly electron gain enthalpies and readily add one or two electrons respectively to attain stable noble gas configuration.