Question

An element ${\text{X}}$ combines with hydrogen to form a compound ${\text{X}}{{\text{H}}_3}$ . The element ${\text{X}}$ is placed on the right side of the periodic table. What is true about the element ${\text{X}}$ ?
(A) Has $3$ valency
(B) Is a metal and is solid.
(C) Is a non- metal and is a gas.
(D) Has $5$ valence electrons
${\text{X}}{{\text{H}}_3}$ reacts with water to form a basic compound.
(a) A, B and C
(b) A, C and D
(c) C, D and E
(d) E, A and B

Answer 1

Hint : Non- metals are reacted with hydrogen to form covalent hydrides. Where covalent hydrides are compounds of hydrogen with similar electronegative elements formed by covalent bonds. Covalent bond involves sharing of electron pairs between atoms. The pair of electrons that participate in this type of bonding is called a shared pair or electron pair.

Complete Step By Step Answer:
Here it is given that the element ${\text{X}}$ is placed in the right side of the periodic table. So that it is clear that the element is non metal & most of them exist in a gaseous state. because in periodic table nonmetals occupy the upper right side, so that elements occupying the right side of the periodic table have higher effective nuclear charge and thereby stabilize electrons more effectively and this will lead to localized covalent bonding and formation of molecules.
The element ${\text{X}}$ is said to combine with $3$ hydrogen atoms to form ${\text{X}}{{\text{H}}_3}$ . The chemical reaction can be written as
${{\text{X}}_2}\,{\text{ + }}\,{\text{3}}\,{{\text{H}}_2}\, \to \,2\,{\text{X}}{{\text{H}}_3}\,$
For one ${\text{X}}$ atom $3$ hydrogen atom needed for the formation of one molecule of ${\text{X}}{{\text{H}}_3}$ . Since the compound is formed by covalent bonding, 1 electron is shared by each hydrogen atom to one ${\text{X}}$ atom to achieve $8$ electrons in its valence shell thereby achieving stability. From the above explanation it is clear that the element ${\text{X}}$ has valency 3. That is the element ${\text{X}}$ needs $3$ electrons in order to achieve stable noble gas configuration. So It is clear the valence electrons that the element ${\text{X}}$ contains is $5$ . Hence the element comes under a nitrogen group that is a group $15$ which has $5$ valence electrons. So here we will take ${\text{N}}{{\text{H}}_3}$ as an example. In ${\text{N}}{{\text{H}}_3}$ the three hydrogen atoms form a covalent bond with nitrogen. Due to the difference in electronegativity between nitrogen and hydrogen atoms , more electronegative nitrogen atoms attract the electrons of hydrogen atoms there by completing the octet. After the three valence electrons of nitrogen form a covalent bond with three hydrogen , nitrogen still has two valence electrons left , which is called a lone pair of nitrogen.
When ${\text{X}}{{\text{H}}_3}$ react with water ${\text{X}}{{\text{H}}_3} + {{\text{H}}_2}{\text{O}} \rightleftharpoons {\text{X}}{{\text{H}}_4}^ + {\text{O}}{{\text{H}}^ - }$
Here ${\text{X}}{{\text{H}}_4}{\text{OH}}$ act as a base , when ${\text{X}}{{\text{H}}_3}$ react with water , water molecule donate a ${{\text{H}}^ + }$ ion to the ${\text{X}}{{\text{H}}_3}$ molecule and forms ${\text{X}}{{\text{H}}_4}^ + {\text{O}}{{\text{H}}^ - }$ . According to the Bronsted- Lowry theory an acid is a proton donor and a base is proton acceptor. So here ${\text{X}}{{\text{H}}_3}$ act as a base and ${{\text{H}}_2}{\text{O}}$ act as a acid. ${\text{X}}{{\text{H}}_4}^ +$ is the conjugate acid of ${\text{X}}{{\text{H}}_3}$ and ${\text{O}}{{\text{H}}^ - }$ be the conjugate base of ${{\text{H}}_2}{\text{O}}$ .
So here the correct statements about element ${\text{X}}$ is
It is non metal and is gas
It has valency of $3$
It has $5$ valence electrons
So the correct answer is option (b).

Note :
The hydrides of non- metals on the periodic table are electronegative characters. Which means that they are less capable of donating an electron and what to keep them because their electron orbital becomes fuller. Instead of donating ${{\text{H}}^ - }$ ions they donate ${{\text{H}}^ + }$ ions.