Question:
Compound A reacts with $NH _4 Cl$ and forms a compound B.
Compound B reacts with $H _2 O$ and excess of $CO _2$ to form compound $C$ which on passing through or reaction with saturated $NaCl$ solution forms sodium hydrogen carbonate Compound $A , B$ and $C$, are respectively.
Compound A reacts with $NH _4 Cl$ and forms a compound B.
Compound B reacts with $H _2 O$ and excess of $CO _2$ to form compound $C$ which on passing through or reaction with saturated $NaCl$ solution forms sodium hydrogen carbonate Compound $A , B$ and $C$, are respectively.
Updated On: Oct 1, 2024
- $CaCl _2, NH _4^{+},\left( NH _4\right)_2 CO _3$
- $CaCl _2, NH _3, NH _4 HCO _3$
- $Ca ( OH )_2, NH _3, NH _4 HCO _3$
- $Ca ( OH )_2, NH _4^{+},\left( NH _4\right)_2 CO _3$
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The Correct Option is C
Solution and Explanation
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Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
- P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.
P block elements consist of:
- Group 13 Elements: Boron family
- Group 14 Elements: Carbon family
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- Group 18 Elements: Neon family