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

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

Compound A reacts with NH4Cl and forms a compound B. Compound B reacts with H2O

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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: