Chemistry

Chemical Compounds and Reactions

329 Questions

Chemical compounds and reactions involve understanding the properties, colors, and formation of various chemical substances. Questions focus on identifying precipitates, observing color changes during reactions, and naming common compounds. This topic is a fundamental part of the chemistry syllabus for competitive exams.

identifying precipitateschemical reaction colorscompound namingcolloidal particle propertiesacid base reactions

Chemical Compounds and Reactions Questions

Multiple choice chemistry substances in common use chemical from common salt compounds of carbon some useful compounds in everyday life

Bleaching powder has a strong bleaching action due to:

  1. calcium

  2. chlorine molecule

  3. nascent oxygen

  4. chloride ion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The bleaching powder has a strong bleaching action due to the formation of nascent oxygen.

When bleaching powder reacts with a small amount of acid, it gives nascent oxygen. 
Thus, it acts as an oxidizing agent and thus acts as a bleaching agent.

$2CaOCl _2 +H _2SO _4 \rightarrow CaCl _2+CaSO _4+2HOCl$

$2HOCl \rightarrow 2HCl+2[O]$

Multiple choice chemistry substances in common use chemical from common salt compounds of carbon some useful compounds in everyday life

The real bleaching agent present in bleaching powder is:

  1. chlorine

  2. oxygen

  3. CaO

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Calcium oxychloride (CaOCl2) is known as bleaching powder. In the presence of small amount of dilute acid, it gives nascent oxygen which acts as an oxidising and bleaching agent. When it is treated with excess of dilute acids, chlorine is evolved which is responsible for the bleaching action of bleaching power.

Multiple choice chemistry substances in common use chemical from common salt compounds of carbon some useful compounds in everyday life

On long standing, bleaching powder undergoes auto-oxidation into:

  1. $Ca(ClO _3) _2, CaCl _2$
  2. $CaO, CaCl _2$
  3. $CaClO, CaClO _3$
  4. $CaClO _2, CaCl _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

On long standing, bleaching powder undergoes auto-oxidation into calcium chlorate $Ca(ClO _3) _2$ and $ CaCl _2$.
 $6CaOCl _2 \rightarrow 6 Ca(ClO _3) _2  + 5 CaCl _2$.
The bleaching powder is $CaOCl _2$

Multiple choice chemistry substances in common use chemical from common salt compounds of carbon some useful compounds in everyday life

On long standing, bleaching powder undergoes auto oxidation.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

On long standing, bleaching powder undergoes auto-oxidation which results in the formation of calcium chloride and calcium chlorate.


$6 CaOCl _2 \rightarrow5CaCl _2   + 6 Ca(ClO3) _2 $

Hence, the given statement is $true$

Multiple choice chemistry substances in common use chemical from common salt compounds of carbon some useful compounds in everyday life

Substance $X$:
(I) reacts with $H _2S$ to produce white turbidity.
(II) changes light green solution of $FeSO _4$ into yellow colour.
(III) reacts with moisture to give pungent smelling gas.
Hence $X$ is:

  1. $KMnO _4$
  2. $K _2Cr _2O _7$
  3. $SO _2$
  4. $CaOCl _2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$CaOCl _2+H _2O\rightarrow Ca(OH) _2+\underset {(\text {Pungent smelling gas})}{Cl _2}\uparrow$
$H _2S+Cl _2\rightarrow 2HCl+\underset {(\text {White turbidity})}{S}$
$\underset {(\text {Light green})}{2FeSO _4}+H _2SO _4+Cl _2\rightarrow Fe _2(SO _4) _3+2HCl$

Multiple choice chemistry substances in common use chemical from common salt compounds of carbon some useful compounds in everyday life

The bleaching property of chlorine water is due to:

  1. formation of nascent oxygen

  2. formation of nascent chlorine
  3. formation of ${O} _{3}$
  4. reducing property of ${Cl} _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ Cl } _{ 2 }+{ H } _{ 2 }O\rightarrow HCl+HOCl\rightarrow 2HCl+\left[ 0 \right] $
$\left[ 0 \right] $-Nascent oxygen
Chlorine water act as strong oxidizing agent because of nascent oxygen produced in water.
Multiple choice chemistry p- block elements-ii anomalous properties of nitrogen nitrogen - 15 group group 15 elements

Which of the following pairs is obtained on heating ammonium dichromate?

  1. $\displaystyle { N } _{ 2 }$ and $\displaystyle { H } _{ 2 }O$
  2. $\displaystyle { N } _{ 2 }O$ and $\displaystyle { H } _{ 2 }O$
  3. $\displaystyle { NO } _{ 2 }$ and $\displaystyle { H } _{ 2 }O$
  4. $\displaystyle NO$ and $\displaystyle { NO } _{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ammonium dichromate on heating gives nitrogen, chromic oxide and water.
$\displaystyle { \left( { NH } _{ 4 } \right)  } _{ 2 }{ Cr } _{ 2 }{ O } _{ 7 }\underrightarrow { \Delta  } { N } _{ 2 }+{ Cr } _{ 2 }{ O } _{ 3 }+4{ H } _{ 2 }O$

Multiple choice chemistry p- block elements-ii anomalous properties of nitrogen nitrogen - 15 group group 15 elements

Select the correct statement for brown ring test.

  1. Nitrates are converted to $NO$
  2. $NO$ forms complex with $Fe$
  3. $Fe^{2+}$ acts as reducing agent here.

  4. Brown colour is due to the complex formation.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation
$ 2HNO _3+ 3H _2SO _4 + 6FeSO _4 \rightarrow 3Fe _2(SO _4) _3 + 2NO + 4H _2O $ $[Fe(H _2O) _6]SO _4+NO \rightarrow [Fe(H _2O) _5(NO)]SO _4+H _2O$ Nitrates are converted into nitric oxide. $NO$ forms complex with $Fe$. $Fe^{2+}$ acts as reducing agent. $[Fe(H _2O) _5(NO)]SO _4$ is brown coloured complex formed.
Multiple choice chemistry p- block elements-ii anomalous properties of nitrogen nitrogen - 15 group group 15 elements

Alkali metals dissolve in liquid ammonia and produces _______.

  1. deep blue solution

  2. a solution which are conducting in nature

  3. red coloured solution
  4. Both $1$ and $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Alkali metal when reacts with liquid ammonia reaction is given by,


$ M +(x+y) NH _3\rightarrow M^+(NH _3)x + e^{-}(NH _3) _y$

 this electrons is free and gives very good conductivity as well blue colour.