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 separation of matter crystals and crystallisation filtration and crystallisation other methods of separation

While obtaining pure copper sulphate by crystallization, the solution is saturated is indicated by :

  1. change in color

  2. change in consistency

  3. formation of bubbles

  4. formation of crystals near rod
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

During purification of copper sulphate by crystallization, the impure sample is mixed with water and then heated to saturation. The saturation is indicated by dipping glass rod in solution from time to time. Small crystals are formed near the glass rods which indicates saturation.

Multiple choice chemistry separation of matter crystals and crystallisation filtration and crystallisation other methods of separation

Copper sulphate crystals are of which color after purification?

  1. White

  2. Black

  3. Blue

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

To crystallise copper sulphate, it is dissolved in water and a small quantity of dilute sulphuric acid is added to prevent the hydrolysis of copper sulphte. The impurities left behind in the solution are removed by filtration. The filtrate is concentrated to the crystallisation point and then cooled. On cooling, transparent blue crystals of copper sulphate separate.

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

When three parts of conc. HCl and one part of conc. $HNO _{3}$ is mixed, a compound 'X' is formed. The correct option related to 'X' is:

  1. 'X' is known as aqua-regia

  2. 'X' is used for dissolving gold

  3. 'X' is used for decomposition of salts of weaker acids

  4. both (a) and (b)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Aqua regia is also called as royal water is a mixture of nitric acid and hydrochloric acid in the  molar ratio of 1:3. Aqua regia is a yellow-orange fuming liquid.Aqua regia is used for the decomposition of weak acids.

Hene option D is correct.

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

There are two colourless solutions (A) and (B), both give white ppt with $AgNO _3$ which dissolves in $aq \, NH _3$. Also (A) on reaction with $H _2O$ gives orange turbidity while (B) gives white turbidity. Identify (A) and (B). 

  1. (A) $SbCI _3 ;\, \, \,$ (B) $BiCI _3$
  2. (A) $BiCI _3 ;\, \, \,$ (B) $SbCI _3$
  3. (A) $SeCI _3; \, \, \,$ (B) $BiCI _3$
  4. (A) $BiCI _3; \, \, \,$ (B) $SeCI _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$SbCI _3$ or $BiCI _3 + AgNO _3 \longrightarrow \underset {white\, \, ppt }{ AgCI\downarrow}$ (due to $CI^-$)

              $AgCI + 2NH _3 (aq) \longrightarrow [\underset{soluble}{Ag(NH _3) _2}] CI$

                      $SbCI _3 + H _2O \longrightarrow \underset{orange}{SbOCI}\downarrow + 2HCI$

                  $SbOCl + 2HCI \longrightarrow \underset{soluble}{SbCI _3}+H _2O$

                      $BiCI _3 + H _2O \longrightarrow \underset{white}{BiOCI\downarrow}+ 2HCI$

                  $BiOCI + 2HCI \longrightarrow \underset{soluble}{BiCI _3}+H _2O$
Turbidity can be dissolved in dil. $HCl$.

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

Aqua-regia dissolves gold. 
which is correct?

  1. Aqua regia, or nitro-hydrochloric acid is a highly corrosive mixture of acids , a fuming yellow or red solution.

  2. The mixtures is formed by freshly mixing concentrated nitric acid and hydrochloric acid hoptimally in a volume ratio of 1:3.

  3. it can dissolve the noble metalsgold and platinum

  4. all of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Aqua regia, or nitro-hydrochloric acid is a highly corrosive mixture of acids, a fuming yellow or red solution. The mixtures is formed by freshly mixing concentrated nitric acid and hydrochloric acid hoptimally in a volume ratio of 1:3. It was named so because it can dissolve the noble metalsgold and platinum
Hence answer 1 is correct.

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

    Appearance  Reaction with dilute $HCl$  Reaction with dilute $HNO _3$
 Unknown metal #1  Dull gray solid with white oxide coating Dissolved with bubbles of clear gas   Dissolved with bubbles of clear gas
 Unknown metal #2  Solid; lustrous, smooth silver-gray surface  No reaction  Dissolved with bubbles of orange gas

Identify metal #1.

  1. Mercury

  2. Copper

  3. Zinc

  4. Iron

  5. Silver
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 Zinc being an active metal readily reacts with hydrochloric acid at room temperature to form soluble zinc chloride and hydrogen.

   $Zn+2{ H }Cl\rightarrow Zn{ Cl } _{ 2 }+{ H } _{ 2 }$
Zinc reacts with dilute nitric acid  and libreates nitric oxie.
$3Zn+{ 8HNO } _{ 3 }\rightarrow { 3Zn(NO _{ 3 }) } _{ 2 }+4{ H } _{ 2 }O+2NO$

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

| |  Appearance |  Reaction with dilute $HCl$ |  Reaction with dilute $HNO _3$ | | --- | --- | --- | --- | |  Unknown metal #1 |  Dull gray solid with white oxide coating | Dissolved with bubbles of clear gas  |  Dissolved with bubbles of clear gas | |  Unknown metal #2 |  Solid; lustrous, smooth silver-gray surface |  No reaction |  Dissolved with bubbles of orange gas |

Identify metal #2.

  1. Carbon

  2. Copper

  3. Zinc

  4. Sodium

  5. Silver
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Silver has a lustrous smooth gray surface.

It reacts with $HNO _3$ as:
$3Ag+2{ HNO } _{ 3 }\rightarrow { 4AgNO } _{ 3 }+{ NO } _{ 2 }+{ H } _{ 2 }O$.
The orange gas obtained is $NO _2$.

Multiple choice chemistry substances in common use preparation, properties and uses of baking soda chemical from common salt compounds of carbon

In Solvay ammonia process, sodium bicarbonate is precipitated due to :

  1. presence of $NH _3$
  2. reaction with $CO _2$

  3. reaction with brine solution

  4. reaction with NaOH
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In Solvay process, sodium bicarbonate is precipitated due to common ion$(Na^+)$ effect provided by brine (concentrated NaCl Solution).

Multiple choice conductivity and its types electrochemistry

Which of the following statements is true?

  1. When an aqueous solution of NaCl is electrolysed, sodium metal is deposited at cathode

  2. There is no difference between specific conductivity and molar conductivity

  3. Silver nitrate solution can be stored in a copper container

  4. The addition of liquid bromine to iodide solution turns it violet
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When an aqueous solution of NaCl is electrolysed, hydrogen is liberated at cathode. Specific conductivity and molar conductivity are different terms. Silver nitrate solution cannot be stored in a copper container as silver will get precipitated because of high reactivity of Cu than Ag.
The addition of liquid bromine to iodide solution turns it violet.

$Br _{2(l)}+2I^{-} _{(aq)}\rightarrow 2Br^{-} _{(aq)}+I _{2(aq)}$
Here, $Br _{2(l)}$ is Reddish-brown in color and $I _{2(aq)}$is violet in color

Multiple choice chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state? 

  1. $Ag _2SO _4$
  2. $CuF _2$
  3. $ZnF _2$
  4. $Cu _2Cl _2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The respective transition elements exists in
$Ag^{+}-{4d}^{10}5s^0$, (have completely filled d orbital)
$Cu^{+2}-3d^{9}4s^0$, (have incompletely filled d orbital)
$Zn^{+2}-3d^{10}4s^0$ , (have completely filled d orbital)
$Cu^{+1}-3d^{10}4s^0$ (have completely filled d orbital),
Since, $Cu^{+2}$ has unpaired electron it will show electron transitions. 
Hence will be coloured.
Multiple choice chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

In which of the following ions, the colour is not due to $d-d$ transition?

  1. $[Ti(H _2O) _6]^{3+}$
  2. $[Cu(NH _3) _4]^{2+}$
  3. $[CoF _6]^{3-}$
  4. $CrO _4^{2-}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Colour is due to d-d transtition ,then such complexes have colour due to d-d transition are surely octahedral complexes.
Among the given options,                                                   
                                         ${ \left[ Ti{ \left( { H } _{ 2 }O \right)  } _{ 6 } \right]  }^{ 3+ }$
                                        ${ \left[ Cu{ \left( { NH } _{ 3 } \right)  } _{ 4 } \right]  }^{ 2+ }$
                                           ${ \left[ { COF } _{ 6 } \right]  }^{ 3- }$
This are octahedral complexes.
$CrO _4^{2-}$ is a salt,it colour is not due to d-d transition.
Multiple choice chemistry d and f block elements coloured complexes magnetic nature of transition metals general characteristics of first transition series

For $Zn^{2+}$, $Ni^{2+}$, $Cu$, and $Cr^{2+}$ which of the following statements is correct?

  1. Only $Zn^{2+}$ is colourless and $Ni^{2+}$, $Cu^{2+}$ and $Cr$ are coloured

  2. All the ions are coloured

  3. All the ions are colourless
  4. $Zn^{2+}$and $Cu^{2+}$ are colourless while $Ni^{2+}$ and $Cr^{2+}$ are coloured
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $Zn^{+2}$ have fully filled $d$ orbital having no unpaired electrons,it does not undergo electronic transitions,hence colourless.However,rest of the elements given as $Ni^{+2}$ , $Cu^{+2}$ and $Cr^{+2}$ has unpaired electrons which can undergo electronic transitions and hence show colours.