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 further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

In the third group of qualitative analysis, the precipitating reagent is $NH {4}Cl + NH _{4} OH$. The function of $NH _{4}Cl$ is to______

  1. increase the ionization of $NH _{4} OH$
  2. suppress the ionization of $NH _{4} OH$

  3. stabilise the hydroxides of group cations

  4. convert the ions of group third into their respective chlorides
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte, which provides an ion common to that provided by weak electrolyte.


The NH4OH is weak base it does not ionises completely. Thus due to presence of common ion NH4+ in NH4Cl, it supresses the ionisation of weak base NH4OH in order to decrease the OH- concentration so that higher group cations will not get precipitated.

Thus the pair $NH _{4} OH + NH _{4} Cl$ shows common ion effect. 

Ammonium chloride suppresses the ionization of ammonium hydroxide.

Option B is correct.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

Observe the following statements.
I .Bleaching powder is used in the preparation of chloroform.
II. Bleaching powder decomposes in the presence of $\mathrm{C}\mathrm{o}\mathrm{C}1 _{2}$ to liberate $\mathrm{O} _{2}$.
III. Aqueous $\mathrm{K}\mathrm{H}\mathrm{F} _{2}$ is used in the preparation of fluorine.

  1. I,II and III are correct

  2. Only II is correct.

  3. Only I and III are correct

  4. Only I and II are correct.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Molten $KHF _{2}$ is used in the preparation of fluorine.

Rest two statements are true.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

Which of the following will displace the halogen from the solution of of halide
a $) Br _{2}$ added to an $NaCl$ solution
$\mathrm{b}) Cl _{2}$ added to a $KBr$ solution
$\mathrm{c}) Cl _{2}$ added to an $NaF$ solution
$\mathrm{d}) Br _{2}$ added to to a $KI$ solution
Find the correct answer :

  1. a, b, c

  2. b, c

  3. b, d

  4. all are correct
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Br^{-}$ can reduce $Cl _{2}$ to $Cl^{-}$ $(E _{cell}=+ve)$

$I^{-}$ can reduce $Br _{2}$ to $Br^{-}$ $(E _{cell}=+ve)$

$E _{cell}=+ve$ in both case

$\therefore$ Reaction is feasible

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

Iodine can be obtained from NaI solution by the action of:

  1. chlorine

  2. bromine

  3. soluble chloride

  4. soluble bromine
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The iodine can be obtained by reacting chlorine or bromine by NaI. This is because the electronegativity of chlorine and bromine is more than the iodine and also the reactivity of chlorine and bromine is more than the iodine.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

When a dry solid is treated with a mild oxidizing agent, a purple solid is produced. What is the dry solid?

  1. $N _2$
  2. $KI$
  3. $CCl _4$
  4. $AgNO _3$
  5. $CaCO _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A purple solid (iodine) is produced by treatment of the dry solid $KI$ with a mild oxidizing agent. The mild oxidising agent oxidizes iodide ion to iodine.

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

Treatment of the dry solid with a mild oxidizing agent produces a purple solid having formula _______.

  1. $N _2$
  2. $KI$
  3. $CCl _4$
  4. $AgNO _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Iodine can be displaced from potassium iodide if a more reactive element is made to react with potassium iodide. In the presence of a mild oxidising agent, the iodine ions are oxidised to $I _2$. Iodine is purple in colour.

Multiple choice imperfections in solids solid state the solid state chemistry

Silver halides generally show ?

  1. Schottky defect

  2. Frenkel defect

  3. both Frenkel and Schottky defects

  4. cation excess defect
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Silver halides show both Frenkel and Schotkky defects. For Frenkel defect the reason is that there is size difference between the sizes of silver and halide.

Schottky defect is possible because silver halides are highly ionic.
So, correct answer is option C.

Multiple choice imperfections in solids solid state the solid state chemistry

An excess of potassium ions makes KCl crystals appear violet or Lilac in colour since

  1. some of the anionic sites are occupied by an unpaired electron

  2. some of the anionic sites are occupied by a pair of electrons

  3. there are vacancies at someanionic sites

  4. F-centres are created which impart colour to the crystals
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

When KC1 is heated in vapour of K, some of the Cl leave their lattice site and create anion vacancies. This chloride ion wants to combine with K vapour to form potassium chloride. For doing so K atom loses electrons form K ions. This released electron diffuses into the crystal to get entrapped in the anion vacancy called F-centre. When visible light falls on the crystal, this entrapped electron gains energy, goes to the higher level when it comes back to the ground state, energy is released in the form of light.

Multiple choice imperfections in solids solid state the solid state chemistry

The appearance of colour in solid alkali metal halides is generally due to: 

  1. schottky defect

  2. frenkel defect

  3. interstitial position

  4. F-centre
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In F-centre defect in which an anionic vacancy in a crystal is filled by one or more unpaired electron. These electron absorb light in the visible region and emits colour. So, appearance of colour in solid alkali metal halides is generally due to F-centre.

Multiple choice imperfections in solids solid state the solid state chemistry

Statement-I:  Solids having more $F-$centres possess intense colours.

Statement-II:  Excess of $Na^{+}$ in $NaCl$ solid having F-centres makes it appear to pink.

  1. Statement-I is correct but Statement-II is wrong

  2. Statement-I is wrong but Statement-II is correct

  3. Both Statement-I and Statement-II are correct and Statement-II is correct explanation of Statement-I

  4. Both Statement-I and Statement-II are correct but Statement-II is not correct explanation of Statement-I
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

F-centres are indeed responsible for the intense colors in alkali halides due to the excitation of trapped electrons. However, the color of NaCl with F-centres is yellow, not pink, making statement II incorrect.

Multiple choice imperfections in solids solid state the solid state chemistry

Crystals have 'vacant sites' or 'defects' in them. When light strikes a photographic silver bromide paper, silver atoms move in through these defects to:

  1. develop the film

  2. form tiny clumps of silver atoms

  3. form negative images

  4. form a colour image
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When light strikes a photographer $(AgBr)$ paper, it gives energy to the electrons present in the film. These energetic electrons when strike silver ions turn them to silver atoms. So eventually,  ions leave their lattice site and occupy interstitial sites. Since silver atoms are black in color so whenever light strikes a silver ion the photographic film will turn black.

Multiple choice imperfections in solids solid state the solid state chemistry

The pink colour of lithium chloride crystal is due to:

  1. frenkel defect

  2. metal excess defect

  3. metal deficiency defect

  4. impurity defect
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$LiCl$ has non-stochiometric metal excess defect due to anion vacancies. The negative ions $(Cl^-)$ are missing from their lattice sites leaving the holes in which electrons are entrapped so that electrical neutrality is maintained.

When $LiCl$ is heated, $Li$ atoms gets deposited on the surface of the crystal. The $Cl^-$ ions diffuse into the surface and combine with $Li$ atoms to give $LiCl$. This is so because of loss of electrons by $Li$ atoms to form $Li^+$. The released electrons diffuse excess into crystal and occupy anionic sites. As a result, there is an excess of $Li$. The anionic sites occupied by unpaired electrons are F-centers which imparts a pink color to $LiCl$ crystals. The color is observed as a result of excitation of these electrons when they absorb energy from visible light falling on crystals.

Multiple choice chemistry general principles of metallurgy concentration of ore concentration of ores principles of metallurgy

When $ZnS$ and $PbS$ minerals are present together, then $NaCN$ is added to separate them in the froth flotation process as a depressant, because:

  1. $Pb(CN) _2$ precipitated while no effect on $ZnS$
  2. $ZnS$ forms soluble complex $Na _2 Zn(CN) _4$
  3. $PbS$ forms soluble complex $Na _2Pb(CN) _4$
  4. both $(a)$ and $(b)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$NaCN$ prevents $ZnS$ from forming the froth by reacting with it to from complex $Na _2[Zn(CN) _4]$ and acts as a depressant. While it does not prevent $PbS$ from forming froth and both the ores can easily be separated.