Chemistry

Chemical Reactions and Equations

684 Questions

Chemical reactions and equations form a foundational chemistry topic where students identify reaction products, balance chemical formulas, and classify reaction types. It covers critical mechanisms like Markovnikov's rule, precipitation, and endothermic or exothermic processes. These questions are highly common in general science sections of state and central government competitive exams.

Endothermic reactionsPrecipitation reactionsDehydrohalogenation of alkyl halidesChemical equation balancingReaction product identification

Chemical Reactions and Equations Questions

Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

In the conversion of $BrO _3 \ ^- $ to $ BrO _4 \ ^- $:

  1. there is no change in oxidation number

  2. reduction takes place in basic solution
  3. reduction also takes place by $XeF _2$
  4. equivalent mass of $ Br{ O }^- _3$ is one-half of ionic mass
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A. During the conversion of $BrO^- _3\ to\ BrO^- _4$, the oxidation number of bromine changes from +5 to +7. Hence, the option A is incorrect.
B. and C.  Since there is an increase in the oxidation number, the net reaction is the oxidation of bromine. Hence, the options B and C are incorrect.
D. The equivalent mass of $BrO^- _3$ is the ratio of the ionic mass to the change in the oxidation number. It is tone half of the ionic mass as the change in the oxidation number is 2. Thus, the option D is correct.

Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

On heating $KClO _3$, we get :

  1. $KClO _2+O _2$
  2. $KCl+O _2$
  3. $KCl+O _3$
  4. $KCl+O _2+O _3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
On heating $KClO _3$, we get $KCl+O _2$.

The reaction is as follows:
$2KClO _3 \longrightarrow 2KCl+ 3O _2$

When potassium chlorate is heated in the presence of manganese dioxide catalyst, it decomposes to form potassium chloride and oxygen gas. 
Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

Which products are expected from the disproportionation of hypochlorous acid?

  1. $HClO _3$ and $Cl _2O$
  2. $HClO _2$ and $HClO _4$
  3. $HCl $and $Cl _2O$
  4. $HCl $ and $HClO _3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\displaystyle 3HClO \rightleftharpoons 2HCl + HClO _3$

It is a disproportionation reaction of hypochlorous acid where the oxidation number of chlorine from $+1$ to $-1$

Higher is Oxidation no. more is effective nuclear charge, lesser is size.

The disproportionation of hypochlorous acid gives $\displaystyle HCl $ and $\displaystyle HClO _3 $.

Hence option D is correct.

Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

$NaOCl$ is used as a bleaching agent and sterilising agent. It can be prepared by the action of:

  1. $NaCl$ with $H _2O$
  2. $NH _4Cl$ with $NaOH$
  3. $Cl _2$ with cold and dil. $NaOH$
  4. $Cl _2$ with hot and conc. $NaOH$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Chlorine with cold and dilute sodium hydroxide give mixture of chloride salt, hypochlorite and water.

$Cl _2 (l) + 2NaOH (aq) \rightarrow NaCl+ NaClO (aq) + H _2O(l) $

Chlorine with hot and dilute sodium hydroxide give mixture of chloride salt, hypochlorite and water.

$3 Cl _2 (l) + 6 NaOH (aq) \rightarrow 5 NaCl + NaClO _3 (aq) + 3 H _2O (l) $

Hence, the correct option is $C$
Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

Choose the correct statements:

  1. The anyhdride of $HOCl$ is $Cl _2O$.
  2. The anyhdride of $HClO _2$ is $ClO _2$.
  3. The anyhdride of $HClO _3$ is $Cl _2O _6$.
  4. The anyhdride of $HClO _4$ is $Cl _2O _7$.
  5. The anyhdride of $HClO _3$ is $Cl _2O _3$.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

The anyhdride of $HOCl$ is $Cl _2O$.
The anyhdride of $HClO _2$ is $ClO _2$.
The anyhdride of $HClO _3$ is $Cl _2O _6$.
The anyhdride of $HClO _4$ is $Cl _2O _7$.
Anhydrides on reaction with water will give you corresponding acid.

Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

The reaction, $3ClO _{(aq.)}^-\rightarrow ClO _{3(aq.)}^-+2Cl^- _{(aq.)}$ is an example of :

  1. oxidation reaction.

  2. reduction reaction.

  3. disproportionation.

  4. decomposition reaction.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
  1. Disproportionation is a specific type of redox reaction in which a species is simultaneously reduced and oxidised to form two different products.
    Here the $ClO^{-}$ is simultaneously reduced and oxidised to form two different products.

    Hence option C is correct.
Multiple choice chemistry occurrence of carbon compounds in nature substances, objects and energy alternative fuels and energy sources alternative sources of energy

 Methylbutane on reacting with bromine in the presence of sunlight gives mainly:

  1. 2-bromo 3-methylbutane

  2. 1-bromo 3-methylbutane

  3. 2-bromo 2-methylbutane

  4. 1-bromo 2-methylbutane
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Bromination of alkanes is selective. The tertiary hydrogen is the most reactive, leading to the formation of the most stable radical intermediate, which results in 2-bromo-2-methylbutane.

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

When ${ NH } _{ 4 }Cl$ is added to ${ NH } _{ 4 }OH$ solution, the dissociation of ammonium hydroxide is reduced. It is due to:

  1. common ion effect

  2. hydrolysis

  3. oxidation

  4. reduction
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When $NH _4Cl$ is added to $NH _4OH$ solution, concentration of $NH _4^{+}$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced. 

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 dissociation of $NH _4OH$, if excess if $NH _4Cl$ is added before adding $NH _4OH$, the concentration of:

  1. $NH _4^+$ ions increases and $OH^-$ ions decreases
  2. both $NH _4^+$ ions and $OH^-$ ions increases
  3. $NH _4^+$ ions decreases and $OH^-$ ions increases
  4. both $NH _4^+$ ions and $OH^-$ ions decreases
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the law of Mass action, the dissociation of $NH _4OH$ takes place and we have,
$\dfrac{[NH _4^+][OH^-]}{[NH _4OH]} =K$
Ammonium chloride, a strong electrolyte, ionises almost completely as follows:
$NH _4Cl \leftrightarrow NH _4^+ + Cl^-$
So, if excess of $NH _4Cl$ is added before adding $NH _4OH$, the concentration of $NH _4^+$ ions is increased and consequently the concentration of $OH^-$ ions is decreased.

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

Bromine is formed when concentrated $HBr$ is heated with ?

  1. $KMn{O} _{4}$
  2. ${K} _{2}{Cr} _{2}{O} _{7}$
  3. $Mn{O} _{2}$
  4. $All\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When concentrated $HBr$ is heated with $KMnO _4$ then bromine is liberated.The reaction involved is $2KMnO _4+16HBr\longrightarrow 2MnBr _2=2KBr+8H _2O+5Br _2$

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

Which of the statements is/are true about the reactivity order is $F _2\, >\, Cl _2\, >\, Br _2\, >\, I _2.$
I. Lower the activation energy for the chain initiation step, more reactive is the halogen.
II. Lower the activation energy for the first chain propagation step, more reactive is the halogen.
III. More negative is the overall heat of the reaction $(\Delta H _{\zeta}^{})$ of halogenation of alkane, more reactive is the halogen.
IV. Lower the activation energy for the second chain-propagation step, more reactive is the halogen.

  1. (I), (II)

  2. (I), (II), (III)

  3. (II), (III)

  4. (II), (III), (IV)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There are two factors that determine the reactivity of the halogenation of alkanes.
1. Lesser the positive value of $E _{act}$ of the first chain propagating step, more reactive is the halogen.
2. Higher the negative value of overall heat of reaction $(\Delta H _{\zeta}^{})$ more reactive is the halogen. This explains high reactivity and explosive violence with which $F _2$ reacts with $CH _4$.

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

Iodine reacts with hot NaOH solution giving the products as-___?

  1. $NaI$
  2. $ \displaystyle NaIO _{3}$.

  3. A and B

  4. non of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Iodine reacts with hot concentrated NaOH solution giving the products sodium iodide (NaI) and  sodium iodate $\displaystyle  (NaIO _3)$.
With cold and dil NaOH, iodine gives sodium hypo iodite $\displaystyle  (NaOI)$
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 reactions are not possible?

  1. $2NaCl+F _2$ $\longrightarrow \ $ $2NaF+Cl _2$
  2. $2NaCl+Br _2$ $\longrightarrow \ $ $2NaBr+Cl _2$
  3. $2NaF+Cl _2$ $\longrightarrow \ $ $2NaCl+F _2$
  4. $2NaBr+Cl _2$ $\longrightarrow \ $ $2NaCl+Br _2$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Option B and C reactions are not possible because the bromine do not displace chlorine and chlorine do not displace florine as chlorine and florine are most stable than bromine and chorine in their compounds respectively.

Hence option B,C are correct.