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 chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

$B _{2}H _{6}$ undergoes substitution reaction with:

  1. $Br _{2}$ at $100^{o}C$
  2. $HCl$ $/$anhydrous $AlCl _{3}$

  3. both A and B
  4. $Cl _{2}$ at $100^{o}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$B _2H _6 + HCl \rightarrow B _2H _5Cl + H _2$
This reaction takes place in the presence of anhydrous $AlCl _3$.
$B _2H _6 + Br _2 \rightarrow B _2H _5Br + HBr$
This is a slow reaction taking place at $100^oC$.
$B _2H _6 + 6Cl _2 \rightarrow 2BCl _3 + 6HCl$
This is a vigorous reaction taking place at $25^oC$. Diborane does not react with $Cl _2$ at ${100}^{o}C$. This reaction is not reported.

Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

From $B _2H _6$, all the following can be prepared except :

  1. $B _2O _3$
  2. $H _3BO _3$
  3. $B _2(CH _3) _6$
  4. $NaBH _4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$B _2H _6+3O _2\rightarrow B _2O _3+3H _2O+Heat$

$B _2H _6+6H _2O\rightarrow H _3BO _3+6H _2$

$2NaH+B _2H _6\rightarrow etherNaBH _4$

From $B _2H _6$, all can be prepared except $B _2(CH _3) _6$

Hence option $C$ is correct.
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

All the products formed in the oxidation of $NaBH _{4}$ by $I _{2}$, are:

  1. $B _{2}H _{6}$ and $NaI$
  2. $B _{2}H _{6}, H _{2}$ and $NaI$
  3. $BI _{3}$ and $NaH$
  4. $NaBI _{4}$ and $HI$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The reaction is as follows:

$2NaBH _4(s)+I _2(s)\rightarrow B _2H _6(g)+2NaI(s)+H _2(g)$

Hence, the correct option is $\text{B}$
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

Compound $(X)$ on reduction with $LiAlH _4$ gives a hydride $(Y)$ containing 21.72% hydrogen along with other products. The compound $(Y)$ react with air explosively resulting in boron trioxide. Compounds $X$ and $Y$ are respectively:

  1. $BCl _3, B _2H _6$
  2. $B _2H _6,BCl _3$
  3. $BF _3,Al _2O _6$
  4. $B _2H _6,BF _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(A)$ $B{Cl} _{3}, \; {B} _{2}{H} _{6}$
Since ${B} _{2}{O} _{3}$ is formed by reaction of $(Y)$ with air, $(Y)$ therefore should be ${B} _{2}{H} _{6}$ in which % of hydrogen is 21.72. 
${B} _{2}{H} _{6} + 3{O} _{2} \; \longrightarrow \; {B} _{2}{O} _{6} + 3{H} _{2}O + heat$
The compound $(X)$ on reduction with $LiAI{H} _{4}$ gives ${B} _{2}{H} _{6}$. Thus it is boron trihalide.
$4B{X} _{3} + 3LiAl{H} _{4} \; \longrightarrow \; 2{B} _{2}{H} _{6} + 3LiX + 3Al{X} _{3}$ ($X = CI$ or $Br$)
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

In the reaction, 
$2X+B _2H _6 \rightarrow [BH _2(X _2)]^+[BH _4]^-$
'X' cannot be ?

  1. $NH _3$
  2. $CH _3NH _2$
  3. $(CH) _3) _2NH$
  4. $(CH _3) _3N$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Answer:- (D) ${(C{H} _{3})} _{3}N$
${B} _{2}{H} _{6}$ reacts with primary(1º)  and secondary(2º)  amine and form an ionic compound and gives unsymmetrical cleavage of diborane. However with tertiary(3º) amine, ${B} _{2}{H} _{6}$ shows it's symmetrical cleavage and forms an adduct as:-
${B} _{2}{H} _{6} + 2N{(C{H} _{3})} _{3} \; \longrightarrow \; 2B{H} _{3}.N{(C{H} _{3})} _{3}$
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

In the reaction $2X+B _2H _6\rightarrow [BH _2X _2]^+[BH _4]^-$ the amine(s) X is (are) :

  1. $NH _3$
  2. $CH _2NH _2$
  3. $(CH _3) _2NH$
  4. $(CH _3) _3N$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation
Diborane forms addition compounds with lewis bases as it is electron deficient compound act as lewis acid. If the lewis base is small without steric strain unsymmetric cleavage takes place. If the base is large with steric strain symmetric cleavage takes place. 
$B _2H _6 + NH _3 \rightarrow [BH _2(NH _3) _2]^+[BH _4]^- \rightarrow B _3N _3H _6 (Borazol)$
$B _2H _6 + CH _3NH _2 \rightarrow [BH _2(CH _3NH _2) _2]^+[BH _4]^-$
$B _2H _6 + (CH _3) _2NH \rightarrow [B^-H _3-N^+H(CH _3) _2] \rightarrow [BH _2N(CH _3) _2]$
$B _2H _6 + (CH _3) _3N \rightarrow [(CH _3) _3N^+-B^-H _3]$

Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

The reaction which gives Borazole as a major product is ?

  1. $LiH+B _{2}H _{6}\overset{2moles LiH.1moleB _{2}H _{6}}{\rightarrow}$
  2. $B _{2}H _{6}+NH _{3}\xrightarrow[low temp]{2molesB _{2}H _{6}1moleNH _{3}}$
  3. $B _{2}H _{6}+NH _{3}\xrightarrow[low temp]{1moB _{2}H _{6}.2molesNH _{3}}$
  4. $B _{2}H _{6}+NH _{3}\xrightarrow[high\, temp]{1moB _{2}H _{6}.2molesNH _{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$B _{2} H _{6} + NH _{3} \xrightarrow [high\, temp]{1moB _{2} H _{6} .2molesNH _{3}} B _{3} N _{3} H _{6} $

Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.

Compound $C$ is :

  1. diborane

  2. boric acid

  3. borate salt

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

     $B _2H _6+2NMe _3\longrightarrow 2BH _3.NMe _3$

Diborane (A)                        Adduct (B)

$BH _3.NMe _3+3H _2O\longrightarrow H _3BO _3+NMe _3+3H _2$
                                                Boric acid                     $\downarrow$
                                                      (C)                          (D) gas

Compound "C" is boric acid $-H _3BO _3$

Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

A compound $A$ of boron reacts with $NMe _3$ to give an adduct $B$ which on hydrolysis gives a compound $C$ and a gas $D$. Compound $C$ is an acid.


Compound $A$ is :

  1. diborane

  2. boric acid

  3. borate salt

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

     $B _2H _6+2NMe _3\longrightarrow2BH _3.NMe _3$

Diborane (A)                        Adduct (B)

$BH _3.NMe _3+3H _2O\longrightarrow H _3BO _3+NMe _3+3H _2$
                                               Boric acid                      $\downarrow$
                                                    (C)                            (D) gas
Compound $A$ is diborane.

Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

An alkali metal hydride (NaH) react with diborane in 'A' to give a tetrahedral compound 'B' which is extensively used as reducing agent in organic synthesis. The compounds 'A' and 'B' respectively are :

  1. $CH _3COCH _3\,\,\, and \,\,\,B _3N _3H _6$
  2. $(C _2H _5) _2O \,\,\,and\,\,\, NaBH _4$
  3. $C _2H _6\,\,\,and\,\,C _2H _5Na$
  4. $C _6H _6\,\,\,and\,\,\,NaBH _4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Answer:-
When an alkali metal hydride $(NaH)$ react with diborane $({B} _{2}{H} _{6})$ in the presence of ether $({({C} _{2}{H} _{5})} _{2}O)$, a tetrahedral compound (Metal borohydride) is formed which act as a reducing agent in organic synthesis.
$2NaH + {B} _{2}{H} _{6} \; \xrightarrow{{({C} _{2}{H} _{5})} _{2}O} \; \underset{\text{sodium borohydride}}{2NaB{H} _{4}}$
Thus, A is  ${({C} _{2}{H} _{5})} _{2}O$ and B is $NaB{H} _{4}$.
Multiple choice chemistry the p-block elements - group 13 study of diborane some important compounds of boron study of boron

A compound $X$, of boron reacts with $NH _3$ on heating to give another compound $Y$ which is called inorganic benzene. The compound $X$ can be prepared by treating $BF _3$ with lithium aluminum hydride. The compounds $X$ and $Y$ are represented by formula:

  1. $B _2H _6, B _3N _3H _6$
  2. $B _2O _3, B _3N _3H _6$
  3. $BF _3, B _3N _3H _6$
  4. $B _3N _3H _6, B _2H _6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A compound $X$, of boron, reacts with $N{H} _{3}$ on heating to give another compound $Y$ which is called inorganic benzene.
$\underset{X \; (Diborane)}{3{B} _{2}{H} _{6}} + 6N{H} _{3} \longrightarrow 3{[B{H} _{2}{(N{H} _{3})} _{2}]}^{+}{[B{H} _{4}]}^{-} \xrightarrow{heat} \underset{\text{Y (Borazole/Inorganic Benzene)}}{2{B} _{3}{N} _{3}{H} _{6}} + 12{H} _{2}$
$4B{F} _{3} + 3LiAl{H} _{4} \longrightarrow \underset{X}{2{B} _{2}{H} _{6}} +3LiF + 3Al{F} _{3}$