Chemistry Test

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Chemistry Test - 6

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Weekly Quiz Competition

Question 1
3 / -1

The geometry of $BF_{3}$, will be:

A
Linear

B
Pyramidal

C
Trigonal planar

D
Trigonal pyramidal

Solution

The geometry of $BF_{3}$, will be trigonal planer.

The Structure of $BF_{3}$:

The electronegative Fluorine atom surrounded the central boron atom.

The number of electrons used in the skeletal structure is $6$ ( $3$ from boron and $1$ from each three fluorine atoms). i.e., $3$ electron pairs and no lone pairs.

There are $3$ bond pairs, hence according to VSEPR theory, the structure of $BF_{3}$ is trigonal planar.
Question 2
3 / -1

In the case of alkali metals, the covalent character decreases in the order:

A
${MI}>{MBr}>{MCl}>{MF}$

B
${MCl}>{MI}>{MBr}>{MF}$

C
${MF}>{MCl}>{MBr}>{MI}$

D
${MF}>{MCl}>{MI}>{MBr}$

Solution

Fajan's rule states that the covalent character increases with the decrease in the size of the cation and increases with the increase in the size of the anion.
Since we know that the size increases on going down the group, the order of size will be:
${I}^{-}>{Br}^{-}>{Cl}^{-}>{F}^{-}$
According to Fajan's rule, the covalent character increases with the increase in the size of the anion.
$\therefore$ The order of covalent character will be: ${MI}>{MBr}>{MC1}>{MF}$
Question 3
3 / -1

How many stereoisomers are there for the following structure?

A
1

B
2

C
3

D
4

Solution

The IUPAC name of the folllwing figure is 2,3-dihydroxybutanedioic acid, known as tartaric acid, the two chiral centers have the same four substituents and are equivalent. As a result, two of the four possible stereoisomers of this compound are identical due to a plane of symmetry, so there are only 3 stereoisomeric tartaric acids.
Question 4
3 / -1

In a $13 \mathrm{~L}$ vessel initially following reaction occur $\mathrm{C}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{~g})$ by $12 \mathrm{~g} \mathrm{C}, 64 \mathrm{~g}$ $\mathrm{S}_{2}, 76 \mathrm{~g} \mathrm{CS}_{2}$ at $1027^{\circ} \mathrm{C}$ temperature then total pressure is:

A
$200 R$

B

$158 R$

C

$100 R$

D

$79 R$

Solution

$\mathrm{C}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{~g})$
Number of moles of $\mathrm{C}=\frac{12 \mathrm{~g}}{12 \mathrm{~g}}=1$
Number of moles of $\mathrm{SO}_{2}=\frac{64}{64}=1$
Number of moles of $\mathrm{CS}_{2}=\frac{76}{76}=1$
Total number of gaseous moles $=1\left(\mathrm{~S}_{2}\right)+1\left(\mathrm{CS}_{2}\right)=2$ moles
From ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$
$\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$
$=\frac{2 \times \mathrm{R} \times(273+1027)}{13}$
$\Rightarrow \frac{2 \mathrm{R}(1300)}{13}$
$=200 \mathrm{R}$
Question 5
3 / -1

Which of the following is a tautomer of phenol?

A

B

C

D

Solution

According to the tautomerism the tautomer of phenol is:
Question 6
3 / -1

Increasing order of electronegativity is:

A
$\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$

B
$\mathrm{P}<\mathrm{Si}<\mathrm{S}<\mathrm{Cl}$

C
$\mathrm{S}<\mathrm{Si}<\mathrm{P}<\mathrm{Cl}$

D
$\mathrm{Cl}<\mathrm{S}<\mathrm{Si}<\mathrm{P}$

Solution

Increasing order of electronegativity is$\mathrm{Si}<\mathrm{P}<\mathrm{S}<\mathrm{Cl}$.
We know that as we move from left to right across a period, the electronegativity increases till the halogens and them becomes zero for inert gases.
Out of $\mathrm{P}, \mathrm{Cl}, \mathrm{Si}$, and $\mathrm{S}$, all of them are belonging it on the 3rd period and Chlorine is the most electronegative with a value of $3.16$.
So moving from Silicon to Chlorine, the electronegativity increases. Silicon has electronegativity $1.90 .$Phosphorus has electronegativity $2.19$.Sulphur has electronegativity higher than Phosphorus with a value of $2.58$.
Question 7
3 / -1

Benzene forms Nitrobenzene on reaction with conc. HNO₃/H₂SO₄, in the reaction HNO₃ acts as a:

A
Catalyst

B
Reducing agent

C
Base

D
Acid

Solution

Benzene forms Nitrobenzene on reaction with conc. HNO₃/H₂SO₄, in the reaction HNO₃acts as a base.Proton donor is acids and proton acceptor is bases. Conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ and conc. $\mathrm{HNO}_{3}$ react in the following manner:
$\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{H}_{2} \mathrm{NO}_{3}^{+}+\mathrm{HSO}_{4}^{-}$
$\mathrm{H}_{2} \mathrm{NO}_{3}^{+} \rightarrow \mathrm{NO}_{2}^{+}+\mathrm{H}_{2} \mathrm{O}$
Question 8
3 / -1

Which of the following metal ions precipitates as a sulphide upon treatment with ammoniacal H₂S?

A
$F e (I I I)$

B
$A l (I I I)$

C
$M g (I I)$

D
$Z n (I I)$

Solution

The basic reaction of ammoniacal $H_{2} S$ is also called a group reagent. $K s p (Z n S)$ is very high and $Z n^{2 +}$ is precipitated as $Z n S$ by a high concentration of $S^{2 -}$ formed when $H_{2} S$ is passed in ammoniacal solution.
$H_{2} S ⇌ Z n + S^{_{2} -}$ (I)
$H^{+} + O H^{-} ⇌ H_{2} O$ (II)
Reaction (I) is favoured in forward side if $H^{+}$ is removed immediately by $O H^{-}$ $(N H_{4} O H)$ .
$Z n^{2 +} + S^{2 -} ⟶ Z n S ↓$ Whiteppt
$F e^{3 +}$ and $A I^{3 +}$ are precipitated as hydroxide.
Question 9
3 / -1

Equal masses of $H_{2},O_{2}$ and methane have been taken in a container of volume $V$ at temperature $27^{\circ} C$ under identical conditions. The ratio of the volumes of gases $H_{2}, O_{2}$ and $CH_{4}$ would be

A
$8 :16 :1$

B
$16 : 8 : 1$

C
$16 :1 : 2$

D
$8 : 1 : 2$

Solution

According to Avogadro's hypothesis,
Volume of a gas $(\mathrm{V}) \propto$ number of moles $(\mathrm{n})$
Therefore, the ratio of the volumes of gases can be determined in terms of their moles.
$\therefore$ The ratio of volumes of $H_{2}: O_{2}:$ methane $\left(C H_{4}\right)$ is given by
$V_{H_{2}}: V_{O_{2}}: V_{C H_{4}}=n_{H_{2}}: n_{O_{2}}: n_{C H_{4}}$
$\Rightarrow V_{H_{2}}: V_{O_{2}}: V_{C H_{4}}=\frac{m_{H_{2}}}{M_{H_{2}}}: \frac{m_{O_{2}}}{M_{O_{2}}}: \frac{m_{C H_{4}}}{M_{C H_{4}}}$
Given, $ m_{H_{2}}=m_{O_{2}}=m_{C H_{4}}=m$
$\left[\because n=\frac{\text { mass }}{\text { molar mass }}\right]$
Thus, $V_{H_{2}}: V_{O_{2}}: V_{C H_{4}}:=\frac{m}{2}: \frac{m}{32}: \frac{m}{16}$
$=16: 1: 2$
Hence, the correctoption is (C).
Question 10
3 / -1

Diborane combines with ammonia at $120^{\circ} \mathrm{C}$ to give:

A
${B}_{2} {H}_{6} \cdot {NH}_{3}$

B
${B}_{2} {H}_{6} \cdot 2 {NH}_{3}$

C
${B}_{2} {H}_{6} .3 {NH}_{3}$

D
${B}_{2} {H}_{6} \cdot 4 {NH}_{3}$

Solution

The reaction of diborane with ammonia gives borazines. the reaction is:
$3 {~B}_{2} {H}_{6}+2 {NH}_{3} \rightarrow {B}_{2} {H}_{6} \cdot 2 {NH}_{3}+12 {H}_{2}$
Aminoboranes contain B-N bonds in which the boron is trigonally hybridized. The lone pair on nitrogen may be donated to a vacant p orbital on boron, giving rise to $B-N$ bond order greater than one.
Aminoboranes are prototypes of alkenes. It can be prepared from amine boranes.