MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 6

Nitric acid is manufactured by the Ostwald process, in which nitrogen oxide reacts with water.

$3NO_3(g)+H_2O(l) \rightarrow 2HNO_3(aq)+NO(g)$ How many grams of nitrogen dioxide are required in this reaction to produce 25.2 g of

$HNO_3$ ?

0%
21.6 g
0%
27.6 g
0%
46.8 g
0%
16.0 g

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can obtained starting only with 10.00g of ammonia and 20.00g of oxygen?

0%
15 g
0%
10 g
0%
2 g
0%
20 g

When we inhale helium gas funny sound appear which is due to:

0%
high density of helium than air
0%
low density of air than helium
0%
low density of helium than air
0%
reaction between air and helium

Ammonia is produced when a mixture of ammonium chloride and substance X is heated. What is substance X?

0%
Ammonium sulphate
0%
Barium chloride
0%
Calcium hydroxide
0%
Silver nitrate

Large scale manufacturing of nitric acid by the Ostwald process utilizes the reaction-

0%
$2NaNO_3+H_2SO_4 \rightarrow Na_2SO_4 + 2HNO_3$
0%
$4NH_3+5O_2 \rightarrow 4NO+6H_2O$
0%
$NO^{+}_{2}+NO^−_3 + H_2O \rightarrow 2HNO_3$
0%
$2NO+O_2+H_2O \rightarrow HNO_3+HNO_2$

The correct statement among the following is:

0%
$(SiH_3)_3 N$ is pyramidal and more basic than $(CH_3)_3 N$
0%
$(SiH_3)_3 N$ is planar and more basic than $(CH_3)_3N$
0%
$(SiH_3)_3 N$ is pyramidal and less basic than $(CH_3)_3 N$
0%
$(SiH_3)_3 N$ is planar and less basic than $(CH_3)_3 N$

Explanation

In case of

$(CH_3)_3N$ , nitrogen is

$sp^{3}$ hybrid and pyramidal shape and no back-bonding i.e. more basic.

In case of

$(SiH_3)_3N$ , nitrogen is

$sp^{2}$ hybrid and planar due to back bonding and less basic because lone pair is not available for donation.

1248824_1614836_ans_dbbeb5bbd41447f88b8b002e32daa65c.png

White phosphorus has:

0%
Six $P-P$ single bonds
0%
Four $P-P$ single bonds
0%
Four lone pairs of electrons
0%
$P-P-P$ angle of $60^{\circ}$

In the preparation of

$HNO_3$ by Ostwald process ammonia is:

0%
Oxidized
0%
Reduced
0%
Neither oxidized nor reduced
0%
Both oxidized and reduced

Nitric Acid is manufactured by:

0%
Ostwald's Process
0%
Haber's Process
0%
Contact Process
0%
None of these

In ostwald process of manufacturing of

$HNO_3$ , catalyst used is

0%
Mo
0%
Fe
0%
Pt
0%
Na

In white phosphorous

$(P_4)$ molecule which one is not correct?

0%
Six $P-P$ single bonds are present
0%
Four $P-P$ single bonds are present
0%
Four lone pairs of electron are present
0%
$P\ P\ P$ bond angle is $60^o$

Explanation

$P_4$ +

$3NaOH$ +

$3H_2O$

$\rightarrow$

$3NaH_2PO_2$ +

$PH_3$

The correct Answer is options D.

1623553_1728323_ans_e92ff312b6574220bbcaf32bf66d9390.jpg

Oxygen can be obtained from bleaching powder by:

0%
adding dilute acid
0%
passing carbon dioxide
0%
heating with a cobalt salt
0%
adding alkalies

Explanation

$\text{Oxygen can be obtained from bleaching powder by heating with a cobalt salt. Bleaching powder loses oxygen}$

$\text{in the presence of cobalt chloride catalyst.}$

$Reaction:$

$2CaOCl_2\xrightarrow{Co^{2+}}2CaCl_2+O_2$

$\text{Option C is correct.}$

White phosphorous has :

0%
six $P-P$ single bonds
0%
four $P-P$ single bonds
0%
four lone pairs of electrons
0%
$P\ P\ P$ angle of $60^o$

The catalyst used in the manufacture of

$NO$ by Ostwald's process is:

0%
$Pt$
0%
$Fe$
0%
$Mo$
0%
$Cu$

The electron affinity values

$(kJmol^{−1})$ of three halogens X, Y, and Z respectively are −349, −333, and −Then X, Y, and Z respectively are :

0%
$F_2 , CI_2, \ and\ Br_2$
0%
$CI_2, F_2 \ and \ Br_2$
0%
$CI_2, Br_2,\ and\ F_2$
0%
$Br_2, CI_2, \ and \ F_2$

______ is the by-product obtained in the Serpek's process.

0%
Oxygen
0%
Ammonia
0%
Nitrogen dioxide
0%
Nitric oxide

Which one of the following is not used for the manufacture of ammonia?

0%
Haber process
0%
Cyanamide process
0%
Ammonia mother liquor
0%
Solvay ammonia process

Which one of the compound is not known?

0%
$NCl_5$
0%
$PCl_5$
0%
$AsCl_5$
0%
$NCl_3$

Explanation

Phosphorous and arsenic have d-orbitals and hence can accommodate the electrons from 5 Cl atoms into them. In the case of N, the d-orbitals are absent and electrons can be accommodated only in p-orbitals of 5 Cl atoms. Hence,

$PCl_5$ does not exist.

Hence, the correct option is (A).

Which of the following exhibits the weakest intermolecular forces?

0%
$\mathrm{H}_{2}\mathrm{O}$
0%
$\mathrm{N}\mathrm{H}_{3}$
0%
He
0%
HCl

Explanation

Hint: London dispersion force is the weakest intermolecular force.

Explanation:

There are three major intermolecular forces that act on the compound such as;

London dispersion forces, Dipole-Dipole interactions, and hydrogen bonding.

Among those, the London dispersion forces are the weakest while hydrogen bonding is the strongest of all.
Hydrogen bonding is formed by the elements $N, O,F$ when attached to $H$ .
So, here in the case of $H_2O$ and $NH_3$ hydrogen bonding is present.
In the case of $HCl$ , we have a dipole-dipole interaction present due to the electronegativity difference between $H$ and $Cl$ .
Helium is the noble gas element and there are no attractive forces that exist in them except London dispersion. So, $He$ has London dispersion force which is the weakest intermolecular force.

Correct Option: $C$

The solubility of noble gases in water shows the order :

0%
$He > Ar > Kr > Ne > Xe$
0%
$He > Ne > Ar > Kr > Xe$
0%
$Xe > Kr > Ar > Ne > He$
0%
$Xe > Ar > Ne > Kr > He$

What is the atomic number (Z) of the noble gas that reacts with fluorine?

0%
54
0%
10
0%
18
0%
2

Explanation

He(2)
Ne(10)
Ar(18)
Kr(36)
Xe(54)

Xe reacts with

$F_{2}$ .

Its atomic no. is 54

Option A is correct.

Which of the following has the highest density?

0%
$Ne$
0%
$Ar$
0%
$He$
0%
$Xe$

Explanation

Hint: Density is the mass per unit volume which increases down the group.

Explanation:

The density of elements increases down the group. So, the density of xenon is the highest. The density of $Xe$ is the highest

Final answer: The correct answer is option (D)

Assertion (A): Solubility of noble gases in water decreases with increase in atomic size.
Reason (R): Solubility is due to dipole-induced dipole interaction.

0%
Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$ .
0%
Both $A$ and $R$ are correct and $R$ is not the correct explanation of $A$ .
0%
$A$ is true but $R$ is false.
0%
$A$ is false but $R$ is true.

Explanation

Solubility

$\alpha$ dipole-induced dipole interaction

$\alpha$ Surface area
More surface area more interaction more solubility. Surface area increases down the group. Thus, solubility increases down the group.

Statement - I: He -II has high viscosity and flows downward.
Statement - II: Liquid helium is used as cryogenic liquid.

0%
Both I and II are true. II is correct

explanation of I.
0%
Both I and II are true. II is not correct

explanation of I.
0%
I is true but II is false
0%
I is false but II is true.

In clathrate atoms or molecules the bond formed is :

0%
Metallic
0%
Covalent
0%
Ionic
0%
They do not form bond

Which of the following is a superfluid?

0%
Krypton
0%
Argon II
0%
Helium II
0%
Helium I

Explanation

$He(II)$ is a superfluid.

Total number of covalent bonds in a white phosphorous molecule is:

0%
1
0%
4
0%
3
0%
6

Explanation

Number covalent bonds present in white phosphorous

$P_4$ is 6.

Hence opion D is correct.

The anomalous behaviour of nitrogen is due to:

0%
small size and high electronegativity
0%
non availability of d-orbitals in valency shell
0%
ease of multiple bond formation
0%
all are correct

Explanation

Nitrogen differs from other members of the group due to its smaller size, high electronegativity, high IE and non availability of d-orbitals. Nitrogen has unique ability to form

$p\pi - p\pi$ multiple bonds with itself and with other elements having small size and high electronegativity.

Hydrolysis of

$\mathrm{M}\mathrm{g}_{3}\mathrm{N}_{2}$ gives :

0%
$\mathrm{N}\mathrm{H}_{3}$
0%
$\mathrm{N}\mathrm{O}$
0%
$\mathrm{H}_{2}$
0%
$\mathrm{C}\mathrm{O}_{2}$

Explanation

Hydrolysis of

$Mg_3N_2$ is explained by following reaction.

$Mg_3N_2 + 6H_2O \rightarrow 3Mg{(OH)}_{3} + 2NH_3$
So,

$NH_3$ is formed as product.

Clathrates are_________

0%
Inert gases
0%
Inert gases $+$ organic or inorganic compounds
0%
Halogen $+$ carbon black
0%
Halogen $+$ transition elements

Which compound does not give ammonia gas on reaction with water ?

0%
$\mathrm{M}\mathrm{g}_{3}\mathrm{N}_{2}$
0%
$\mathrm{A}\mathrm{l}\mathrm{N}$
0%
$\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{N}_{2}$
0%
$\mathrm{C}\mathrm{a}(\mathrm{C}\mathrm{N})_{2}$

Explanation

$Mg_3N_2 + 6H_2O \longrightarrow3Mg(OH)_2 + 2NH_3$

$AlN + 3H_2O \longrightarrow NH_3 + Al(OH)_3$

$2H_2O + Ca(CN)_2 \longrightarrow Ca(OH)_2 + 2HCN$

$CaCN_{2} + 3H_{2}O\longrightarrow$

$CaCO_{3} + 2NH_{3}$

$Ca(CN)_2$ does not produce ammonia on reacting with water.

The most abundant element in the earth's crust among the following is_________.

0%
$\mathrm{P}$
0%
$\mathrm{A}\mathrm{s}$
0%
$\mathrm{S}\mathrm{b}$
0%
$\mathrm{B}\mathrm{i}$

Explanation

The most abundant element among the given elements in the earth's crust is phosphorus. It occurs in the minerals of the apatite family. An example is a fluorapatite

$\displaystyle Ca_9[PO_4]_6 \cdot CaF_2$

Phosphorus is the most abundant element of 15th group, accounting for 0.10% of the mass of the earth's crust.

Option A is correct.

Assertion (A): In the synthesis of ammonia by Haber's process, mixture of potassium and aluminium oxides can be used as promoter.
Reason (R): Promoter enhances the activity of catalyst.

0%
Both (A) and (R) are true and (R) is the correct explanation of (A)
0%
Both (A) and (R) are true and (R) is not the correct explanation of (A)
0%
(A) is true and (R) is false
0%
(A) is false but (R) is true

Gas obtained by heating a mixture of ammonium chloride and slaked lime is:

0%
$\mathrm{N}\mathrm{H}_{3}$
0%
$\mathrm{N}_{2}$
0%
$\mathrm{N}_{2}\mathrm{O}$
0%
$\mathrm{N}\mathrm{O}_{2}$

Explanation

$2\ NH_4Cl+Ca(OH)_2\rightarrow CaCl_2^{}+2NH_3^{}+2H_2O$

The gas obtained is ammonia gas.

One mole of magnesium nitride on the reaction with an excess of water gives :

0%
one mole of ammonia
0%
one mole of nitric acid
0%
two moles of ammonia
0%
two moles of nitric acid

Explanation

Hint: Nitride on reaction with water gives ammonia

Step 1: Formula of magnesium nitride

The formula for magnesium nitride will be

$Mg_3N_2$

Step 2: Recation of magnesium nitride with excess of water

$Mg_3N_2 + 6H_2O \rightarrow 3Mg(OH)_2 + 2NH_3$

Hence one mole of magnesium nitride on reaction with excess water gives two moles of ammonia.

Final Step: So the correct answer is option $C$ .

Hydrolysis of

$\mathrm{N}\mathrm{Cl}_{3}$ gives

$\mathrm{N}\mathrm{H}_{3}$ and

$\mathrm{X}$ . Which of the following is X?

0%
$\mathrm{H}\mathrm{Cl}\mathrm{O}_{4}$
0%
$\mathrm{H}\mathrm{Cl}\mathrm{O}_{3}$
0%
$\mathrm{HO}\mathrm{Cl}$
0%
$\mathrm{H}\mathrm{Cl}\mathrm{O}_{2}$

Explanation

Hydrolysis of

$NCl_3$ can be explained by the following reaction

$NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl$
So, X is

$HOCl$ .

Ammonia is not a product in the :

0%
hydrolysis of nitrolim
0%
hydrolysis of aluminium nitride
0%
decomposition of ammonium nitrite
0%
hydrolysis of urea

Explanation

$NH_3NO_3\rightarrow N_2O+2H_2O$
Nitrous oxide is produced after decomposition of ammonium nitrite instead of ammonia.

Assertion (A): A mixture of

$CaCN_2$ and graphite is known as nitration mixture.
Reason (R) :

$NH_3$ is collected by downward displacement of air.

0%
Both (A) and (R) are true and (R) is the correct explanation of (A)
0%
Both (A) and (R) are true and (R) is not the correct explanation of (A)
0%
(A) is true and (R) is false
0%
(A) is false but (R) is true

Explanation

$HNO_{3}+H_{2}SO_{4}$ is known as nitration mixture. So, A is wrong.

$NH_{3}$ is collected by downward displacement of air as

$NH_{3}$ is lighter than air.

Nitrogen form

$N_2$ but phosphorous when forms

$P_2$ gets readily converted into

$P_4$ because : -

0%
Triple bond present between phosphorous atom
0%
$p_\pi- p_\pi$ bonding is weak
0%
$p_\pi- p_\pi$ bonding is strong
0%
Multiple bond form easilly

Explanation

Phosphorus is unable to form

$p\pi-p\pi$ bonds due to larger atomic size. So,

$P$ atom is linked with

$P$ atoms using

$3$ sigma bonds. On the other hand, Nitrogen forms

$1-\sigma$ and

$2-\pi$ bonds i.e., triple bonds and exists as diatomic molecules.

A metal X on heating in nitrogen gas gives Y. Y on treatment with

$H_2O$ gives a colourless gas which when passed through

$CuSO_4$ solution gives a blue colour. Y is :

0%
$\mathrm{Mg(N}\mathrm{O}_{3})_{2}$
0%
$\mathrm{M}\mathrm{g}_{3}\mathrm{N}_{2}$
0%
$\mathrm{N}\mathrm{H}_{3}$
0%
$\mathrm{M}\mathrm{g}\mathrm{O}$

Explanation

Colorless gas is

$NH_{3}$ .

$Mg+N_2 \rightarrow Mg_3N_2$

X Y

$Mg_{3}N_{2}+3H_{2}O \rightarrow 3MgO+2NH_{3}$

Ammonia when passed through copper sulphate forms a deep blue colored complex.

Option B is correct.

Ammonia gas is prepared by :

0%
reduction of sodium nitrite or sodium nitrate by the reaction with zinc dust and sodium hydroxide.
0%
hydrolysis of calcium cyanamide.
0%
heating of $\left ( NH_{4} \right )_{2}Cr_{2}O_{7}$
0%
heating ammonium chloride with slaked lime.

White phosphorus

$P_{4}$ has:

0%
six $P-P$ single bonds
0%
four $P-P$ single bonds
0%
four lone pairs of electrons
0%
$P-P-P$ angle of 60 degerees

The ease of liquefaction of noble gases increases in the order

0%
Ar > Kr > Xe > Ne > He
0%
Xe > Kr > Ne > Ar > He
0%
Kr > Xe > He > Ne > Ar
0%
He > Ne > Ar > Kr > Xe

Ammonia gas is prepared by :

0%
reduction of sodium nitrite or sodium nitrate by the reaction with zinc dust and sodium hydroxide.
0%
hydrolysis of calcium cyanamide.
0%
heating of $(NH_{4})_{2}Cr_{2}O_{7}$
0%
heating ammonium chloride with slaked lime.

The following acids have been arranged in order of decreasing acid strength. Identify the correct order.

$ClOH (I),\ BrOH (II),\ IOH (III)$

0%
$I>II>III$
0%
$II>I>III$
0%
$III>II>I$
0%
$I>III>II$

Which of the following statements is/are incorrect ?

0%
In $B_2H_6(s)$ there are four $2c-2e^-$ bonds
0%
In $(SiH_3)_3 \ddot {P}$ there is back bonding
0%
$(CH_3)_3\ddot{N}$ and $(SiH_3)_3\ddot{N}$ are pyramidal
0%
$Al_2Cl_6(s)$ has banana bonding

Explanation

Only statement A is correct. All terminal

$B-H$ bonds are 2c-2e bonds.
In

${(SiH_3)}_3N$ ,

$Si$ has vacant d-orbital which forms

$p\pi-d\pi$ bonding. So there is no back bonding in

${(SiH_3)}_3N$ .

${(CH_3)}_3N$ has

$sp^3$ hybridization on nitrogen with pyramidal shape due to

$p\pi-p\pi$ bonding whereas in

${(SiH_3)}_3N$ ,

$Si$ has vacant d orbitals which forms

$p\pi-d\pi$ bonding. So,

${(SiH_3)}_3N$ has planar structure whereas

${(CH_3)}_3N$ has pyramidal structure.

$Al_2Cl_6$ is a dimer of

$AlCl_3$ in which

$Cl$ forms coordinate bond with

$Al$ as shown in figure.

When

$MnO_{2}$ is added to

$KClO_{3}$ in the preparation of oxygen the decomposition temperature of

$KClO_{3}$ is reduced from

$350^{o}C$ to ......

0%
$300^{o}C$
0%
$150^{o}C$
0%
$250^{o}C$
0%
room temperature

Explanation

When

$MnO_2$ is added to

$KClO_3$ in the preparation of oxygen the decomposition temperature of

$KClO_3$ is reduced from

$350^oC$ to

$250^oC$ .

As the role of

$MnO_2$ in the preparation of

$O_2$ is that of a catalyst. It does not take part in the reaction.

$O_2$ is prepared in the laboratory by heating

$KClO_3$ which requires to be heated up to a temperature up to

$350^oC$ . It is dangerous too. But

$MnO_2$ acts as a

$positive \ catalyst$ which enables the reaction to take place at

$250^oC$ at a much faster rate.

Option C is correct.

In which of the following reactions,

$O_{2}$ is not formed as one of the product?

0%
$KClO_{3}\xrightarrow [ Heat ]{ MnO_{2} }$
0%
$SnCl_{2}+HCl+O_{3}\rightarrow$
0%
$FeSO_{4}+H_{2}SO_{4}+O_{3}\rightarrow$
0%
$PbS+O_{3}\rightarrow$

Consider the following reaction,

$NO_2 +NaOH \to (A)+(B)+H_2O$
which of the following is correct?

0%
both $(A)$ and $(B)$ on heating with $Na$ from $N_2$
0%
both $(A)$ and $(B)$ on treatment with $Zn$ and $NaOH$ produce $NH_3$
0%
both are correct
0%
none is correct

What is formed xenon combines with fluorine in the pressure of mercury vapour?

0%
$XeF$
0%
$Xe_2F$
0%
$XeF_2$
0%
$XeF_4$

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 6 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 6 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.