MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 11

For the manufacture of ammonia by the reaction

$N_{2}+3H_{2} \rightleftharpoons 2NH_{3}+21.9\ K cal$ , the favorable conditions are:

0%
low temperature, low pressure & catalyst
0%
low temperature, high pressure & catalyst
0%
high temperature, low pressure & catalyst
0%
low temperature, high pressure

Explanation

The equation representing habers process of

${ NH }_{ 3 }$

synthesis is-

${ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\rightleftharpoons { 2NH }_{ 3 }(g)\\$

pressure: 150-200 atm

temperature :

$450℃-500℃$

catalyst : Iron

Among

$BF_{3},N(SiH_{3})_{3},N(CH_{3})_{3}, O(SiH_{3})_{2},CF_{3},CCl_{3}^{-}$ the number of molecules or molecular ions which have back bonding is:

0%
$1$
0%
$3$
0%
$4$
0%
$all$

Explanation

$BF_{ 3 }$ has trigonal planner structure in which B has empty P-orbital.Hue,F acts as lewis base and donater its lone pair to boron and it results in back bonding.
1013367_1072100_ans_395da5abd14d4762822454e79353b93b.PNG

1013367_1072100_ans_395da5abd14d4762822454e79353b93b.PNG

Which of the following does not react with xenon to form compunds?

0%
$OSeF_{5}$
0%
$OTeF_{5}$
0%
$F_{2}$
0%
$N_{2}$

The catalyst and promoter respectively used in the Haber's process of industrial synthesis of ammonia are:

0%
$Mo,{ V }_{ 2 }{ O }_{ 5 }$
0%
${ V }_{ 2 }{ O }_{ 5 },Fe$
0%
$Fe,Mo$
0%
$Mo,Fe$

Explanation

Correct Answer: Option

$C$

Explanation of correct answer:

A catalyst is a substance that boosts up the reaction and helps out to form maximum product in the minimum time period, at the same time the promoter improves its performance of chemical reaction by interacting with active components of catalysts.
Ammonia formation process is exothermic and reversible that is why we use catalyst to increase the speed of the reaction.

In ammonia formation by Haber's process,

$Fe$ acts as a catalyst and

$Mo$ as a promoter.

Explanation of incorrect answer:

$V_{2}O_{5}$ act as a catalyst but not in ammonia formation, only

$Fe$ act as a catalyst and

$Mo$ as a promoter, hence other options are incorrect.

Which of the following elements occur free in nature?

0%
Nitrogen
0%
Phosphorus
0%
Arsenic
0%
Antimony

When two atoms of chlorine combine to form one molecule of chlorine gas, the energy of the molecule:

0%
Greater than that of separate atom
0%
Equal to that of separate atom
0%
Lower than that of separate atoms
0%
None of the above

Explanation

The energy of the molecule is less than that of a separate atom because the

$Cl_2$ molecule is more stable than the

$Cl$ atom.

$Cl$ atom is more reactive and it has more energy and less stability. It combines with other

$Cl$ atom to form

$Cl_2$ which has less energy and more stability than an individual atom.

Hence, option

$C$ is correct.

Which of the following has

$p\pi -d\pi$ bonding?

0%
${ NO }_{ 3 }^{ - }$
0%
${ CO }_{ 3 }^{ -2 }$
0%
${ BO }_{ 3 }^{ -3 }$
0%
${ SO }_{ 3 }^{ -2 }$

Explanation

$p\pi-d\pi$ bonding includes d-orbitals,if you observe the options only S belongs to

$3^{rd}$ period and has d-orbitals(though unoccupied in ground state).

Other options

$C$ and

$N$ belongs to

$2^{nd}$ period while B belongs to

$1^{st}$ period and doesn't have d-orbitals.

$p\pi -d\pi \quad$ bonding is present in:

0%
${ NO }_{ 3 }^{ - }$
0%
${ SO }_{ 3 }^{ 2- }$
0%
${ BO }_{ 3 }^{ 3- }$
0%
${ CO }_{ 3 }^{ 2- }$

Explanation

So,

$1p\pi - d\pi$ bonding is present is it.

The structure of

$SO_3^2-$ is
1155574_1098236_ans_c225430483bd466c9545f71220197bec.png

$NH_3$ is formed in the following steps
I.

$Ca+2C\rightarrow CaC_2$ 50% yield
II.

$CaC_2+N_2\rightarrow CaCN_2+C$ 100% yield
III.

$CaCN_2+3H_2O\rightarrow 2NH_3+CaCO_3$ 50% yield
To obtain 2 moles

$NH_3$ , calcium required is:

0%
1 mol
0%
2 mol
0%
3 mol
0%
4 mol

Explanation

It can be easily observed from the reaction given that 1mole of Ca is required to get 2 moles of

$NH_3$

Which compound of xenon is not possible ?

0%
$XeF_{2}$
0%
$XeF_{4}$
0%
$XeF_{5}$
0%
$XeF_{6}$

Explanation

Xe has a complete filled 5p configuration. As a result when it undergoes bonding with an odd number (3 or 5) of F atoms it leaves behind one unpaired electron. This causes the molecule to become unstable. As a result

$XeF_3$ and $XeF_5$ do not exist.

Hence option C is correct answer.

As per Braun's principle, yield of Ammonia will be more in Haber's process under ______, conditions (

$L=$ Low;

$H=$ high;

$T=$ Temp;

$P=$ Pressure)

0%
$LT;LP$
0%
$LT;HP$
0%
$HT;HP$
0%
$HT;LP$

Explanation

The manufacture of

$NH_3$ by Haber's process is given as

${ N }_{ 2 }(g)+{ 3H }_{ 2 }(g)\rightleftharpoons { 2NH }_{ 3 }(g)\\ \triangle 6H°=-92.4\\$

so, it is an exothermic reversible reaction and occurs with decrease in volume.

so, the low temperature will favour the forward reaction which is exothermic in nature

and since the reaction occurs with a decrease in volume .so, high pressure will favour the formation of of

$NH_3$ .

In Carius method, of estimation of halogens,

$250mg$ of an organic compound gave

$141mg$ of

$AgBr$ . The percentage of bromine in the compound is:

[Atomic mass

$Ag=108,Br=80$ amu]

0%
$24$
0%
$36$
0%
$8$
0%
$60$

Explanation

Hence, the correct option is

$A$

1411770_1142426_ans_6ef781fad8f14671a998b6a03a421643.PNG

The role of zinc stearate in the process of vulcanisation is:

0%
to accelerate the process
0%
to slow down the process
0%
to stop the process
0%
to initiate the process

Which of the following characteristics regarding 7 halogens is not correct?

0%
Ionization energy decreases with increase in atomic number.
0%
Electronegativity decreases with increase in atomic number.
0%
Electron affinity decreases with increase in atomic number.
0%
Enthalpy of fusion increases with increase in atomic number

Which of the following alkyl halides is used as a methylating agent?

0%
${C}_{2}{H}_{5}Br$
0%
${C}_{6}{H}_{5}Cl$
0%
${CH}_{3}Cl$
0%
${C}_{2}{H}_{5}Cl$

Explanation

methylation is a form of alkylation with a methyl group replacing a hydrogen atoms .

So,

$CH_3Cl$ provides that methyl group wich replaces the hydrogen atom of the substrate.

An alkyl halide reacts with alcoholic ammonia in a sealed tube, the product formed will be

0%
a primary amine
0%
a secondary amine
0%
a tertiary amine
0%
All of the above

Stability of tetrabromides follows the order :

0%
${ SeBr }_{ 4 } < { TeBr }_{ 4 } < { PoBr }_{ 4 }$
0%
${ SeBr }_{ 4 } < { PoBr }_{ 4 } < { TeBr }_{ 4 }$
0%
${PoBr }_{ 4 } < { TeBr }_{ 4 } < { SeBr}_{ 4 }$
0%
None of the these

Explanation

Stability of higher oxidation state increase as we go down the group - Tetrabromides of sulphur and selenium. i.e,

$SBr_4,SeBr_4$ are unstable where as

$TeBr_4, PoBr_4$ are quite stable.

Stability :

$SeBr_4<TeBr_4<PoBr_4$

In which reaction ammonia acts as an acid?

0%
${NH}_{3}+HCl\rightarrow {NH}_{4}Cl$
0%
${NH}_{3}+{H}^{+}\rightarrow {NH}_{4}^{+}$
0%
${NH}_{3}+Na\rightarrow Na{NH}_{2}+\cfrac{1}{2}{H}_{2}$
0%
${NH}_{3}$ cannot act as an acid

Explanation

$NH_3$ is a strong base due to high electron density on nitrogen

So, it cannot act as an acid .

One of the following combinations which illustrates the law of reciprocal proportion?

0%
N $_{2}$ O $_{3}$ ,N $_{2}$ O $_{4}$ ,N $_{2}$ O $_{5}$
0%
NaCl ,NaBr ,Nal
0%
CS $_{2}$ ,CO $_{2}$ , SO $_{2}$
0%
MgO,Mg(OH) $_{2}$

$" N < P < As"$
Above order is incorrect for:

0%
Electropositivity
0%
Covalent radii
0%
Size of $M^{3-}$
0%
Acidic nature of $M_2O_3$ type oxide

Explanation

$N, \ P \ and \ As$ belong to the same group. On moving down the group the atomic radii of atoms increase. Due to the increase in atomic size, the covalent radii and ionic radii also increases. Also with an increase in the size of atom, the nuclear pull on the outermost electrons as well as the ability to attract electrons decreases, resulting in decrease in electronegativity. Hence, electropositivity increases.

$M_2O_3$ type of oxides have central atom in +3 oxidation state. On moving down the group the ability of atoms to loose electrons increases due to the weakening of nuclear pull on the outermost electrons, hence, basicity increases. Thus, the acidic nature of oxides decreases.

The correct order of stability is

0%
${ N }_{ 2 }>{ N }_{ 2 }^{ + }>{ N }_{ 2 }^{ - }$
0%
${ N }_{ 2 }>{ N }^{ - }_{ 2 }>{ N }_{ 2 }^{ + }$
0%
${ N }_{ 2 }^{ + }>{ N }^{ - }_{ 2 }>{ N }_{ 2 }$
0%
None of these

Explanation

Bond order for

$N_2=3$

Bond order for

$N_2^+=2.5$

Bond order for

$N_2^-=2$

Bond order

$\propto$ Stability of molecule.

Therefore, the correct order of stability is:

$N_2>N_2^+>N_2^-$

A piece of magnesium ribbon was heated to redness in an atmosphere of nitrogen and on cooling water was added the gas evolved was:

0%
ammonia
0%
hydrogen
0%
nitrogen
0%
oxygen

Explanation

Magnesium metal reacts with nitrogen in air to form magnesium nitride which on hydrolysis gives ammonia gas.

$3Mg+N_2\xrightarrow{Burning} Mg_3N_2$

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

An element X forms an oxide

$XO_3$ . What is the valency of X?

0%
1
0%
2
0%
3
0%
6

When a mixture of carbon monoxide and chlorine is exposed to sunlight the product formed is:

0%
carbon tetrachloride
0%
phosgene
0%
phosphine
0%
carbon chloride

A mixture of

$NH_4Cl$ and

$NaCl$ can be separated by

0%
filtration
0%
Distillation
0%
Sublimation
0%
Decantation

Explanation

$NH_4$ undergoes sublimation while NaCI does not.

In which one of the following processes is iron used as a catalyst in the preparation of ammonia gas?

0%
Bayer's process
0%
Haber's process
0%
Froth floatation process
0%
None of the above

Which element forms maximum multiple bonds?

0%
N
0%
P
0%
As
0%
Bi

Explanation

Nitrogen forms maximum multiple bonds.
For example, it forms

$N \equiv N$ triple bond in dinitrogen.

The bond in compound formed from combination of 14 group and 17 group elements of Periodic table will be ..............

0%
Electrovalent bond
0%
Co-ordinate bond
0%
Van der Walls bond
0%
Covalent bond

Explanation

Groups

$14$ is carbon family. Common characteristics of groups 14 elements are they form covalent bonds. Because carbon's electronic configuration is

$1s^2\,2s^2\,2p^2$ . If forms 4 covalent bonds in compounds. Groups

$17$ is Halogen family. The bond formed by

$17$ element

$(Cl,I,Br\,etc)$ is covalent bond.

The most favourable condition for the manufacture of

$NH_3$ is:

0%
high temperature and high pressure
0%
low temperature and low pressure
0%
high temperature and low pressure
0%
low temperature and high pressure

Explanation

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

$\Delta H = - 92.38$ kj/mol

(i) The reaction is exothermic, so according to Le -Chatelier's principle it is favoured at low temperature.

(ii)The number of moles are decreasing in the reaction. According to Le-Chateliers principle such reactions are favoured at high pressure.

$\therefore$ Low temperature and high pressure are favourable conditions for formation of NH

$_3$ .

In Haber process of Ammonia production the element used as catalytic promoter to increase the activity of Iron catalyst is .......................

0%
NI (Nickel)
0%
W (Tungston)
0%
V (Vanadium)
0%
Mo (Molybednum)

Explanation

In Haber process of ammonia production the element used as catalytic promoter to increase the activity of Iron catalyst is Mo (Molybednum).

$N_2+3H_2 \rightleftharpoons 2NH_3 + 24 \quad kcal$
The temperature is

$450^oC$ and the pressure is 200-250 atm.

Generally non-metals are non-lustrous. Exception to this property is-

0%
Flourine
0%
Chlorine
0%
Bromine
0%
Iodine

Which of the following has more acidic in nature?

0%
HClO
0%
HClO $_2$
0%
HClO $_3$
0%
HClO $_4$

Explanation

Acidity of oxyacids of halogen increases with increase in oxidation number of halogen Oxidation number of CI in HCIO is + 1, HCIO

$_2$ is +3, HClO

$_3$ is + 5 and HCIO

$_4$ is + 7.

$\therefore$ HClO

$_4$ is most acidic among given choices.

An element X combines with hydrogen to form a compound

$XH_3$ . The element X is placed on the right side ofthe periodic table. What is true about the element X?
A. Has 3 valence electrons.
B. Is a metal and is solid.
C. Is a non-metal and is a gas.
D. Has 5 valence electrons.
E.

$XH_3$ reacts with water to form a basic compound?

0%
A, B and C
0%
B,C and D
0%
C, D and E
0%
E, A and B

Explanation

Here,

$X$ is nitrogen with atomic number 7 and electronic configuration

${ 1s }^{ 2 }{ 2s }^{ 2 }{ 2p }^{ 3 }$

It is a non metal belonging to group 15.

Solution of ammonia In water is alkaline in nature and is highly soluble in water.

Which molecule is relatively more stable?

0%
$O_2$
0%
$H_2$
0%
$Li_2$
0%
$N_2$

Explanation

Nitrogen molecule (

$\displaystyle N_2$ ) is relatively more stable. Each

$N$ atom has completed its octet and there is a triple bond between two

$N$ atoms. In other molecules, there is either a single bond or a double bond between two atoms. A triple bond is stronger than a single or double bond.

Which of the following chalcogen does not make bonds with germanium?

0%
Sulfur
0%
Selenium
0%
Tellurium
0%
Polonium

Noble gases have __________ electron affinity.

0%
high
0%
low
0%
zero
0%
very low

The wrong statements about

$\displaystyle NH_{3}$ is :

0%
it is oxidised with oxygen at $\displaystyle 700^{\circ}C$ in the presence of platinum
0%
it gives black precipitate with calomel
0%
it can be dried by $\displaystyle P_{2}O_{5},\ H_{2}SO_{4}\,$ and $\, CaCl_{2}$
0%
it gives white fumes with HCI

Explanation

$NH_3$ is oxidised to nitric oxide and water with oxygen at

$700^oC$ in the presence of platinum.

Aqueous ammonia reacts with calomel

$Hg_2Cl_2$ to produce metallic mercury, a black precipitate.

$CaO$ calcium oxide is used for drying of ammonia. Because

$CaO$ being basic in nature does not react with

$NH_3$ while

$P_2O_5,$ conc.

$H_2SO_4 \$ and

$\ CaCl_2$ reacts with it.

$CaO + H_2O + NH_3 \rightarrow NH_3 + Ca(OH)_2$

$CaCl_2 + 8NH_3 \rightarrow CaCl_2.8NH_3$

$P_2O_5 + 3H_2O + 6NH_3 \rightarrow 2(NH_4)_3PO_4$

$H_2SO_4 + NH_3 \rightarrow (NH_4)_2SO_4$

$NH_3$ gives White fumes of

$NH_4Cl$ with

$HCl$ .

The gas oxygen from thermal decomposition of

$\displaystyle (NH_4)_2 \, Cr_2 O_7$ :

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

Explanation

$\displaystyle (NH_4)_2 \, Cr_2O_7 \, \overset{\Delta}{\longrightarrow} Cr_2O_3 + N_2 + 4H_2O$

Thermodynamically most stable allotrope of phosphorus is_____

0%
red
0%
white
0%
black
0%
yellow

Why does one who drinks alcohol get tempted to drink more alcohol?

0%
Because $CO_2$ increases in liver
0%
Because formaldehyde increases in liver
0%
Because amount of enzyme P- $450$ increases in liver
0%
Because amount of disulphiram increases in liver

Explanation

Because the amount of enzyme P-

$450$ increases in the liver.
In the liver of alcohol drinkers, the amount of the P-

$450$ enzyme increases, and thus, the person is tempted to drink more alcohol.

In Ostwald process of manufacturing of

$HNO_3$ catalyst used is:

0%
Mo
0%
Fe
0%
Mn
0%
Pt

Explanation

Ostwald process for manufacture of

$HNO_3$ ,

$4 NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O+ \Delta$

$2NO + O_2 \xrightarrow{50^o C} 2 NO_{2(g)}$

$4NO_2 + 2H_2O + O_2(g)\rightarrow 4HNO_3$

$\therefore$ Pt is used as catalyst in Ostwald process.

Option D is correct.

'Kajal' is produced from:

0%
stibnite
0%
charcoal
0%
carbon black
0%
asphalt

Explanation

Kohl/Kajal is a dark powder used as eye makeup. It is made from Stibinite

$(Sb_2S_3)$ which is the main use of Antimony,

$Sb_2S_3$ is used in Kohl/Kajal due to its soft material and dark colour.

The halogen

$($ Group

$17$ elements

$)$ :

0%
have an electron configuration of a noble gas less seven electrons
0%
are highly electro positive
0%
show variable oxidation state of $-1, +1, +3, +5$ and $+7$ in their various compounds
0%
forms a volatile, covalent hydrides $HX$ in which the halogen $(X)$ shows an oxidation state of $+1$

Which noble gas is most soluble in water?

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

Explanation

Solubility of noble gases in water increase down the group due to increase in dipole interaction between

$H_2O$ molecule and noble gas atom down the group .

Which group of periodic table contains no metal?

0%
IA
0%
IIIA
0%
VIIA
0%
VIII

Explanation

Group IA and III A contain mostly metals. Group VIII contains transition elements which are metals. Group VII A,contains mostly non-metals

$(F,Cl, Br)$ .

VIIA group is the group of halogens.

It contains no metal.

The word halogen means salt producing so, halogens reacts with metals to form salts.

The halogen family is as follows:

Flourine

Chlorine

Bromine

Iodine

Astatine.

The Group VA element that makes up part of bones and teeth is:

0%
N
0%
P
0%
As
0%
Sb

Which one of the following compounds is heated with slaked lime to produce ammonia?

0%
${NH}_{4}Cl$
0%
$Na{NO}_{3}$
0%
$K{NO}_{3}$
0%
$Ca{({NO}_{3})}_{2}$

Explanation

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

Noble gases are absorbed by :

0%
Anhydrous calcium chloride
0%
Ferric hydroxide
0%
Conc. $H_2SO_4$
0%
Activated coconut charcoal

Explanation

Activated charcoal is a very good absorbent of gas, whereas anhydrous

$CaCl_2$ ,

$Fe(OH)_3$ and conc.

$H_2SO_4$ are dehydrating agents.

Why are some elements chemically inert?

0%
Because their outermost shell is completely empty
0%
Because their outermost shell is completely filled
0%
Because their innermost shell is completely empty
0%
Because their innermost shell is completely filled

Explanation

Due to the fully filled outermost shell, the elements are stable. Hence they do not take part in the reaction.

OPtion

$B$ is correct.

The comparatively high b.pt of

$HF$ is due to:

0%
high reactivity of fluoring
0%
small size of hydrogen atom
0%
formation of hydrogen bonds and consequent association
0%
high IE of flourine

Explanation

Intermolecular hydrogen bonding is found in

${(HF)}_{n}$ due to higher electronegativity of fluorine atoms

$\underset { \downarrow \\ hydrogen\quad bonding }{ \cdot \cdot \cdot H-F\cdot \cdot \cdot H-F\cdot \cdot \cdot H-F\cdot \cdot \cdot }$

Hydrogen bonding is helpful in the association of

$HF$ molecule, so

$HF$ is found in liquid form.

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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