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Test: 35 Year JEE Previous Year Questions: The p-Block Elements

41 Questions MCQ Test Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper | Test: 35 Year JEE Previous Year Questions: The p-Block Elements

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Attempt Test: 35 Year JEE Previous Year Questions: The p-Block Elements | 41 questions in 80 minutes | Mock test for JEE preparation | Free important questions MCQ to study Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper for JEE Exam | Download free PDF with solutions

QUESTION: 1

Alum helps in purifying water by

A.
forming Si complex with clay partiles
B.
sulphate part which combines with the dirt and removes it
C.
coagulaing the mud particles
D.
making mud water soluble.

Solution:

Alum furnishes Al³⁺ ions which bring about coagulation of negatively charged clay particles, bacteria etc.

QUESTION: 2

In XeF₂, XeF₄, XeF₆ the number of lone pairs on Xe are respectively

A.
2, 3, 1
B.
1, 2, 3
C.
4, 1, 2
D.
3, 2, 1.

Solution:

In the formation of XeF₂, sp³d hybridisation occurs which gives the molecule a trigonal bipyramidal structure.

In the formation of XeF₄, sp³d² hybridization occurs which gives the molecule an octahedral structure.

In the formation of XeF₆, sp³d³ hybridization occurs which gives the molecule a pentagonal bipyramidal structure.

QUESTION: 3

In case of nitrogen, NCl₃ is possible but not NCl₅ while in case of phosphorous, PCl₃ as well as PCl₅ are possible. It is due to

A.
availability of vacant d orbitals in P but not in N
B.
lower electronegativity of P than N
C.
lower tendency of H-bond formation in P than N
D.
occurrence of P in solid while N in gaseous state at room temperature.

Solution:

NOTE : In phosphorous the 3d- orbitals are available.
Hence phosphorous can from pentahalides also but nitrogen can not form pentahalide due to absence of dorbitals

QUESTION: 4

Which of the following statements is true?

A.
HF is less polar than HBr
B.
absolutely pure water does not contain any ions
C.
chemical bond formation take place when forces of attraction overcome the forces of repulsion
D.
in covalency transference of electron takes place.

Solution:

QUESTION: 5

Number of sigma bonds in P₄O₁₀ is

A.
6
B.
7
C.
17
D.
16.

Solution:

QUESTION: 6

Oxidation number of Cl in CaOCl₂ (bleaching power) is:

A.
zero, since it contains Cl₂
B.
– 1, since it contains Cl^–
C.
+ 1, since it contains ClO^–
D.
+ 1 and – 1 since it contains ClO^– and Cl^–

Solution:

hence oxidation no of Cl in OCl– is
–2 + x₂ = –1
x₂ = 2–1 = +1
now oxidation no. of another Cl is –1 as it is present as Cl^–.

QUESTION: 7

What may be expected to happen when phosphine gas is mixed with chlorine gas ?

A.
PCl₃ and HCl are formed and the mixture warms up
B.
The mixture cools down only
C.
PH₃. Cl₂ is formed with warming up
D.
PCl₅ and HCl are formed and the mixture cools down

Solution:

On mixing phosphine with chlorine gas PCl₅ and HCl forms. The mixture cools down.
PH₃ + 4Cl₂ –––––→ PCl₅ + 3HCl

QUESTION: 8

Concentrated hydrochlor ic acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that

A.
oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas
B.
strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke.
C.
due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself.This moisture forms droplets of water and hence the cloud.
D.
concentrated hydrochloric acid emits strongly smelling HCl gas all the time.

Solution:

QUESTION: 9

Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite

A.
is an allotropic form of diamond
B.
h as molecules of var iable molecular masses like polymers
C.
has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds
D.
is a non-crystalline substance

Solution:

In graphite, carbon is sp² hybridized. Each carbon is thus linked to three other carbon atoms forming hexagonal rings. Since only three electrons of each carbon are used in making hexagonal ring, fourth electron of each carbon is free to move. This makes graphite a good conductors of heat and electricity.
Further graphite has a two dimensional sheet like structure. These various sheets are held together by weak van der Waal’s force of attraction. due to these weak forces of attraction, one layer can slip over the other. This makes graphite soft and a good lubricating agent.

QUESTION: 10

Glass is a

A.
super-cooled liquid
B.
gel
C.
polymeric mixture
D.
micro-crystalline solid

Solution:

Glass is a translucent or transparent amorphous supercooled solid solution or we can say super cooled liquid of silicates and borats having a general formula R₂O. MO. 6 SiO₂. where R = Na or K and M = Ca, Ba, Zn or Pb.

QUESTION: 11

Which one of the following substan ces has the highest proton affinity ?

A.
H₂S
B.
NH₃
C.
PH₃
D.
H₂O

Solution:

Among the given compounds, the is most basic. Hence has highest proton affinity

QUESTION: 12

For making good quality mirrors, plates of float glass are used. These are obtained by floating molten glass over a liquid metal which does not solidify before glass. The metal used can be

A.
tin
B.
sodium
C.
magnesium
D.
mercury

Solution:

It is mer cury because it exists as liquid at room temperature.

QUESTION: 13

Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?

A.
Bond dissociation energy
B.
Ionization enthalpy
C.
Electron affinity
D.
Hydration enthalpy

Solution:

Fluorine is the strongest oxidizing agent due to low heat of dissociation and high heat of hydration which compensate the lower value of electron affinity.

QUESTION: 14

Which one of the following statement regarding helium is incorrect ?

A.
It is used to produce and sustain powerful superconducting magnets
B.
It is used as a cryogenic agent for carrying out experiments at low temperatures
C.
It is used to fill gas balloons instead of hydrogen because it is lighter and non-inflammable
D.
It is used in gas-cooled nuclear reactors

Solution:

Property of He -It is used in weather balloons and tires of airplanes as He is non inflammable. He is heavier than H. So the correct option is c.

QUESTION: 15

Beyllium and aluminium exhibit many properties which are similar. But, the two elements differ in

A.
forming covalent halides
B.
forming polymeric hydrides
C.
exhibiting maximum covalency in compounds
D.
exhibiting amphoteric nature in their oxides

Solution:

The maximum valency of beryllium is +2 while that of aluminium is +3.

QUESTION: 16

Aluminium chloride exists as dimer, Al₂Cl₆ in solid state as well as in solution of non-polar solvents such as benzene.
When dissolved in water, it gives

A.
[Al(OH)₆]^3- + 3HCl
B.
[Al(H₂O)₆]³⁺ + 3Cl^-
C.
Al³⁺ + 3Cl^-
D.
Al₂ O₃ + 6HCl

Solution:

QUESTION: 17

Excess of KI reacts with CuSO₄ solution and then Na₂S₂O₃ solution is added to it. Which of the statements is incorrect for this reaction ?

A.
Na₂S₂O₃ is oxidised
B.
CuI₂ is formed
C.
Cu₂I₂ is formed
D.
Evolved I₂ is reduced

Solution:

QUESTION: 18

The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is

A.
three
B.
one
C.
two
D.
zero

Solution:

Hypophosphorous acid

QUESTION: 19

The correct order of the thermal stability of hydrogen halides (H – X) is

A.
HI > HCI < HF > HBr
B.
HCI< HF > HBr < HI
C.
HF > HCI > HBr > HI
D.
HI < HBr > HCI < HF

Solution:

The H–X bond strength decreases from HF to HI. i.e.
HF > HCl > HBr > HI. Thus HF is most stable while HI is least stable. This is evident from their decomposition reaction: HF and HCl are stable upto 1473K, HBr decreases slightly and HI dissociates considerably at 713K. The decreasing stability of the hydrogen halide is also reflected in the values of dissociation energy of the H–X bond

QUESTION: 20

Heating an aqueous solution of alumin ium chloride to dryness will give

A.
Al(OH)Cl₂
B.
Al2O₃
C.
Al₂Cl₆
D.
AlCl₃

Solution:

The solution of aluminium chloride in water is acidic due to hydrolysis.

AlCl₃ + 3H₂O –––––→ Al(OH)₃ + 3HCl.
On heating it till dryness Al(OH)₃ is converted into Al₂O₃

QUESTION: 21

In silicon dioxide

A.
there are double bonds between silicon and oxygen atoms
B.
silicon atom is bonded to two oxygen atoms
C.
each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms
D.
each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms.

Solution:

In SiO₂ (quartz), each of O-atom is shared between two SiO₄^4– tetrahedra

QUESTION: 22

The structure of diborane (B₂H₆) contains

A.
four 2c-2e bonds and four 3c-2e bonds
B.
two 2c-2e bonds and two 3c-3e bonds
C.
two 2c-2e bonds and four 3c-2e bonds
D.
four 2c-2e bonds and two 3c-2e bonds

Solution:

In diborane structure B₂H₆ there are two 2c-2e bonds and two 3c–2e bonds (see structure of diborane).
Structure of B₂H₆ :

QUESTION: 23

Which of the following statements is true?

A.
HClO₄ is a weaker acid than HClO₃
B.
HNO₃ is a stronger acid than HNO₂
C.
H₃PO₃ is a stronger acid than H₂SO₃
D.
In aqueous medium HF is a stronger acid than HCl

Solution:

The is stronger than The more the oxidation state of N, the more is the acid character.

QUESTION: 24

The increasing order of the first ionization enthalpies of the elements B, P, S and F (Lowest first) is

A.
B < P < S < F
B.
B < S < P < F
C.
F < S < P < B
D.
P < S < B < F

Solution:

The correct order of ionisation enthalpies is

F > P > S > B

NOTE : On moving along a period ionization enthalapy increases from left to right and decreases from top to bottom in a group. But this trend breaks up in case of atom having fully or half filled stable orbitals.
In this case P has a stable half filled electronic configuration hence its ionisation enthalapy is greater in comparision to S. Hence the correct order is

B < S < P < F.

QUESTION: 25

What products are expected from the disproportionation reaction of hypochlorous acid?

A.
HCl and Cl₂O
B.
HCl and HClO₃
C.
HClO₃ and Cl₂O
D.
HClO₂ and HClO₄

Solution:

During disproportionation same compound undergo simultaneous oxidation and reduction.

QUESTION: 26

Identify the incorrect statement among the following.

A.
Br₂ reacts with hot and strong NaOH solution to give NaBr and H₂O.
B.
Ozone reacts with SO₂ to give SO₃.
C.
Silicon reacts with NaOH_(aq) in the presence of air to give Na₂SiO₃ and H₂O.
D.
Cl₂ reacts with excess of NH₃ to give N₂ and HCl.

Solution:

Chlorine reacts with excess of ammonia to produce ammonium chloride and nitrogen.

QUESTION: 27

Regular use of the following fertilizers increases the acidity of soil?

A.
Ammonium sulphate
B.
Potassium nitrate
C.
Urea
D.
Superphosphate of lime.

Solution:

(NH₄)₂SO₄ + 2H₂O → 2H₂SO₄ + NH₄OH H₂SO₄ is strong acid and increases the acidity of soil.

QUESTION: 28

Which one of the following is the correct statement?

A.
Boric acid is a protonic acid
B.
Beryllium exhibits coordination number of six
C.
Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase
D.
B₂H₆.2NH₃ is known as ‘inorganic benzene’

Solution:

The correct formula of inorganic benzene is B₃N₃H₆ so (d) is incorrect statement

Boric acid is a lewis acid so (a) is
incorrect statement.
The coordination number exhibited by beryllium is 4 and not 6 so statement (b) is incorrect.
Both BeCl₂ and AlCl₃ exhibit bridged structures in solid state so (c) is correct statement.

QUESTION: 29

Which one of the following reactions of xenon compounds is not feasible?

A.
3Xe F₄ + 6H₂O → 2 Xe + XeO₃ + 12HF + 1.5O₂
B.
2Xe F₂ + 2H₂O → 2Xe + 4HF + O₂
C.
X_e F₆ + RbF → Rb[Xe F₇]
D.
XeO₃ + 6HF → Xe F₆ + 3H₂O

Solution:

The products of the concerned reaction react each other forming back the reactants.

XeF₆ + 3H₂O → XeO₃+ 6HF .

QUESTION: 30

Which of the following statement is wrong?

A.
The stability of hydride increases from NH₃ to BiH₃ in group 15 of the periodic table.
B.
Nitrogen cannot form dπ – pπ bond.
C.
Single N – N bond is weaker than the single P – P bond.
D.
N₂O₄ has two resonance structures.

Solution:

The ease of for mation an d stability of hydrides decreases rapidly from NH3 to BiH₃. This is evident from their dissociation temperature which decreases from NH₃ to BiH₃. As we go down the group the size of central atom increases and thus metal-hydrogen bond becomes weaker due to decreased overlap between the large central atom and small hydrogen atom.

QUESTION: 31

Which of the following statements regarding sulphur is incorrect?

A.
S₂ molecule is paramagnetic.
B.
The vapour at 200°C consists mostly of S8 rings.
C.
At 600°C the gas mainly consists of S₂ molecules.
D.
The oxidation state of sulphur is never less than +4 in its compounds.

Solution:

Oxidation of sulphur varies from – 2 to + 6 in its various compounds

QUESTION: 32

Boron cannot form which one of the following anions?

A.
BF₆^3–
B.
BH₄^–
C.
B (OH)^–₄
D.
BO₂^–

Solution:

Boron cannot form BF₆^3– due to non-availability of d-orbitals.

QUESTION: 33

The molecule having smallest bond angle is :

A.
NCl₃
B.
AsCl₃
C.
SbCl₃
D.
PCl₃

Solution:

All the members form volatile halides of the type AX₃. All halides are pyramidal in shape. The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.

QUESTION: 34

Among the following oxoacids, the correct decreasing order of acid strength is:

A.
HOCl > HClO₂ > HClO₃> HClO₄
B.
HClO₄ > HOCl > HClO₂> HClO₃
C.
HClO₄ > HClO₃ > HClO₂> HOCl
D.
HClO₂ > HClO₄ > HClO₃> HOCl

Solution:

Acidic strength increases as the oxidation number of central atom increases.
Hence acidic strength order is

(+7) (+5) (+3) (+1)
HClO4 > HClO3 > HClO2 > HClO

QUESTION: 35

Which one of the following properties is not shown by NO?

A.
It is diamagnetic in gaseous state
B.
It is neutral oxide
C.
It combines with oxygen to form nitrogen dioxide
D.
It’s bond order is 2.5

Solution:

Nitric oxide is par amagn etic in th e gaseous state because of the presence of one unpaired electron in its outermost shell.
The electronic configuration of NO is

QUESTION: 36

The correct statement for the molecule, CsI₃ is:

A.
It is a covalent molecule.
B.
It contains Cs⁺ and I₃^- ions.
C.
It contains Cs³⁺ and I^– ions.
D.
It contains Cs⁺, I^– and lattice I₂ molecule.

Solution:

CsI3 dissociates as

CsI₃ → Cs⁺ + I₃^–

QUESTION: 37

Which among the following is the most reactive ?

A.
I₂
B.
ICI
C.
Cl₂
D.
Br₂

Solution:

ICl

Order of reactivity of halogens
Cl₂ > Br₂> I₂

But, the interhalogen compounds are generally more reactive than halogens (except F₂), since the bond between two dissimilar electronegative elements is weaker than the bond between two similar atoms i.e, X – X

QUESTION: 38

Assertion: Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
Reason: The reaction between nitrogen and oxygen requires high temperature.

A.
The assertion is incorrect, but the reason is correct
B.
Both the assertion and reason are incorrect
C.
Both assertion and reason are correct, and the reason is the correct explanation for the assertion
D.
Both assertion and reason are correct, but the

Solution:

Nitrogen and oxgen in air do not react to form oxides of nitrogen in atmosphere because the reaction between nitrogen and oxygen requires high temperature.

QUESTION: 39

Which one has the highest boiling point ?

A.
Kr
B.
Xe
C.
He
D.
Ne

Solution:

Xe. As we move down the group, the melting and boiling points show a regular increase due to corresponding increase in the magnitude of their van der waal forces of attraction as the size of the atom increases.

QUESTION: 40

The pair in which phosphorous atoms have a formal oxidation state of + 3 is :