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JEE Main Test Series- p-Block Elements, Equations and Formulas, Previous Year Questions & Solution
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The p- Block elements is one of the most important and among the topics with the highest weightage in JEE Main Chemistry Syllabus. The chapter carries a weightage of around 6% in the Chemistry section. 1-2 questions from p-Block Elements are mandatorily present in JEE Main Question Paper every year. Check JEE Main Chemistry Syllabus

Questions from p-Block Elements topic in JEE Main can be asked from sub-topics such as general trends in physical and chemical properties of elements and compounds like Boron, Silicon, Nitrogen, Phosphorus, Sulphur, and Halogens.

On a general note, p-Block Elements refer to elements that have a position within group 13 (group IIIA) to group 17 (group VIIA) of the periodic table along with group 18 which is the zero group elements. These are shiny and a good conductor of electricity, they also tend to lose electrons. Check Study Notes on p-Block Elements

JEE Main p-Block Elements Test series - Previous Year Asked Questions and Solutions

Ques. The amorphous form of silica is

(A) Cristobalite

(B) Kieselguhr

(C) Tridymite

(D) Quartz

Solution: Kieselguhr is the amorphous form of silica. Therefore correct answer is (B) Kieselguhr

Ques. From the following statements regarding H2O2, choose the incorrect statement :

(A) It has to be stored in plastic or wax-lined glass bottles in the dark

(B) It has to be kept away from dust

(C) It can act only as an oxidizing agent

(D) It decomposes on exposure to light

Solution- Na2SO3 + H2O2 → Na2SO4 + H2O

Hydrogen peroxide can act as both oxidising and reducing agent.

Therefore, correct answer is (C) It can act only as an oxidizing agent

Ques. Glass is a 

(A)super -cooled liquid 

(B) gel

(C)polymeric mixture

(D)micro-crystalline solid

Solution: Correct Answer is (A) Glass is a transparent or translucent supercooled liquid.

Ques. Which one has the highest boiling point?

(A) Kr

(B) Xe

(C) He

(D) Ne

Solution- The boiling point increases down the group, hence Xe has the highest boiling point in the given elements. Correct answer therefore is (B) Xe

Ques. Which one of the following statements regarding helium is incorrect?

(A) It is used to fill gas balloons instead of H2 Because it is lighter and non-inflammable

(B)It is used in gas-cooled nuclear reactors

(C)It is used to produce and sustain powerful superconducting reagents

(D)It is used as a cryogenic agent for carrying out experiments at low temperatures

Solution: (A) Hydrogen is lightest and is non-inflammable.

Ques. The soldiers of Napoleon army while at the Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transforma­tion is related to

(A)interaction with the nitrogen of the air at very low temperatures

(B)interaction with water vapour contained in the humid air

(C)a change in the partial pressure of oxygen in the air

(D)a change in the crystalline structure of tin

Solution- (D) Transformation of the crystal structure.

Ques. Which of the following are Lewis acids?

(A) AlCl3 and SiCl4

(B) PH3 and SiCl4

(C) BCl3 and AlCl3

(D) PH3 and BCl3

Solution- Lewis acid is an electron pair acceptor. Both BCl3 and AlCl3 have vacant p- orbital and thus incomplete octet. Hence they will act as Lewis acid. Therefore, correct answer is (C) BCl3 and AlCl3

Ques. Boron cannot form which one of the following anions?

(A) B(OH)4–

(B) BO2–

(C) BF63-

(D) BH4–

Solution- Because of the non-availability of d-orbitals, boron is unable to expand its octet. Hence the maximum covalency of boron cannot exceed 4. Therefore correct answer is (C) BF63-

Ques. The decreasing values of bond angles from NH3 (106°) to SbH3 (101°) down group-15 of the periodic table is due to

(A)increasing bp-bp repulsion

(B)increasing p-orbital character in sp3

(C)decreasing lp-bp repulsion

(D)decreasing electronegativity

Solution- Repulsions will be less with lower electronegativity. Therefore the correct answer is (D) decreasing electronegativity

Ques. The gas evolved on heating CaF2 and SiO2 with concentrated H2SO4, on hydrolysis gives a white gelatinous precipitate. The precipitate is:

(A) silica gel

(B) silicic acid

(C) hydrofluosilicic acid

(D) calcium fluorosilicate

Solution-

H2SO4 + CaF2 → CaSO4 + 2HF

4HF + SiO2 → SiF4 +2H2O

3SiF4 +2H2O → 2H2SiF6 + SiO2

Hydrofluosilicic acid is H2SiF6

Therefore, the correct answer is (C) hydrofluosilicic acid

Ques. Example of a three-dimensional silicate is :

(A) Beryls

(B) Zeolites

(C) Feldspars

(D) Ultramarines

Solution- Feldspars are an example of three-dimensional silicate. Therefore the correct answer is (C) Feldspars

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

(A) The mixture only cools down

(B) PCl3 and HCl are formed and the mixture warms up

(C) PCl5 and HCl are formed and the mixture cools down

(D) PH3·Cl2 is formed by warming up.

Solution- 

PH3 + 4Cl2 → PCl5 + 3HCl

Phosphine gas when mixed with chlorine gas, gives phosphorus pentachloride and HCl.

Therefore the correct answer is (C) PCl5 and HCl are formed and the mixture cools down

Ques. Chlorine water on standing loses its colour and forms:-

(A) HCl and HClO2

(B) HCl only

(C) HOCl and HOCl2

(D) HCl and HOCl

Solution- Cl2 + H2O → HCl + HOCl

Therefore the correct answer is (D) HCl and HOCl

Ques. Which one of the following properties is not shown by NO? (A)It’s bond order is 2.5.
(B)It is diamagnetic in a gaseous state
(C)It is a neutral oxide
(D)It combines with oxygen to form nitrogen dioxide

Solution- NO is paramagnetic. Therefore the correct answer is (B) It is diamagnetic in a gaseous state

Ques. Which one has the highest boiling point? 

(a) He 

b) Ne

c) Kr 

d) Xe

Solution- The boiling point increases with an increase in Molecular mass. Therefore the correct answer is (D) Xe

Check  JEE Main Sample Papers

JEE Main p-Block Elements Quick Formulas

Here is a list of formulas for the P-Block Elements Test for JEE mains, which can be followed by the students while preparing for their examination-

Silicones from the hydrolysis of (CH3)3SiCl:

begin mathsize 12px style 2 left parenthesis CH subscript 3 right parenthesis subscript 3 SiCl rightwards arrow with straight H subscript 2 straight O on top 2 left parenthesis CH subscript 3 right parenthesis Si text   end text left parenthesis OH right parenthesis rightwards arrow with blank on top end style

Silicones from the hydrolysis of a mixture of (CH3)3SiCl and (CH2)2SiCl2:

COMPOUNDS OF TIN:

NITROGEN (N) AND ITS COMPOUNDS:

Oxides of Nitrogen

Name Formula Oxidation methods of preparation Common methods of preparation Physical appearance and chemical nature
Dinitrogen oxides[Nitrogen(I) oxide] N2O +1 begin mathsize 12px style NH subscript 4 NO subscript 3 rightwards arrow with Heat on top straight N subscript 2 straight O plus 2 straight H subscript 2 text   end text end subscript straight O end style Colourless gas, neutral
Nitrogen monoxide[Nitrogen(II) oxide](Nitric acid) NO +2 2NaNo2 + 2FeSO4 + 3H2SO4→ Fe2(SO4)2 + 2NaHSO4 + 2H2O + 2NO Colourless gas, neutral
Dinitrogen trioxide [Nitrogen(III)oxide(Nitrogen sesquioxide) N2O2 +3 begin mathsize 12px style 2 NO plus straight N subscript 2 text   end text end subscript straight O subscript 2 text   end text end subscript rightwards arrow with 250 text   end text straight k on top 2 text   end text straight N subscript 2 text   end text end subscript straight O subscript 3 text   end text end subscript end style Blue solid, acidic
Nitrogen dioxide[Nitrogen (IV)oxide] NO2 +4 begin mathsize 12px style 2 Pb left parenthesis NO subscript 3 right parenthesis subscript 2 rightwards arrow with 637 text   end text straight K on top 4 NO subscript 2 plus 2 PbO plus straight O subscript 2 end style brown gas, acidic
Dinitrogen tetroxide [Nitrogen(IV)oxide] N2O4 +4 Colourless solid/ liquid, acidic -
Dinitrogen pentoxide[Nitrogen (IV)oxide] N2O­5 +5 Colourless solid, acidic -

Oxoacids of Phosphorus 

Name Formula The oxidation state of phosphorus Characteristic bonds and their number Preparation
Hypophosphorous H2PO2 +1 One P–OH Two P–H One P=O White P4 + alkali
Orthophosphoric H3PO3 +3 Two P–OH One P–H One P=O P2O3 + H2O
Pyrophosphorus H4P2O5 +3 Two P–OH Two P–H Two P–H PCl3 + H3PO3
Hypophosphoric H4P2O6 +4 Four P–OH Two P=O One P–P Red P4 + alkali
Orthophosphoric H3PO4 +5 Three P–OH One P=O P4O10 + H2O
Pyrophosphoric H4P2 O7 +5 Four P–OH One P=O One P–O–P Heat phosphorus acid
Metaphosphoric (HPO3)3 +5 Three P–OH Three P=O Three P–O–P Phosphorus acid + Br2, heat in a sealed tube

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