Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements)


29 Questions MCQ Test Chemistry for JEE Advanced | Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements)


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This mock test of Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements) for JEE helps you for every JEE entrance exam. This contains 29 Multiple Choice Questions for JEE Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements) (mcq) to study with solutions a complete question bank. The solved questions answers in this Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements) exercise for a better result in the exam. You can find other Retro (Past 13 Year) IIT-JEE Advanced (P Block Elements) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Under am bient conditions, the total number of gases released as products in the final step of the reaction scheme, shown below, is

Solution:

Plan This problem can be solved by using concept involved in chemical properties of xenon oxide and xenon fluoride.
XeF6 on complete hydrolysis produces XeO3.
XeO3 on reaction with OH- produces   4 which on further treatment with OH- undergo slow disproportionation reaction and produce  along with Xe(g), H2O (/) and O2 (g) as a by-product. 
Oxidation half-cell in basic aqueous solution

Reduction half-cell in basic aqueous solution

Balanced overall disproportionation reaction is
Complete sequence of reaction can be shown as

QUESTION: 2

The product formed in the reaction of SOCI2 with white phosphorus is

(2014 Adv., Only One Option Correct Type)

Solution:

This problem is based on chemical properties of phosphorus.
White phosphorus on reaction with thionyl chloride (SOCI2) produces phosphorus trichloride.


But if amount of thionyl chloride (SOCI2) is in excess then it produces phosphorus pentachloride.

*Multiple options can be correct
QUESTION: 3

The correct statement(s) about O3 is(are)

(2013 Adv., One or More than One Options Correct Type)

Solution:

Plan Due to resonance, bond lengths between two atoms are equal. Species is said to be diamagnetic if all electrons are paired.

O3 is bent molecule. All electrons paired thus, it is diamagnetic.

Thermal decomposition of O3 is exothermic.

QUESTION: 4

The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas in the presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus T.

Q. 

P and Q respectively, are the sodium salts of

(2013 Adv., Comprehension Type)

Solution:




QUESTION: 5

The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas in the presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus T.

Q. 

R, S and T respectively, are

Solution:




QUESTION: 6

The unbalanced chemical reactions given in Column I show missing reagent or condition (?) which are provided in Column II. Match Column I with Column II and select the correct answer using the codes given below.

Solution:




QUESTION: 7

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.

Q. 

25 mL of household bleach solution was mixed with 30 mL of 0.50 M Kland 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is

(2012, Comprehension Type)

Solution:

 The involved redox reactions are



Molarity of Na2S2O3 = 0.25 N x 1 = 0.25 M
mmol of Na2S2O3 used up = 0.25 x 48 = 12
Now, from stoichiometry of reaction (ii) 12 mmol of   would have reduced 6 mmol of I2.
From stoichiometry of reaction (i)
mmol of OCI- reduced = mmol in I2 produced = 6
Molarity of household bleach solution = 
Alternate Method

mmol of   mmol. Remaining part is solved in the same manner.

QUESTION: 8

Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.

Q. 

Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is

Solution:

Bleaching powder is Ca(OCI) Cl. Therefore, the oxoacid whose salt is present in bleaching powder is HOCI. Anhydride of HOCI is Cl2O as

Note The oxidation number of element in anhydride and oxoacid remains the same.

QUESTION: 9

Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

(2012, Only One Option Correct Type)

Solution:

Let oxidation number of N be x.

QUESTION: 10

The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product respectively are 

Solution:

The reaction of white phosphorus with aqueous alkali is

In the above reaction, phosphorus is simultaneously oxidised 

Therefore, this is an example of disproportionation reaction. Oxidation number of phosphorus in PH3 is - 3 and in NaH2PO2 is +1. However, +1 oxidation number is not given in any option, one might think that NaH2PO2 has gone to further decom positio n on heating.

QUESTION: 11

The shape of XeO2F2 molecules is 

Solution:

In XeO2F2, the bonding arrangement around the central atom Xe is

4σ-bonds+ 1./p= 5
Hybridisation of Xe = sp3d
sp3d-hybridisation corresponds to trigonal bipyramidal geometry. Also, in trigonal bipyramidal geometry, lone pairs remain present on equatorial positions in order to give less electronic repulsion.

Note According to Bent's rule, the more electronegative atoms must be present on axial position. Hence, F are kept on axial positions.

QUESTION: 12

Extra pure N2 can be obtained by heating

Solution:


Azide salt of barium can be obtained in purest form as well as the decomposition product contain solid Ba as by product alongwith gaseous nitrogen, hence no additional step of separation is required. Other reactions are

*Answer can only contain numeric values
QUESTION: 13

Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO2 gas. The number of sodium bromide molecule involved in the balanced chemical equation is 

(2011, Interger Type)


Solution:

Br2 is disproportionated in basic medium as

*Answer can only contain numeric values
QUESTION: 14

Among the following, the number of compounds that can react with PCI5 to give POCl3 is O2, CO2, SO2, H2O, H2SO4, P4O10


Solution:

PCI5 produces POCI3 with the following reagents

QUESTION: 15

All the compounds listed in Column I react with water. Match the result of the respective reactions with the appronate options listed in Column II.

Solution:




QUESTION: 16

The reaction of P4 with X leads selective to P4O6. The X is

Solution:

In limited supply of oxygen, phosphorus is oxidised to its lower oxide P4O6 while excess of oxygen gives P4O10. A mixture of O2 and N2 is used for controlled oxidation of phosphorus intoP4O6.

*Multiple options can be correct
QUESTION: 17

The nitrogen oxide(s) that contain(s) N—N bond(s) is(are)

Solution:

The structures of these oxides are


Hence, options (a), (b) and (c) have N—N bonds.

QUESTION: 18

Aqueous solution of Na2S2O3 on reaction with Cl2 gives

(2008, Only One Option Correct Type)

Solution:

Sodium thiosulphate, Na2S2O3 gets oxidised by chlorine water as

FeCI3 oxidises Na2S2O3 to Na2S4O6

QUESTION: 19

Passage for Q. Nos. (19-21)

There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorus.

Q. 

Among the following, the correct statement is

 (2008, Comprehension Type)

Solution:

Due to greater solubility in water and prone to microbial attack, nitrates are less abundant in earth's crust.

QUESTION: 20

There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorus.

Q. 

 Among the following, the correct statement is

Solution:

NH3 is stronger Lewis base than PH3. In a group of hydrides, basic strength decreases down the group.

QUESTION: 21

There are some deposits of nitrates and phosphates in earth’s crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorus.

Q. 

White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a

Solution:

White phosphorus undergo disproportionation in alkaline medium.

QUESTION: 22

The percentage of p-character in the orbitals form ing P—P bonds in P4 is

(2007, Only One Option Correct Type)

Solution:

In P4, all phosphorus are sp3-hybridised and has 75% p-character.

QUESTION: 23

Passage for Q. Nos. (23-25)

The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers + 2, + 4 and + 6. XeF4 reacts violently with water to give XeO3. The compounds of xenon exhibit rich stereochemisty and their geometries can be deduced considering the total- number of electron pairs in the valence shell.

Q. 

Argon is used in arc welding because of its

Solution:

Ar, being inert, provide inert atmosphere in arc welding and prevent from undesired oxidation.

QUESTION: 24

The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers + 2, + 4 and + 6. XeF4 reacts violently with water to give XeO3. The compounds of xenon exhibit rich stereochemisty and their geometries can be deduced considering the total- number of electron pairs in the valence shell.

Q. 

The structure of XeO3 is

Solution:


Xe is sp3-hybridised with one lone pair. Hence, molecule of XeO3 has pyramidal shape

QUESTION: 25

The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers + 2, + 4 and + 6. XeF4 reacts violently with water to give XeO3. The compounds of xenon exhibit rich stereochemisty and their geometries can be deduced considering the total- number of electron pairs in the valence shell.

Q. 

XeF4 and XeF6 are expected to be

Solution:

Both XeF4 and XeF6 are strong oxidising agent.

QUESTION: 26

Which of the following will not be oxidised by O3

Solution:
QUESTION: 27

Which gas is evolved when PbO2 is treated with conc. HNO3

(2005, Only One Option Correct Type)

Solution:

QUESTION: 28

A pale blue liquid which obtained by equimolar mixture of two gases at - 30°C is

(2005, Only One Option Correct Type)

Solution:

Equimolar amounts of NO and NO2 at -30°C gives N2O3(/) which is a blue liquid.

QUESTION: 29

Which of the following isomers of phosphorus is thermodynamically most stable?

Solution:

Black phosphorus is thermodynamically most stable allotrope of phosphorus. It is due to three dimensional network structure of polymeric black phosphorus.