Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding)


30 Questions MCQ Test Chemistry for JEE Advanced | Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding)


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This mock test of Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding) for JEE helps you for every JEE entrance exam. This contains 30 Multiple Choice Questions for JEE Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding) (mcq) to study with solutions a complete question bank. The solved questions answers in this Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding) exercise for a better result in the exam. You can find other Revisal Problems (Past 13 Years) JEE Advanced (Chemical Bonding) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Only One Option Correct Type

Direction (Q, Nos. 1-5) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c), and (d), out of which ONLY ONE option is correct.

Q. 

Select correct matching about [Ni(CN)4]2- , Ni(CO)4, [NI(H2O)4)]2+

Solution:

CN- and CO are strong ligands. Thus, unpaired electrons in Ni or Ni2+ are paired. H2O is a week ligand, thus unpaired electrons are not affected.
Ni [Ar]



Note There is no unpaired electron in [Ni(CO)4]
— diamagnetic
Also, [Ni (CN)4]2- - diamagnetic but [Ni(H2O)4] has unpaired electrons. Thus, it is paramagnetic.

QUESTION: 2

Energy gap between the highest filled bands and the lowest empty bands in elements are given :

Thus,

Solution:

Based on Band-theory of metallic bonding, as energy increases, conducting nature decreases and element changes from conductor to insulator.

QUESTION: 3

Match Column I with Column II and select the correct answer from the codes given below the table.

Solution:


QUESTION: 4

Hybridisation of the underlined atoms is/are affected in the following transformations :

Solution:



QUESTION: 5

In BrF3 molecule, the lone pairs occupy equatorial positions to minimise

Solution:

(p - Ip) and (Ip - bp) repulsions are m inim ised by taking lone-pairs on equatorial position.

QUESTION: 6

Statement Type

Direction (Q. Nos. 6 - 8) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Q. 

Statement I : Two different bond lengths are observed in PF5 molecule but only one bond length is observed in SF6 molecule.

Statement II : One lone pair exists in PF5

Solution:


In PF5, there are two types of angles thus has two types of bond lengths.
In SF6, there is only one type of bond angle thus, only one type of bond.
There is no lone pair on P in PF5.
Thus, statement I is correct and statement II is incorrect.

QUESTION: 7

Statement I : The ion CIF2- is linear but the ion CIF2+ is bent.

Statement II : Cl-atom is sp2-hybridised in each.

Solution:


Lone pairs on Cl are at the equatorial positions to minimise repulsion.

Two F-atoms are at the vertices thus linear structure.
Two lone pairs on Cl—sp3 hybridisation thus bent.
Thus, statement I is correct but statement II is incorrect.

QUESTION: 8

Statement I : Carbon and silicon are in group 14 in but ( Si=Si) bond is not formed but (C=C) bond is formed.
Statement II :  Larger size of Si results in poor sideway overlap of p-orbitals to form π-bond.

Solution:

C,Si-—group 14. Thus, Statement I is correct larger size of Si results in poor sideway overlap of s-orbitals to form π-bond.
Thus, both statements I and II are correct and statement II is the correct explanation of statement I.

*Multiple options can be correct
QUESTION: 9

One or More than One Options Correct Type

Direction (Q. Nos. 9-21) This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c), and (d), out of which ONE or MORE THAN ONE are correct.

Q. 

Select correct statement(s)

Solution:



Maleic acid (cis-isomer) is more polar than its frans-isomer. Hence, maleic acid is more ionised than fumaric acid.
(b) Maleate ion (I) is stable due to intramolecular H-bonding. Hence, of maleic acid is smaller than that of fumaric acid in which intramolecular H-bonding is absent (II).

= First ionisation constant
= Second ionisation constant 


Nickel dimethyl glyoximate complex is stable due to intramolecular H-bonding.

*Multiple options can be correct
QUESTION: 10

Select the correct variation of solubility of different salts in water.

Solution:

By the coulombic force of attraction (F) between cation and anion is

z1 = z2 = (+ 1 or -1) charges on cation or anion.
r = internuclear distance
D = Dielectric constant (same for H2O solvent).
If F is small, ions goes into solution and thus makes salt soluble in water.

Larger (r) means smaller (F).
(a) r increases hence solubility increases true.
(b) as in (a) true.
(c) If one of the ions is very big (as Cs), the solubility decreases with increasing size of the other ion.
F < C I < Br < I,
size
Thus, Csl < CsBr < ,CsCI < CsF
Thus, true, solubility
(d) NO3- ion is larger.
size of Li+ < Na+ < K+ < Cs+
Hence, solubility decreases with increase in size of M+
CsNO3 < KNO3 < NaNO3 < LiNO3- true.

*Multiple options can be correct
QUESTION: 11

Which of the following molecules is expected to exhibit diamagnetic behaviour?

Solution:

Based on MO theory
(a) C2 (12 electrons) - all paired electrons
(b) N2 (14 electrons) - all paired electrons
(c) O2 (16 electrons) - two electrons unpaired
(d) S2 (32 electrons) - two electrons unpaired
Thus, C2 or N2 are diamagnetic.

*Multiple options can be correct
QUESTION: 12

Which is/are correct variation (s) for

Solution:




N = Number of unpaired electrons


Bond energy as bond order magnetic moment  

*Multiple options can be correct
QUESTION: 13

Which is/are not correct variation

Solution:




*Multiple options can be correct
QUESTION: 14

The hybridisation scheme for the central atom includes ad-orbital contribution in

Solution:

(a) l3- sp3d
(b) PCl3 sp3
(c) XeF2 sp3d
(d) SF4 sp3d

*Multiple options can be correct
QUESTION: 15

Select the correct statement(s).

Solution:

(a) Smaller the energy gap, larger the conduction.
(b) Unpaired electrons in graphite is responsible for conduction used in bonding in hexagonal layer.


Bonding electrons = 4
Antibonding electrons = 2

Thus, Li2 is stable.
(d) Diamond is non-conducting graphite is conducting.

*Multiple options can be correct
QUESTION: 16

One or more species out of the following uses sp2-hybrid orbitals in bonding is/are

Solution:

due to one lone pair


It has three (C—H) α-bonds and singly occupied p-orbital. Thus, sp2.

*Multiple options can be correct
QUESTION: 17

Consider the following species :






Sets of species with same type of hybridisation on underlined atom is/are

Solution:


sp2 (one lone pair is a part of hybrid orbital).


(one unpaired electron is a part of hybrid orbtial).


 (one lone pair is a part of hybrid orbital)

*Multiple options can be correct
QUESTION: 18

In which of the following cases, driving force is to complete the octet of B-atom permanently

Solution:

(a) 
(B—H) bond has only one electron thus B is electron deficient.
(b) Octet of B is complete.
(c) Octet of B is complete.

Octet of B is incomplete.

*Multiple options can be correct
QUESTION: 19

Which of the follow ing m olecules has the m axim um num ber of A—X bonds of identical bond length when A is the central atom and X is the surrounding atom?

Solution:


In IF7, there are five equatorial bonds of equal bond length. 

*Multiple options can be correct
QUESTION: 20

Select the correct statement(s) about PCI5,

Solution:

P-atoms are sp3d -hybridised.

A, B and C are inclined at 120° with one each other form equatorial bonds. D and E at 90° with (ABC) plane. D is above and E is below this plane forming axial bonds. Axial bonds suffer more repulsive interaction from equatorial bonds, hence axial bonds are longer and sligtly weaker than the equatorial bonds.

*Multiple options can be correct
QUESTION: 21

Which of the following structures assume linear structure?

Solution:



QUESTION: 22

Comprehension Type

Direction (Q. Nos. 22-25) This section contains 2 paragraphs, each describing theory, experiments, data, etc. four questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage l

Homolytic and heterolytic bond cleavage of organic compounds may result in short lived intermediates.

Q.

Hybridisation of C-atom in the species formed is of the type

Solution:

sp2-hybridised carbon bond angle =120°

sp3-hybridised carbon bond angle < 109°28'

Unpaired electron is in unhybridised p2-orbital.

Thus sp2-hybridised bond angle = 120°.

Thus (b), (c)

QUESTION: 23

Passage l

Homolytic and heterolytic bond cleavage of organic compounds may result in short lived intermediates.

Q. 

Bond angle  in the increasing order is 

Solution:

sp2-hybridised carbon bond angle =120°

sp3-hybridised carbon bond angle < 109°28'

Unpaired electron is in unhybridised p2-orbital.

Thus sp2-hybridised bond angle = 120°.

​Thus (b), (c)

QUESTION: 24

Passage II

Consider the following compounds of nitrogen and answer the given questions : 

Q. 

Hybridisation of nitrogen atom can be

Solution:

Each N-atom has one lone pair and three α-bonds —sp3 hybridised.

(Ip - bp) repulsion makes it. Trigonal pyramidal.
Y : N[Si(CH3)3]3 Lone pair on N-atom is transferred to the empty d-orbitals of silicon (pπ - dπ overlapping) and its Lewis base character is lost.
Thus, y has trigonal planar structure.

as in (X) trigonal pyramidal.

QUESTION: 25

Passage II

Consider the following compounds of nitrogen and answer the given questions : 

Q. 

Structures can be

I. Trigonal planar
II. Trigonal pyramidal
III. Square planar

Select the correct structures

Solution:

Each N-atom has one lone pair and three α-bonds —sp3 hybridised.

(Ip - bp) repulsion makes it. Trigonal pyramidal.
Y : N[Si(CH3)3]3 Lone pair on N-atom is transferred to the empty d-orbitals of silicon (pπ - dπ overlapping) and its Lewis base character is lost.
Thus, y has trigonal planar structure.

as in (X) trigonal pyramidal.

QUESTION: 26

Matching List Type

Direction (Q. Nos. 26 and 27) Choices for the correct combination o f elements from Column I and Column II are given as options.

Q. 

Match the reactions in Column I with the nature of reactions/type of the products in Column II.

[IITJEE2007]

Solution:


O2- undergoes disproportionation.

Cr-metal ion is bridged and has tetrahedral structure, (ii) → (r)

MnO4- is reduced.
NO3- is oxidised thus a redox reaction.
NO3- has triangular planar structure.

Fe2+ is oxidised and NO3- is reduced thus redox reaction.
Thus, (iv) → 4 (p)

QUESTION: 27

Match each of the diatomic molecules in Column I with the property/properties in Column II.

[IITJEE 2009]

Solution:

(i) B2 (10 electrons)

Two unpaired electrons-paramagnetic.
Bonding electrons = 6
Anit-bonding electrons = 4


(B— B) bond is formed by mixing of s and p-orbitals of valence shell.

Thus, (i) → 4 (p, q, r, t)
(ii) N2 (14 electrons)

No unpaired electron - diamagnetic
Bonding electrons = 10
Anti-bonding electrons = 4



One unpaired electrons thus paramagnetic Bonding electrons = 10
Anti-bonding electrons = 7



(iv) O2 (16 electrons)
There are two electrons in anti-bonding
Thus, it is paramagnetic.
Bond order = 2

*Answer can only contain numeric values
QUESTION: 28

One Integer Value Correct Type

Direction (Q. Nos. 28-30) This section contains 3 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Q. 

At 300 K and1.00 atm, density of gaseous HF is 3.17 gL-1. How many HF molecules are associated by H-bonding?


Solution:




m (HF) of single unit = 20 gL-1
Thus, number of HF molecules = 78/20 = 4

*Answer can only contain numeric values
QUESTION: 29

AX4 forms square planar type structure. Distance between A and X-atoms is 4 units. What is distance between two X-atoms in trans-position?


Solution:

 


Distance between A and X = 4 units
Thus, distance between two X -atoms = 4 x 2 = 8 units.

*Answer can only contain numeric values
QUESTION: 30

A total of n x1020 energy levels are present in 3s conduction band of single crystal of sodium weighing 26.8 mg. What is the value of n?


Solution:

Na (11):1s22s22p63s1
Electron in (3s) orbital is a conduction band , Each orbital has one electron, thus one energy level.


= 7.0 x 1020 atoms
= 7.0 x 1020 conduction bands. 
Thus,

n = 7