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JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - JEE MCQ


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26 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2

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JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 1

Number of paired electrons in O2 molecule is : (1995S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 1

O2 = Oxygen (Z = 8) has following molecular orbital configuration of O2.

 i.e., 2 unpaired and 14 paired electrons.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 2

Among the following species, identify the isostructural pairs.    NF3, NO3- , BF3, H3O+, HN3 (1996 - 1 Mark)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 2

Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus

In NF3 :
Thus  N in NF3 is sp3 hybridized as 4 orbitals are involved in bonding.
In NO3 :
Thus N in NO3 is sp2 hybridized as 3 orbitals are involved in bonding

In BF3 :
Thus B in BF3 is sp2 hybridized and 3 orbitals are involved in bonding

In H3O+ :
Thus O in H3O+ is sp3 hybridized as 4 orbitals are involved in bonding.
Thus, isostructural pairs are [NF3, H3O+] and [ NO3-, BF3]

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JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 3

The number and type of bonds between two carbon atoms in CaC2 are : (1996 - 1 Mark)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 3

Calcium carbide is an ionic compound (Ca2+ C2–) which produces acetylene on reacting with water. Thus the structure of C2– is [C ≡ C]2–. It has one s and two p bonds. [∵ A triple bond consists of one s and two p-bonds]

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 4

Which contains both polar and non-polar bonds?

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 4

(a)     It has ionic and non-polar

(b) H – C ≡ N - It has ionic and polar covalent bonds.

(c) 

It has polar and non polar both type of covalent bonds.

(d)  

It has non polar covalent bonds only.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 5

The critical temperature of water is higher than that of O2 because the H2O molecule has (1997 - 1 Mark)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 5

Critical temperature of water is higher than O2 because H2O molecule has dipole moment which is due to its V-shape.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 6

Which one of the following compounds has sp2 hydridization? (1997 - 1 Mark)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 6

TIPS/Formulae :  H = (V + M – C + A)
(i) CO2; H = (4 + 0 – 0 + 0) = 2
∴ sp hybridisation.
(ii) SO2; H =  (6 + 0 – 0 + 0) = 3
∴ sp2 hybridisation.
(iii) CO; H = (4 + 0 – 0 + 0) = 2
∴ sp hybridisation.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 7

The geometry and the type of hybrid orbital present about the central atom in BF3 is (1998 - 2 Marks)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 7

H =(3 + 3 + 0 - 0) = 3 
∴ Boron, in BF3 , is sp2 hybridised leading to trigonal planar shape.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 8

The correct order of increasing C — O bond length of CO, CO2-3, CO2, is             (1999 - 2 Marks)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 8

KEY CONCEPT

(i) Bond length 

(ii) Bond order is calcula ted by eith er the h elp of molecular orbital theory or by resonance.

(i) Bond order of CO as calculated by molecular orbital

theory  = 

(ii) Bond order of CO2 (by resonance method)

(iii) Bond order in CO32– (by resonance method)

∴ Order of bond length of C – O is CO < CO2 < CO32–

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 9

The geometry of H2S and its dipole moment are (1999 - 2 Marks)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 9

Hybridisation of S in H2S =(6 + 2 + 0 – 0) = 4
∴ S has sp3 hybridisation and 2 lone pair of electrons in H2S
∴ It has angular geometry and so it has non-zero value of dipole moment.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 10

Molecular shapes of SF4, CF4 and XeF4 are (2000S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 10

The structure of species can be predicted on the basis of hybridisation which in turn can be known by knowing the number of hybrid orbitals (H) in that species


For SF4 : S is sp3d hybridised in SF4.
Thus SF4 has 5 hybrid orbitals of which only four are used by F, leaving one lone pair of electrons on sulphur.
For CF4 :
  ∴ sp3 hybridisaion
Since all the four orbitals of carbon are involved in bond formation, no lone pair is present on C having four valence electrons
For XeF4 :
 ∴ sp3d2 hybridization  of the six hybrid orbitals, four form bond with F, leaving behind two lone pairs of electrons on Xe.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 11

The hybridisation of atomic orbitals of nitrogen in NO+2, NO3- and NH+4 are (2000S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 11

For  

∴ sp hybridisation

For  

∴ sp2 hybridisation

For 

∴ sp3 hybridisation

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 12

The common features among the species CN, CO and NO+ are (2001S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 12

Number of electrons in each species are CN- = 6 + 7 +1= 14 , CO = 6+ 8= 14
NO+ = 7 +8 - 1=14
Each of the species has 14 electrons which are distributed in MOs as below

Bond order = 

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 13

The correct order of hybridization of the central atom in the following species NH3, [PtCl4]2 , PCl5 and BCl3 is (2001S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 13

TIPS/Formulae : H = [V + M – C + A]
Hybridisation of N in NH3
 [5 + 3 – 0 + 0] = 4     ∴ sp3
Hybridisation of Pt in [PtCl4]2
 [2 + 4 – 0 + 2] = 4      ∴ dsp2
Hybridisation of P in PCl5
 [5 + 5 – 0 + 0] = 5      ∴ sp3d
Hybridisation of B in BCl3
 [3 + 3 – 0 + 0] = 3      ∴ sp2

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 14

Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BF3 and NH3

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 14

H3N → BF3 where both N, B are attaining tetrahedral geomerty.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 15

Identify the least stable ion amongst the following : (2002S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 15

NOTE THIS STEP : Write configuration of all species.
Half filled and full filled orbitals are more stable as compared to nearly half filled and nearly full filled orbitals.

Li = 1s2, 2s2 ; Be = 1s2, 2s2, 2p1
B = 1s2, 2s2, 2p2  ;  C– = 1s2, 2s2, 2p3
∴ Bewill be least stable. It has lowest I.E.  

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 16

Which of the following molecular species has unpaired electron(s) ? (2002S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 16

N2(7+7=14); σ1s2, σ1*s2, σ2s2, σ*2s2

F2(9 + 9 = 18); σ1s2, σ*1s2, σ2s2, σ*2s2, σ2px2

O2 (8 + 8 +1 = 17); σ1s2, σ*1s2, σ2s2, σ*2s2, σ2px2,

O22- (8 + 8 + 2 = 18); σ1s2, σ*1s2, σ2s2, σ*2s2, σ2px2,

∴ O2 is the only species having unpaired electron.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 17

Which of the following are isoelectronic and isostructural?NO3, CO32–, ClO3, SO3 (2003S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 17

NOTE : Isoelectronic species have same number of electrons and isostructural species have same type of hybridisation at central atom.
NO3 ; No. of e = 7 + 8 × 3 + 1 = 32, hybridisation of N in NO3 is sp3
CO32– ; No. of e = 6 + 8 × 3 + 2 = 32, hybridisation of C in
CO32– is sp3 ClO3 ; No. of e= 17 + 8 × 3 + 1 = 42, hybridisation of Cl in
ClO3 is sp3 SO3; No. of e= 16 + 8 × 3 = 40, hybridisation of S in
SO3 is sp2
∴ NO3– and CO32– are isostructural and isoelectronic.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 18

According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding O+2 (2004S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 18

O2 : σ1s2 , σ*1s2 , σ2s2, σ*2s2,σ2px2 ,

Bond order = 

(two unpaired electrons in antibonding molecular orbital)

Bond order = 

(One unpaired electron in antibonding molecular orbital) Hence O2 as well as O+2 is paramagnetic, and bond order of O+2 is greater than that of O2.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 19

Which species has the maximum number of lone pair of electrons on the central atom? (2005S)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 19

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 20

Among the following, the paramagnetic compound is (2007)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 20

(i) In Na2O2, we have O22- ion. Number of valence elctrons of the two oxygen in O22- ion = 8 × 2 + 2 =18 which are present as follows

∴ Number of unpaired electrons = 0, hence, O22-is diamagnetic.

(ii) No. of valence electrons of all atoms in O3 = 6 × 3 = 18.
Thus, it also, does not have any unpaired electron, hence it is diamagnetic. (iii)

No. of valence electrons of all atom in N2O = 2 × 5 + 6  = 16.
Hence, here also all electrons are paired. So it is diamagnetic.

(iv) In KO2, we have O2- No. of valence electrons of all atoms in O2- = 2 × 6 + 1 = 13,
Thus it has one unpaired electron, hence it is paramagnetic.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 21

The species having bond order different from that in CO is (2007)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 21

Molecular electronic configuration of

Therefore, bond order = 

Bond order  = 

Bond order = 

 

Bond order  = 

Bond order  = 

∴ NO has different bond order from that in CO.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 22

Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is (2010)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 22

Molecular orbital configuration of B2(10) as per the condition will be

Bond order of B2 =,B2 will be diamagnetic.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 23

The species having pyramidal shape is : (2010)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 23

OSF2 : .  It has 1 lone pair.

(Shape is trigonal pyramidal)

The shapes of SO3, BrF3 and SiO32- are triangular planar respectively.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 24

Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl , CN and H2O, respectively, are (2011)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 24

[NiCl4]2–. = 3d8 configuration with nickel in +2 oxidation state, Cl being weak field ligand does not compel for pairing of electrons.
So,
 [NiCl4]2–

Hence, complex has tetrahedral geometry

[Ni(CN)4]2– = 3d8 configuration with nickel in + 2 oxidation state, CN being strong field ligand compels for pairing of electrons.
So,

Hence, complex has square planar geometry.

[Ni(H2O)6] = 3d8 configuration with nickel in +2 oxidation state. As with 3d8 configuration two dorbitals are not available for d2sp3 hybridisation. So, hybridisation of Ni (II) is sp3d2 and Ni (II) with six co-ordination will have octahedral geometry.

Note : With water as ligand, Ni (II) forms octahedral complexes.

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 25

Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is (JEE Adv. 2014)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 25

Be2 = σ1s2 σ*1s2 σ2s2 σ*2s2
B2 = σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2
C2 = σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px1 π2py1
N2 = σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 π2py2
Thus only C2 will be paramagnetic

JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 26

The geometries of the ammonia complexes of  Ni2+, Pt2+ and Zn2+ respectively, are (JEE Adv. 2016)

Detailed Solution for JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - Question 26

Ni2+ with NH3 shows CN = 6 forming [Ni(NH3)6]2+ (Octahedral)

Pt2+ with NH3 shows CN = 4 forming [Pt(NH3)4]2+ (5d series CMA, square planner)

Zn2+ with NH3 shows CN = 4 forming [Zn(NH3)4]2+ (3d10 configuration, tetrahedral)

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