NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

JEE Revision Notes

NEET : NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

The document NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev is a part of the NEET Course JEE Revision Notes.
All you need of NEET at this link: NEET

Multiple Choice Questions - I

Q.1. Isostructural species are those which have the same shape and hybridization. Among the given species identify the isostructural pairs.
(a) [NF3 and BF3]
(b) [BF4 and NH+4 ]
(c) [BCl3 and BrCl3]
(d) [NH3 and NO3 ]

Ans. (b)
Solution.

  • NF3 is pyramidal whereas BF3 is planar triangular.NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
  • BF4 and NH+4 ions both are tetrahedral.NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
  • BCl3 is triangular planar whereas BrCl3 is pyramidal.NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
  • NH3 is pyramidal whereas NO3 is triangular planar.NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev


Q.2. Polarity in a molecule and hence the dipole moment depends primarily on the electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?
(a) CO2
(b) HI
(c) H2O
(d) SO2

Ans. (c)
Solution. H2O will have the highest dipole moment due to the maximum difference in electronegativity of H and O atoms.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Q.3. The types of hybrid orbitals of nitrogen in NO+2, NO3, and NH+4 respectively are expected to be
(a) sp, sp3, and sp2
(b) sp, sp2, and sp3
(c) sp2, sp, and sp3
(d) sp2, sp3, and sp

Ans. (b)
Solution. The number of orbitals involved in hybridization can be determined by the application of formula:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
where H = number of orbitals involved in hybridization
► V= valence electrons of central atom
► M= number of monovalent atoms linked with central atom
► C = charge on the cation
► A = charge on the anion

► NO+2 : H = 1/2 [5 + 0  - 1 + 0] = 2 or sp
► NO+3 : H = 1/2 [5 + 0  - 0 + 0] = 3 or sp2
► NO+4 : H = 1/2 [5 + 4  - 1 + 0] = 4 or sp3

Q.4. Hydrogen bonds are formed in many compounds, e.g., H2O, HF, NH3. The boiling point of such compounds depends to a large extent on the strength of hydrogen bonds and the number of hydrogen bonds. The correct decreasing order of the boiling points of the above compounds is:
(a) HF > H2O > NH3

(b) H2O > HF > NH3
(c) NH3 > HF > H2O
(d) NH3 > H2O > HF
Ans. (b)
Solution: The strength of H-bonding depends on the electronegativity of the atom, which follows the order: F > O > N.
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
► But each H2O molecule is linked to 4 other H2O molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, the correct decreasing order of the boiling points is H2O > HF > NH3.

Q.5. In PO3–4 ion, the formal charge on the oxygen atom of  P – O bond is
(a) + 1
(b) – 1
(c) – 0.75
(d) + 0.75
Ans. (b)
Solution: The formal charge of the atom in the molecule or ion = (Number of valence electrons in a free atom) – (Number of lone pair electrons + 1/2 Number of bonding electrons)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► For charge on oxygen = 6 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev= 6 -7 = - 1

Q.6. In NO3 ion, the number of bond pairs and lone pairs of electrons on the nitrogen atom is
(a) 2, 2
(b) 3, 1
(c) 1, 3
(d) 4, 0
Ans. (d)
Solution: In N-atom, number of valence electrons = 5

  • Due to the presence of one negative charge, number of valence electrons = 5 + 1 = 6. 
  • One O-atom forms two bonds (= bond) and two O-atom are shared with two electrons of N-atom.
  • Thus, 3 O-atoms are shared with 8 electrons of N-atom.
  • Number of bond pairs (or shared pairs) = 4
  • Number of lone pairs = 0
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.7. Which of the following species has tetrahedral geometry?
(a) BH4
(b) NH2
(c) CO23
(d) H3O+
Ans. (a)
Solution: 

  • Boron is surrounded by 4 bonded pairs only.
  • In BH-4 = no. of bond pair = 4
  • No. of lone pair = 0
  • sphybridised, so have tetrahedral geometry

► NH-2 = V - shape
► H3O+ = Pyramidal
► CO-23= triangular planar

Q.8. The number of π bonds and σ bonds in the following structure is–
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(a) 6, 19
(b) 4, 20
(c) 5, 19
(d) 5, 20
Ans. (c)
Solution: Each C atom is sp2 hybridized and surrounded by 3 sigmas and 1 pi bond between two C atoms. Structure to be rectified.
► 8 C - H + 11 C - C bonds = 19 σ bonds.
► There are 5π -bonds and 19 σ bonds.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.9. Which molecule/ion out of the following does not contain unpaired electrons?
(a) N+2
(b) O2
(c) O2–2
(d) B
2
Ans. (c)
Solution: 
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Thus, O2-has no unpaired electrons.

Q.10. In which of the following molecule/ion, all the bonds are not equal?
(a) XeF4
(b) BF4
(c) C2H4
(d) SiF
4
Ans. (c)
Solution. C2H4 has one double bond and four single bonds. The bond length of a double bond (C = C) is smaller than a single bond (C – H).
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.11. In which of the following substances will hydrogen bond be strongest?
(a) HCl
(b) H2O
(c) HI
(d) H2S

Ans. (b)
Solution. HCl, HI, and H2S do not form H-bonds. Only H2O forms hydrogen bonds. One H2O molecule forms four H-bonding.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Q.12. If the electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2, the four electrons involved in chemical bond formation will be_____.
(a) 3p6
(b) 3p6, 4s2
(c) 3p6, 3d2
(d) 3d2, 4s
2
Ans. (d)
Solution. In transition elements (n -1) d and ns orbitals take part in bond formation.NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev


Q.13. Which of the following angle corresponds to sp2 hybridization?
(a) 90°
(b) 120°
(c) 180°
(d) 109°
Ans. (b)
Solution. sp2 hybridization gives three sp2 hybrid orbitals which are planar triangular forming an angle of 120° with each other.

Directions: The electronic configurations of three elements A, B, and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.

NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.14. The stable form of A may be represented by the formula :
(a) A
(b) A2
(c) A3
(d) A4

Ans. (a)
Solution. The given electronic configuration shows that A represents noble gas because the octet is complete. A is neon which has 10 atomic numbers.

Q.15. The stable form of C may be represented by the formula :
(a) C
(b) C2
(c) C3
(d) C4
Ans. (b)
Solution. The electronic configuration of C represents chlorine. Its stable form is dichlorine (Cl2), i.e., C2.

Q.16. The molecular formula of the compound formed from B and C will be
(a) BC
(b) B2C
(c) BC2
(d) BC3
Ans. (d)
Solution. B represents P, and C represents Cl. The stable compound is PCl3, i.e., BC3.

Q.17. The bond between B and C will be
(a) Ionic
(b) Covalent
(c) Hydrogen
(d) Coordinate
Ans. (b)
Solution. Both B and C are non-metals, and therefore, the bond formed between them will be covalent.

Q.18. Which of the following order of energies of molecular orbitals of N2 is correct?
(a) (π2py) < (σ2pz) < (π*2px) ≈ (π*2py)
(b) (π2py) > (σ2pz) > (π*2px) ≈ (π*2py)
(c) (π2py) < (σ2pz) > (π*2px) ≈ (π*2py)
(d) (π2py) > (σ2pz) < (π*2px) ≈ (π*2py)
Ans. (a)
Solution. Molecules like B2, C2 & N2 having 1 to 3 electrons in p orbital energy of σ2p molecular orbital is greater than that of π2px and π2py molecular orbitals.

Q.19. Which of the following statement is not correct from the viewpoint of molecular orbital theory?
(a) Be2 is not a stable molecule.
(b) He2 is not stable, but He+2  is expected to exist.
(c) Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(d) The order of energies of molecular orbitals in N2 molecule is
σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz
Ans. (d)
Solution.

  • (a) Be2 (4 + 4 = 8) = σ1s2, σ*1s2 ,σ1s2 ,σ*2s2
    Bond order (BO) = 1/2 [Number of bonding electron (Nb) - Number of anti - bonding electrons Na]
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
  • (b) Here, the bond order of He2 is zero. Thus, it does not exist.
    He2 (2 + 2 = 4) = σ1s2 , σ*1s2
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    Here, the bond order of He2 is zero. Thus, it does not exist.
    He+2 (2 + 2 - 1 = 3) = σ1s2, σ*1s1
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    Since the bond order is not zero, this molecule is expected to exist.
  • (c) N2 (7 + 7 = 14) = σ1s2, σ*1s2 ,σ2s2 ,σ*2s, (π2p2x ≈ π* 2p2y) < σ*2p2z
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    Thus, the dinitrogen (N2) molecule contains a triple bond, and no other molecule of the second period have more than a double bond. Hence, the bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.
  • (d) It is incorrect. The correct order of energies of molecular orbitals in N2 molecule is
    σ2s < σ* 2s < σ2pz < (π2px ≈ π2py) < (π* 2px ≈ π* 2py) < σ*2pz


Q.20. Which of the following options represents the correct bond order : 
(a) O2 > O2 > O+2
(b) O2 < O2 < O+2
(c) O2 > O2 < O+2
(d) O2 < O2 > O+2
Ans. (b)
Solution.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► Bond order, O
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► Bond order, O-
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► Bond order, O+
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.21. The electronic configuration of the outermost shell of the most electronegative element is
(a) 2s22p5
(b) 3s23p5
(c) 4s24p5
(d) 5s25p5
Ans. (a)
Solution: 

The electronic configuration represents:
► 2s22p5= fluorine = most electronegative element
► 3s23p5 = chlorine
► 4s24p5 = bromine
► 5s25p5 = iodine

Q.22. Amongst the following elements whose electronic configurations are given below, the one having the highest ionization enthalpy is
(a) [Ne]3s23p1
(b) [Ne]3s23p3
(c) [Ne]3s23p2
(d) [Ar]3d104s24p3
Ans. (b)
Solution: (i), (iii), and (iv) have exactly half-filled p-orbitals but (ii) is smaller in size. Hence, (ii) has the highest ionization enthalpy.


Multiple Choice Questions - II

In the following questions, two or more options may be correct.
Q.23. Which of the following have identical bond order?
(a) CN
(b) NO+
(c) O2
(d) O2–2

Ans. (a, b)
Solution.

► CN (number of electrons = 6 + 7 + 1 = 14)
► NO+ (number of electrons = 7 + 8 – 1 = 14)
► O2 (number of electrons = 8 + 8 + 1 = 17)
► O22- (number of electrons = 8 + 8 + 2 = 18)
Thus, CN and NO+ because of the presence of same number of electrons, have same bond order.

Q.24. Which of the following attain the linear structure:
(a) BeCl2
(b) NCO+
(c) NO2
(d) CS2

Ans. (a, d)
Solution.

  • BeCl2 (Cl – Be – Cl) and CS2 (S = C = S) both are linear, NCO+ is non-linear. However, remember that NCO(N = C = O) is linear because it is isoelectronic with CO2.
  • NO2 is angular with bond angled 132° and each O – N bond length of 1.20A°(intermediate between single and double bond).

Q.25. CO is isoelectronic with
(a) NO+
(b) N2
(c) SnCl2
(d) NO2
Ans. (a, b)
Solution.

► Number of electrons in CO =14
► Number of electrons in NO+ =14
► Number of electrons in N2 = 14
► Number of electrons in SnCl2 = 84
► Number of electrons in NO2 = 24

Q.26. Which of the following species have the same shape?
(a) CO2
(b) CCl4
(c) O3
(d) NO2

Ans. (c, d)
Solution. CO2 →Linear, CCl4 → Tetrahedral, O3 →Angular (V-shaped), NO2 →Angular (V-shaped)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.27. Which of the following statements are correct about CO2–3?
(a) The hybridization of the central atom is sp3.
(b) Its resonance structure has one C–O single bond and two C=O double bonds.
(c) The average formal charge on each oxygen atom is 0.67 units.
(d) All C–O bond lengths are equal.
Ans. (c, d)
Solution. 
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► Formal charge in O atom (1): NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► Formal charge in O atom (2): NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► Formal charge in O atom (3): NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
► The average formal charge on each oxygen atom: NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
All C - O bonds are equal due to resonance.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.28. Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic?
(a) N2
(b) N2–2
(c) O2
(d) O2–2
Ans. (a, d)
Solution.

  • (a) Electronic configuration of N2 =NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    It has no unpaired electron. This indicates it is a diamagnetic species.
  • (b) Electronic configuration of N2-2 ion =NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    It has two unpaired electrons, so it is paramagnetic in nature.
  • (c) Electronic configuration of O2 = NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    The presence of two unpaired electrons shows its paramagnetic nature.
  • (d) Electronic configuration of O2-2 ion = NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
    It contains no unpaired electron. Therefore, it is diamagnetic in nature.

Q.29. Species having the same bond order are:
(a) N2
(b) N2
(c) F+2
(d) O2
Ans. (c, d)
Solution. The total number of electrons in F+2 and O-2 are the same, i.e. 17 electrons.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.30. Which of the following statements are not correct?
(a) NaCl being an ionic compound is a good conductor of electricity in the solid-state.
(b) In canonical structures, there is a difference in the arrangement of atoms.
(c) Hybrid orbitals form stronger bonds than pure orbitals.
(d) VSEPR Theory can explain the square planar geometry of XeF4.
Ans. (a, b)
Solution.

(a) Ionic compounds are good conductors only in molten state or aqueous solution since ions are not furnished in solid-state.
(b) In canonical structures, there is a difference in the arrangement of electrons.

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Exam

,

ppt

,

Extra Questions

,

shortcuts and tricks

,

Free

,

Previous Year Questions with Solutions

,

Viva Questions

,

NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

,

video lectures

,

NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

,

Sample Paper

,

Semester Notes

,

Summary

,

Objective type Questions

,

mock tests for examination

,

pdf

,

NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

,

study material

,

past year papers

,

MCQs

,

practice quizzes

,

Important questions

;