NCERT Exemplar: Chemical Bonding & Molecular Structure - 1 | Chemistry Class 11 - NEET PDF Download

Multiple Correct Questions (Type-I)

Q1: Isostructural species are those which have the same shape and hybridization. Among the given species identify the isostructural pairs.
(a) [NF₃ and BF₃]
(b) [BF_4^– and NH⁺₄ ]
(c) [BCl₃ and BrCl₃]
(d) [NH₃ and NO₃^– ]

Correct Answer is Option (b)

• NF₃ is pyramidal whereas BF₃ is planar triangular.
• BF_4^– and NH⁺₄ ions both are tetrahedral.
• BCl₃ is triangular planar whereas BrCl₃ is T Shape.
• NH₃ is pyramidal whereas NO₃- is triangular planar.

Q2: 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

Correct Answer is Option (c)

H₂O will have the highest dipole moment due to the maximum difference in electronegativity of H and O atoms.

Q3: The types of hybrid orbitals of nitrogen in NO⁺₂, NO^–_3, and NH⁺₄ respectively are expected to be
(a) sp, sp^3, and sp²
(b) sp, sp^2, and sp³
(c) sp², sp, and sp³
(d) sp², sp^3, and sp

Correct Answer is Option (b)

• The number of orbitals involved in hybridization can be determined by the application of formula:
  
  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
•  NO₄⁺ : H = 1/2 [5 + 4  - 1 + 0] = 4 or sp³

Q4: 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

Correct Answer is Option (b)

• 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.

Q5: In PO^3–₄ ion, the formal charge on the oxygen atom of  P – O bond is
(a) + 1
(b) – 1
(c) – 0.75
(d) + 0.75

Correct Answer is Option (b)

• 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)For charge on oxygen = 6 - = 6 -7 = - 1

Q6: In NO^–₃ 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

Correct Answer is Option (d)

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

Q7: Which of the following species has tetrahedral geometry?
(a) BH^–₄
(b) NH^–₂
(c) CO^2–₃
(d) H₃O⁺

Correct Answer is Option (a)

• Boron is surrounded by 4 bonded pairs only.
• In BH^-₄ = no. of bond pair = 4
• No. of lone pair = 0
• sp³ hybridised, so have tetrahedral geometry

•  NH^-₂ = Bent shape
•  H₃O⁺ = Trigonal Pyramidal
•  CO^-2₃= Trigonal Planar

Q8: Number of π bonds and σ bonds in the following structure is–

(a) 6, 19
(b) 4, 20
(c) 5, 19
(d) 5, 20

Correct Answer is Option (c)

• 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.
  
  

Q9: Which molecule/ion out of the following does not contain unpaired electrons?
(a) N⁺₂
(b) O₂
(c) O^2–₂
(d) B₂

Correct Answer is Option (c)

Thus, O₂^2-has no unpaired electrons.

Q10: In which of the following molecule/ion, all the bonds are not equal?
(a) XeF₄
(b) BF_4^–
(c) C₂H₄
(d) SiF₄

Correct Answer is Option (c)

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).

Q11: In which of the following substances will hydrogen bond be strongest?
(a) HCl
(b) H₂O
(c) HI
(d) H₂S

Correct Answer is Option (b)

HCl, HI, and H₂S do not form H-bonds. Only H₂O forms hydrogen bonds. One H₂O molecule forms four H-bonding.

Q12: 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, 4s²

Correct Answer is Option (d)

The electron involved in the formation of any chemical bond are the valence shell electrons. In the given electronic configuration, the valence shell electrons are in d- orbital and s-orbital.

Q13: Which of the following angle corresponds to sp² hybridization?
(a) 90°
(b) 120°
(c) 180°
(d) 109°

Correct Answer is Option (b)

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.

Q14: Stable form of A may be represented by the formula :
(a) A
(b) A₂
(c) A₃
(d) A₄

Correct Answer is Option (a)

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

Q15: Stable form of C may be represented by the formula :
(a) C
(b) C₂
(c) C₃
(d) C₄

Correct Answer is Option (b)

The electronic configuration of C represents chlorine. Its stable form is dichlorine (Cl₂), i.e., C₂.

Q16: The molecular formula of the compound formed from B and C will be
(a) BC
(b) B₂C
(c) BC₂
(d) BC₃

Correct Answer is Option (d)

B represents P, and C represents Cl. The stable compound is PCl_3, i.e., BC₃.

Q17: The bond between B and C will be
(a) Ionic
(b) Covalent
(c) Hydrogen
(d) Coordinate

Correct Answer is Option (b)

Both B and C are non-metals, and therefore, the bond formed between them will be covalent.

Q18: Which of the following order of energies of molecular orbitals of N₂ is correct?
(a) (π2p_y) < (σ2p_z) < (π*2p_x) ≈ (π*2p_y)
(b) (π2p_y) > (σ2p_z) > (π*2p_x) ≈ (π*2p_y)
(c) (π2p_y) < (σ2p_z) > (π*2p_x) ≈ (π*2p_y)
(d) (π2p_y) > (σ2p_z) < (π*2p_x) ≈ (π*2p_y)

Correct Answer is Option (a)

Molecules like B₂, C₂ & N₂ having 1 to 3 electrons in p orbital energy of σ2p molecular orbital is greater than that of π2p_x and π2p_y molecular orbitals.

Q19: 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 < σ2p_z < (π2p_x = π2p_y) < (π*2p_x = π*2p_y) < σ*2p_z

_{Correct Answer is Option (d)}

• (a) Be₂ (4 + 4 = 8) = σ1s², σ*1s² ,σ1s² ,σ*2s²
  Bond order (BO) = 1/2 [Number of bonding electron (N_b) - Number of anti - bonding electrons N_a]

         = (4-4)/2=0

• (b) Here, the bond order of He₂ is zero. Thus, it does not exist.
  He₂ (2 + 2 = 4) = σ1s² , σ*1s²
  
  Here, the bond order of He₂ is zero. Thus, it does not exist.
  He⁺₂ (2 + 2 - 1 = 3) = σ1s², σ*1s¹
  
  Since the bond order is not zero, this molecule is expected to exist.
• (c) N₂ (7 + 7 = 14) = σ1s², σ*1s² ,σ2s² ,σ*2s² , (π2p²_x ≈ π* 2p²_y) < σ*2p²_z
  
  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 < σ2p_z < (π2p_x ≈ π2p_y) < (π* 2p_x ≈ π* 2p_y) < σ*2p_z

Q20: Which of the following options represents the correct bond order : 
(a) O^–₂ > O₂ > O⁺₂
(b) O^–₂ < O₂ < O⁺₂
(c) O^–₂ > O₂ < O⁺₂
(d) O^–₂ < O₂ > O⁺₂

Correct Answer is Option (b)

• Bond order, O₂ 

•  Bond order, O₂^- 

•  Bond order, O₂ ⁺

∴

Q21: The electronic configuration of the outermost shell of the most electronegative element is
(a) 2s²2p⁵
(b) 3s²3p⁵
(c) 4s²4p⁵
(d) 5s²5p⁵

Correct Answer is Option (a) 

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

Q22: 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

Correct Answer is Option (b)

(i), (iii), and (iv) have exactly half-filled p-orbitals but (ii) is smaller in size. Hence, (ii) has the highest ionization enthalpy.

Multiple Correct Questions (Type - II)

In the following questions, two or more options may be correct.
Q23: Which of the following have an identical bond order?
(a) CN^–
(b) NO⁺
(c) O₂^–
(d) O^2–₂

Correct Answer is Option (a) & (b)

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

Q24: Which of the following attain the linear structure:
(a) BeCl₂
(b) NCO⁺
(c) NO₂
(d) CS₂

Correct Answer is Option (a) & (d)

• 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).

Q25: CO is isoelectronic with
(a) NO⁺
(b) N₂
(c) SnCl₂
(d) NO^–₂

Correct Answer is Option (a) & (b)

•  Number of electrons in NO⁺ =14
•  Number of electrons in N₂ = 14
•  Number of electrons in SnCl₂ = 84
•  Number of electrons in NO₂^– = 24

Q26: Which of the following species have the same shape?
(a) CO₂
(b) CCl₄
(c) O₃
(d) NO₂^–

Correct Answer is Option (c) & (d)

CO₂ →Linear, CCl₄ → Tetrahedral, O₃ →Angular (V-shaped), NO^–₂ →Angular (V-shaped)

Q27: Which of the following statements are correct about CO^2–₃?
(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.

Correct Answer is Option (c) & (d)

•  Formal charge in O atom (1):
• Formal charge in O atom (2):
•  Formal charge in O atom (3):
•  The average formal charge on each oxygen atom:

All C - O bonds are equal due to resonance.

Q28: 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–

Correct Answer is Option (a) & (d)

• (a) Electronic configuration of N₂ =
  It has no unpaired electron. This indicates it is a diamagnetic species.
• (b) Electronic configuration of N^2-₂ ion =
  
  It has two unpaired electrons, so it is paramagnetic in nature.
• (c) Electronic configuration of O₂ =
  
  The presence of two unpaired electrons shows its paramagnetic nature.
• (d) Electronic configuration of O₂^2- ion =
  
  It contains no unpaired electron. Therefore, it is diamagnetic in nature.

Q29: Species having the same bond order are:
(a) N₂
(b) N₂^–
(c) F₂⁺
(d) O₂^–

Correct Answer is Option (c) & (d)

The total number of electrons in F₂⁺ and O₂ ^-are the same, i.e. 17 electrons.

Q30: 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.

Correct Answer is Option (a) & (b)

(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.