CBSE Class 11 Chemistry HOTs Chemical Bonding and Molecular Structure

Question. Among the following, the species with identical bond order are
(a) CO and O₂^2–
(b) O₂^– and CO
(c) O₂^2– and B₂
(d) CO and N⁺₂
Answer : C

Question. The hybridizations of N, C and O shown in the following compound

respectively, are
(a) sp², sp, sp²
(b) sp², sp², sp²
(c) sp², sp, sp
(d) sp, sp, sp²
Answer : A

Question. For AB bond if percent ionic character is plotted against electronegativity difference (X_A – X_B), the shape of the curve would look like

The correct curve is
(a) (A)
(b) (B)
(c) (C)
(d) (D)
Answer : C

Question. The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH₃ (1.5 D) is larger than that of NF₃ (0.2D). This is because
(a) in NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite directions
(b) in NH₃ as well as NF₃ the atomic dipole and bond dipole are in opposite directions
(c) in NH₃ the atomic dipole and bond dipole are in the opposite directions whereas in NF₃ these are in the same direction
(d) in NH₃ as well as in NF₃ the atomic dipole and bond dipole are in the same direction
Answer : A

Question. The most polarizable ion among the following is
(a) F^–
(b) I^–
(c) Na⁺
(d) Cl^–
Answer : B

Question. The number of sigma (s) and pi (p) bonds present in 1,3,5,7 octatetraene respectively are
(a) 14 and 3
(b) 17 and 4
(c) 16 and 5
(d) 15 and 4
Answer : B

Question. The bond dissociation energy of B – F in BF₃ is 646 kJ mol^–1 whereas that of C – F in CF₄ is 515 kJ mol^–1. The correct reason for higher B – F bond dissociation energy as compared to that of C – F is
(a) stronger s bond between B and F in BF₃ as compared to that between C and F in CF₄.
(b) significant pp – pp interaction between B and F in BF₃ whereas there is no possibility of such interaction between C and F in CF_4.
(c) lower degree of pp – pp interaction between B and F in BF₃ than that between C and F in CF_4.
(d) smaller size of B– atom as compared to that of C– atom.
Answer : B

Question. In which of the following species, d-orbitals having xz and yz two nodal planes involved in hybridization of central atom?
(a) IO₂F^–
(b) ClF^–₄
(c) IF₇
(d) None of these
Answer : C

Question. Minimum F – S – F bond angle present in :
(a) SSF₂
(b) SF₆
(c) SF₂
(d) F₃SSF
Answer : D

Question. Which of the following statements is/are true
1. PH₅ and BiCl₅ do not exist
2. pπ – d p bond is present in SO₂
3. I⁺₃ has bent geometry
4. SeF₄ and CH₄ have same shape
(a) 1, 2, 3
(b) 1, 3
(c) 1, 3, 4
(d) 1, 2, 4
Answer : A

Question. Among the following transformations, the hybridization of the central atom remains unchanged in
(a) C₂^2– → HCOOH
(b) BF₃ → BF₄^-
(c) NH₃ → NH₄⁺
(d) PCl₃ → PCl₅
Answer : C

Question. Which of the following statement is correct about I⁺₃ and I^–₃- molecular ions?
(a) Number of lone pairs at central atoms are same in both molecular ions
(b) Hybridization of central atoms in both ions are same
(c) Both are polar species
(d) Both are planar species
Answer : D

Question. The correct statement with regard to H⁺₂ and H^–₂ is
(a) Both H⁺₂ and H^–₂ do not exist
(b) H^–₂ is more stable than H⁺₂
(c) H⁺₂ is more stable than H^–₂
(d) Both H⁺₂ and H^–₂ are equally stable
Answer : C

Question. The correct order of bond energies in NO, NO⁺ and NO^– is:
(a) NO^– > NO > NO⁺
(b) NO > NO^– > NO⁺
(c) NO⁺ > NO > NO^–
(d) NO⁺ > NO^– > NO
Answer : C

Question. Which one of the following molecule will have all equal X—F bond length? (where X = Central atom)
(a) SOCl₂F₂
(b) SeF₄
(c) PBr₂F₃
(d) IF₇
Answer : A

Question. In which species, X—O bond order is 1.5 and contains pπ – dπ bond(s).
(a) I₂^2–F^–₂
(b) HCOO^–
(c) SO^2–₃
(d) XeO₂F₂
Answer : A

Question. The type of bonds present in sulphuric anhydride is,
(a) 3σ and three pπ – dπ
(b) 3σ, one pπ – pπ and two pπ – dπ
(c) 2σ and three pπ – dπ
(d) 2σ and two pπ – dπ
Answer : B

Question. Select the incorrect statement about N₂F₄ and N₂H₄ :
(I) In N₂F₄, d-orbitals are contracted by electronegative fluorine atoms, but dorbital contraction is not possible by Hatom in N₂H₄
(II) The N-N bond energy in N₂F₄ is more than N-N bond energy in N₂H₄
(III) The N-N bond length in N₂F₄ is more than that of in N₂H₄
(IV) The N-N bond length in N₂F₄ is less than that of in N₂H₄
(a) I, II and III
(b) I and III
(c) II and IV
(d) II and III
Answer : B

Question. The correct order of ‘S—O’ bond length is
(a) SO₃^2– > SO₄^2– - > SO₃ > SO₂
(b) SO₃^2– > SO₄^2– - > SO₂ > SO₃
(c) SO₄^2– > SO₄^2– > SO₂ > SO₃
(d) SO₄^2–  > SO₄^2– > SO₃ > SO₂
Answer : B

Numeric Value Answer

Question. Consider the following molecule

Calculate the value of p ÷ q, here p and q are total number of dp–pp bonds and total number of sp³ hybridised atoms respectively in given molecule.
Answer : 1

Question. Total number of species which used all three porbitals in hybridisation of central atoms and should be non-polar also are XeO₂F₂, SnCl₂, IF₅, I₃⁺ , XeO₄, SO₂, XeF₇⁺ , SeF₄
Answer : 2

Question. Consider the following orbitals 3s, 2px, 4d_xy, 4d_z² , 3d_x2 - _y2, 3py, 4s, 4pz and find total number of orbital(s) having even number of nodal plane.
Answer : 5

Question. Calculate the value of “x + y – z” here x, y and z are total number of non-bonded electron pair(s), pie(p) bond(s) and sigma (s) bonds in hydrogen phosphite ion respectively.
Answer : 3

Question. In O^–₂, O₂ and O₂^2– molecular species, the sum of the total number of antibonding electrons is ________
Answer : 21

Question. What is the % of p-character with central atom in SF6 molecule?
Answer : 50

Question. For the following molecules :
PCl₅, BrF₃, ICl^–₂, XeF^–₅- , NO^–₃-, XeO₂F₂, PCl⁺₄ , CH⁺₃
Calculate the value of a + b/c
a = Number of species having sp³ d-hybridisation
b = Number of species which are planar
c = Number of species which are non-planar
Answer : 3

Question. Find total number of orbital which can overlap colaterally, (if inter nuclear axis is z) s, px, p_y, p_z, d_xy, d_yz, d_xz, d_z2, dx²–y²
Answer : 6

Question. Consider the following values for an ionic compound NaCl.
ΔH_f (NaCl) = –200 KJ/mol
ΔH_sub(Na(s)) = 650 KJ/mol
ΔHdiss(Cl₂(g)) = 400 KJ/mol
I.E₁(Na(g)) = 500 KJ/mol
Electron gain enthalpy (Cl(g)) = –350 KJ/mol Using Born Haber Cycle, find the value of lattic energy (U) in KJ/mol.
Answer : 1200