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) [BF–4 and NH+4 ]
(c) [BCl3 and BrCl3]
(d) [NH3 and NO–3 ]
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?
Solution. H2O will have the highest dipole moment due to the maximum difference in electronegativity of H and O atoms.
Q.3. The types of hybrid orbitals of nitrogen in NO+2, NO–3, 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
Solution. 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
► 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
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
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)
► For charge on oxygen = 6 - = 6 -7 = - 1
Q.6. 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
Solution: In N-atom, number of valence electrons = 5
Q.7. Which of the following species has 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–
(a) 6, 19
(b) 4, 20
(c) 5, 19
(d) 5, 20
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.
Q.9. Which molecule/ion out of the following does not contain unpaired electrons?
Thus, O2-2 has no unpaired electrons.
Q.10. In which of the following molecule/ion, all the bonds are not equal?
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).
Q.11. In which of the following substances will hydrogen bond be strongest?
Solution. HCl, HI, and H2S do not form H-bonds. Only H2O forms hydrogen bonds. One H2O molecule forms four H-bonding.
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_____.
(b) 3p6, 4s2
(c) 3p6, 3d2
(d) 3d2, 4s2
Solution. In transition elements (n -1) d and ns orbitals take part in bond formation.
Q.13. Which of the following angle corresponds to sp2 hybridization?
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.
Q.14. The stable form of A may be represented by the formula :
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 :
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
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
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)
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
Q.20. 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
► Bond order, O2
► Bond order, O-2
► Bond order, O+2
Q.21. The electronic configuration of the outermost shell of the most electronegative element is
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
Solution: (i), (iii), and (iv) have exactly half-filled p-orbitals but (ii) is smaller in size. Hence, (ii) has the highest ionization enthalpy.
In the following questions, two or more options may be correct.
Q.23. Which of the following have identical bond order?
Ans. (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)
► 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:
Ans. (a, d)
Q.25. CO is isoelectronic with
Ans. (a, b)
► 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 NO–2 = 24
Q.26. Which of the following species have the same shape?
Ans. (c, d)
Solution. CO2 →Linear, CCl4 → Tetrahedral, O3 →Angular (V-shaped), NO–2 →Angular (V-shaped)
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)
► 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.
Q.28. Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic?
Ans. (a, d)
Q.29. Species having the same bond order are:
Ans. (c, d)
Solution. The total number of electrons in F+2 and O-2 are the same, i.e. 17 electrons.
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)
(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.