Chemistry 02 – Chemical bonding and molecular structure

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1. Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.

i) [NF3 and BF3]

ii)[BF 4 and NH4 +

iii) [BCl 3 and BrCl 3]

iv)[NH3 and NO 3– ]

Ans. (ii)

Explanation: Both the molecules have sp3 hybridisation and shape is tetrahedral.

2. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?

(i) CO2
(ii) HI
(iii) H2O
(iv) SO2

Ans. (iii)

Explanation: Oxygen is highly electronegative and because of two lone pair on oxygen, the dipole moment is high.

3. The types of hybrid orbitals of nitrogen in respectively are expected to be

(i) sp, sp3 and sp2
(ii) sp, sp2 and sp3
(iii) sp2, sp and sp3
(iv) sp2, sp3 and sp

Ans. (ii)

The hybridization in NO is sp and in NO 3 and in NH4 + ,it is sp3

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 bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

i) HF > H2O > NH3

ii) H2O > HF > NH3

(iii) NH3 > HF > H2O

(iv) NH3 > H2O > HF

Ans. (ii)

Explanation: Strength of hydrogen bond depends on electronegativity, size of atom and number of hydrogen bonds per mol. F is most electronegative but, in H2O, the no. of H-Bonds are more and size is smaller than N.

5. In P O 4 3 –  ion the formal charge on the oxygen atom of P–O bond is

(i) +1
(ii) –1
(iii) –0.75
(iv) +0.75

Ans. (ii)
Explanation: The formal charge of an atom in a polyatomic molecule or ion may be defined as the difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure. Formal charge of the atom in the molecule or ion = (Number of valence electron in the free atom) – (Number of lone pair electrons + 1/2 number of bonding electrons)So formal charge on each O-atom

=6 -(6+1/2 × 2 ) =6 –  7 =-1

6. In NO 3 –  ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are

(i) 2, 2
(ii) 3, 1
(iii) 1, 3
(iv) 4, 0

Ans. (iv)
Explanation: Central atom N is surrounded by 2 covalent bonds with 1 oxygen atom and 2 coordinate covalent bonds with 2 oxygen atoms.

7. Which of the following species has tetrahedral geometry?

(i) BH4
(ii) NH2  –
(iii)CO 3 2

(iv) H3O

Ans. (i)
Explanation: 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

NH2 =V – Shape H30 + = pyramidal

CO3  -2= triangular planar

8. Number of π bonds and σ bonds in the following structure is–

(i) 6, 19
(ii) 4, 20
(iii) 5, 19
(iv) 5, 20

Ans. (iii)
Explanation: Each C atom is sp2 hybridized and surrounded by 3 sigma and 1 pi bond between two C atoms. Structure to be rectified.

8 C – H + 11 C – C bonds = 19 σ bonds.
There are 5 π -bonds and 19 σ bonds.

9. Which molecule/ion out of the following does not contain unpaired electrons?

i) N2+

ii)O2

iii)O2 2-

iv)B2

Ans. (iii)
Explanation: In O2 2- antibondingP πx = P πy orbitals have paired electrons.

10.In which of the following molecule/ion all the bonds are not equal?

(i) XeF4
(ii)BF 4
(iii) C2H4
(iv) SiF4

Ans. (iii)

Explanation: Each C atom is surrounded by 3 σ bonds and 1π bond.

11.In which of the following substances will hydrogen bond be strongest?

i)HCl
(ii) H2O
(iii) HI
(iv) H2S

Ans. (ii)

Explanation: Oxygen is more electronegative as compared to others and its size is small.

12. If the electronic configuration of an element is 1s2 2s2 2p6 3s6 3p2 3d2 4s2 , the four electrons involved in chemical bond formation will be_____.

(i) 3p6
(ii) 3p6 , 4s2
(iii) 3p6 , 3d2
(iv) 3d2 , 4s2

Ans. (iv)
Explanation: In bonding, valence electrons and electrons from penultimate shell participate. In transition elements ns electrons and {n -1)d electrons participate.

13. Which of the following angle corresponds to sp2 hybridisation?

(i) 90°
(ii) 120°
(iii) 180°
(iv) 109°

Ans. (ii)
Explanation: As a result of sp2 hybridization geometry is triangular planar forming an angle of 120° with each other.

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

A 1S22S2SP6  
B  1S22S2SP63S23SP3
C 1S22S2SP63S23SP5

14. Stable form of A may be represented by the formula:

(i) A
(ii) A2
(iii) A3
(iv) A4

Ans. (i)
Explanation: Octate of A is complete.

15. Stable form of C may be represented by the formula:

(i) C
(ii) C2
(iii) C3
(iv) C4

Ans. (ii)

Explanation: KK[  σ 2S]2 [σ * 2S]2 [π2Px]2 [π2py]2

Bond order = 1/2 [6 -2 ] = 2

16. The molecular formula of the compound formed from B and C will be

(i) BC
(ii) B2C

(iii) BC2
(iv) BC3

Ans. (iv)
Explanation: B represents phosphorus (P) and C represents Chlorine (Cl). The compound
formed is PCl3 i.e., BC3

17. The bond between B and C will be

(i) Ionic
(ii) Covalent
(iii) Hydrogen
(iv) Coordinate

Ans. (ii)
Explanation: Both B and C are non-metals so, the bond formed between them will be
covalent.

18. Which of the following order of energies of molecular orbitals of N2 is correct?

i)(π2py) <(σ2pz)<(π*2px) ≈ (π*2py)

ii)(π2py) >(σ2pz)>(π*2px) ≈ (π*2py)

iii)(π2py) <(σ2pz)>(π*2px) ≈ (π*2py)

iv)(π2py) >(σ2pz)<(π*2px) ≈ (π*2py)

Explanation: 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.

19. Which of the following statement is not correct from the view point of molecular orbital theory?

(i) Be2 is not a stable molecule.

(ii) He2 is not stable but He+ is expected to exist.

(iii) Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(iv) The order of energies of molecular orbitals in N2 molecule is

σ2s < σ*2s <σ2pz <(π2px = π2py ) < (π*2px = π*2p) < σ *2pz

Ans. (iv)
Explanation: The energy of a σ2p  molecular orbital is greater than that of π2px and π2py  molecular orbitals.

20. Which of the following options represents the correct bond order?

i) O2>O2 >O2+

ii)O2<O2 <O2+

iii)O2>O2 <O2+

iv)O2< O2 >O2+

Ans. (ii)
Explanation: Nb electrons > Na electrons and also depends on bond order.

B.O = (1/2 ) (Nb -Na )

21. The electronic configuration of the outermost shell of the most electronegative element is

i) 2s2 2p5

ii) 3s2 3p5

iii)4s2 4p5

iv) 5s2 5p5

Ans. (i)
Explanation: Elements of Group 17 has ns2np5

electronic configuration. The electronegativity decreases down the group. The elements of this group are most electronegative.

22. Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is

i) [Ne] 3s2 3p1

ii) [Ne] 3s2 3p3

iii)[Ne]3s2 3p2

iv) [Ar] 3d104s2 4p3

Ans. (ii)
Explanation: Since half-filled p-orbital is more stable as compared to partially filled orbital.

In the following questions two or more options may be correct.

23. Which of the following have identical bond order?

i)CN

ii)NO +

iii)O2

iv)O2 2 –

Ans. (i) and (ii)
Explanation: In both the species number of electrons in Nb = number of electrons in Nab

24. Which of the following attain the linear structure:

i) BeCl2

ii)NCO+

iii)NO2

iv) CS2

Ans. (i) and (iv)
Explanation: In both, the central atom is sp hybridized.

25. CO is isoelectronic with

i) NO+

ii)N2

iii)Sncl2

iv) NO 2

Ans. (i) and (ii)
Explanation: In both the species the number of electrons are same but the number of protons are different.

26. Which of the following species have the same shape?

i) CO 2

ii)CCl4

iii)O3

iv) NO2

Ans. (iii) and (iv)
Explanation: Central atom in both the species has same hybridized state and same geometry.

27. Which of the following statements are correct about CO 3 2- ?

(i) The hybridisation of central atom is sp3
(ii) Its resonance structure has one C-O single bond and two C=O double bonds.
(iii) The average formal charge on each oxygen atom is 0.67 units.

(iv) All C-O bond lengths are equal.

Ans. (iii) and (iv)
Explanation: In resonating structures bonds are not fixed and therefore, all bond lengths are equal.

28. Dimagnetic species are those which contain no unpaired electrons. Which among the following are dimagnetic?

i) N2

ii)N22-

iii)O2

iv) O22-

Ans. (i) and (iv)
Explanation: In both the species all orbitals have paired electrons.

29. Species having same bond order are:

i) N2

ii)N2-

iii)F+

iv) O2

Ans. (iii) and (iv)
Explanation: In both the species number of electrons in bonding molecular orbital and in antibonding orbital are same.

30. Which of the following statements are not correct?

(i) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(ii) In canonical structures, there is a difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR Theory can explain the square planar geometry of XeF4

Ans. (i) and (ii)
Explanation: In first statement ions are not free in solid state but they are strongly bonded through electrostatic forces and second statement it does not show canonical structures being ionic compounds.

30. Which of the following statements are not correct?

(i) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(ii) In canonical structures, there is a difference in the arrangement of atoms.
(iii) Hybrid orbitals form stronger bonds than pure orbitals.
(iv) VSEPR Theory can explain the square planar geometry of XeF4

Ans. (i) and (ii)
Explanation: In first statement ions are not free in solid state but they are strongly bonded
through electrostatic forces and second statement it does not show canonical structures eing ionic compounds.

31. Explain the non-linear shape of H2S and non-planar shape of PCl3 using valence shell electron pair repulsion theory.

Ans. In both H2S and PCl3 , lone pairs are present along with bond pairs around the central atom. According to VSEPR, theory L.P-L.P > L.P-B.P > B.P.-B.P. Therefore, shapes are
bdifferent.

32. Using molecular orbital theory, compare the bond energy and magnetic character of  O 2+ andO2species.

Ans. In O2+   B.O = (10 -5) /2 = 2.5

In O2 .B .O = (10 -7) /2 = 1.5

Bond dissociation energy is directly proportional to bond order and paramagnetic character depends on unpaired electrons.O2+  has high dissociation energy and both are paramagnetic.

33. Explain the shape of BrF5

Ans. BrF5

: Br is surrounded by 5 bonded pairs and one lone pair. Shape is square pyramidal.

.

34. Structures of molecules of two compounds are given below:

(a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding.
(b) The melting point of a compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point.
(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it.

Ans. (a) Compound (I) will form intramolecular hydrogen bonding because, NO2 and OH groups are close together in comparison to that in compound(II).The intermolecular hydrogen bonding will be shown by compound (II)

(b) Compound (II) will have higher melting point because, it forms intermolecular H-bonds. Thus, more and more molecules are joined together through H-bonds.

(c) Due to intramolecular H-bonding, compound (I) will not be able to form hydrogen bonds with water thus, will be less soluble in it while compound (II) can form H-bond with water more easily and will be soluble in water.

.

35. Why does type of overlap given in the following figure not result in bond formation?

Ans. Same charges repel and there is zero overlapping hence, no bond formation occurs.

36. Explain why PCl5 is trigonal bipyramidal whereas IF5 is square pyramidal.

Ans. In PCl5 , P is surrounded by 5 bond pairs and no lone pairs, whereas in IF5 , iodine atom is surrounded by 5 bond pairs and one lone pair, therefore the shape is different.

37. In both water and dimethyl ether( CH3..– CH3 ), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.

Ans. Dimethyl ether will have larger bond angle. Because of more repulsion between bond
pairs of CH3 groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in water. The carbon of CH3 in ether is attached to three hydrogen atoms through bonds and electron pairs of these bonds add to the electronic charge density on carbon atom. Hence, repulsion between two —CH3 groups will be more than that between two hydrogen Atoms.

38. Write Lewis structure of the following compounds and show formal charge on each
atom.

39. The energy of σ2P molecular orbital is greater than π2Px and π2py molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species:

N2 , N2+ , N2 , N2 2+

Ans. Configuration of N2  : σ 1S2 < σ * 1S2 < σ2S2<σ *2S2 <[π2Px2 = π2Px2]  < σ 2pz2 <[π*2px2 =π*2px2]<σ*2pz

Bond Order of N2 = (10-4)/2 = 3

N2 is Diamagnetic, as all orbitals has unpaired electron.

N2 + : σ 1S2 < σ * 1S2 < σ2S2<σ *2S2 <[π2Px2 = π2Px2]  < σ 2pz2 <[π*2px2 =π*2px1]<σ*2p1

B.O. of  N2+ = (9-4)/2 = 2.5

N2 –   is Paramagnetic, as one of the orbitals has unpaired electron.

N2 + : σ 1S2 < σ * 1S2 < σ2S2<σ *2S2 <[π2Px2 = π2Px2]  < σ 2pz2 <[π*2px2 =π*2px2]<σ*2p1

B.0 of N2 = (9-5)/2 = 2.5

N2 is Paramagnetic, as one of the orbitals has unpaired electron.

N2 : σ 1S2 < σ * 1S2 < σ2S2<σ *2S2 <[π2Px2 = π2Px2]  < σ 2pz2 <[π*2px2 =π*2px1]<σ*2p

B.O. of N2 2+ = (8-4)/2 = 2

It is Paramagnetic as two, of the orbitals have unpaired electron.
Stability: Stability of species depends on bond order and also depends on number of electrons present in bonding molecular orbital.
Increasing order of stability: N2+ < N2 < N2< N2

N22+=B.O. is lowest.

N2 –   And N2B.O. is same but number of electrons in bonding and in antibonding molecular orbitals are different. In N2 , in antibonding molecular orbitals, electrons are more than N2+

N2 = B.O. is higher.

40. What is the effect of the following processes on the bond order in N2 and O2?

i) N2 → N2+  + e

ii)O2 → O2 + + e

i) N2 → N2+ + e [Bond order of N2 is 3 while of N2+ is 2.5]

Bond order decreases.

ii)O2 →O2+ + e–  [Bond order of O2 is 2 while of O2+is 2.5]

Bond order increases.

41. Give reasons for the following:
(i) Covalent bonds are directional bonds while ionic bonds are non- directional.
(ii) Water molecule has bent structure whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.

Ans. (i) A covalent bond is a directional bond because it is formed by the overlapping of atomic orbitals. Ionic bond is non-directional because it is formed by the transference of
electrons and electrostatic field of an ion is there which is non directional. Positive ion is surrounded by number of anions in any direction depending upon the size and vice-versa.
(ii) In H2O, oxygen atom is sp3hybridized and surrounded by two lone pairs and two bonded pairs. The four sp3 hybrid orbitals give distorted tetrahedral geometry in which the two corners are occupied by hydrogen atoms while other two by the lone pairs. The bond angle is reduced to 104.5° from 109.5° due to greater repulsive forces between lp – lp and molecule acquires a bent or V-shaped structure.

In CO2 , carbon atom is sp hybridized and two sp hybrid orbitals are oriented in opposite direction forming linear shape with an angle of 1800

(iii) In ethyne molecule, both the carbon atoms are sp hybridized and two unhybridized orbitals. Hybridized orbitals of both the carbon atoms are oriented in opposite directions forming a linear structure with an angle of 180°.

42. What is an ionic bond? With two suitable examples explain the difference between an ionic and a covalent bond?

Ionic bond: The bond which is formed by the complete transference of electrons from one atom to other atom as a result positive and negative ions are formed. These ions are held together through electrostatic force of attraction. This force of attraction is called ionic bond, e.g.,

The formation of calcium fluoride,

Ca →Ca2+ + 2e

[Ar]4s   [Ar]

F + e → F

[He]2s2 2p5 [He]2s2 2p6 or[Ne]

Ca2+2 F – → CaF 2 OR Ca2+ (F-)2

The formation of NaCl

Na → Na+  + e

[Ne]3s1    [Ne]

Cl + e–  → Cl

[Ne] 3s2 3sp5 [Ne]3s2 3sp6  or [Ar]

Na+ + Cl → NaCl or Na+Cl

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