NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Chemistry Class 11

JEE : 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 JEE Course Chemistry Class 11.
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MULTIPLE CHOICE QUESTIONS - I

Q.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) [BF4 and NH+4 ]
(iii) [BCl3 and BrCl3]
(iv) [NH3 and NO3 ]

Ans. (ii)
Solution.
NF3 is pyramidal whereas BF3 is planar triangular.
BF4 and NH+4 ions both are tetrahedral.
BCl3 is triangular planar whereas BrCl3 is pyramidal.
NH3 is pyramidal whereas NO3 is triangular planar.

Q.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)
Solution.
H20 will have highest dipole moment due to 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
(i) sp, sp3 and sp2
(ii) sp, sp2 and sp3
(iii) sp2, sp and sp3
(iv) sp2, sp3 and sp

Ans. (ii)
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 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)
Solution.
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 H20 molecule is linked to 4 other H20 molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is HzO > HF > NH3.

Q.5. In PO3–4 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)
Solution.
Formal charge of the atom in the molecule or ion = (Number of valence electrons in 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 nitrogen atom are
(i) 2, 2
(ii) 3, 1
(iii) 1, 3
(iv) 4, 0
Ans. (iv)
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?
(i) BH4
(ii) NH2
(iii) CO23
(iv) H3O+
Ans. (i)
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. Number of π bonds and σ bonds in the following structure is–
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(i) 6, 19
(ii) 4, 20
(iii) 5, 19
(iv) 5, 20
Ans. (iii)
Solution.
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 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?
(i) N+2
(ii) O2
(iii) O2–2
(iv) B
2
Ans. (iii)
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?
(i) XeF4
(ii) BF4
(iii) C2H4
(iv) SiF
4
Ans. (iii)
Solution.
C2H4 has one double bond and four single bonds. Bond length of double bond (C = C) is smaller than 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?
(i) HCl
(ii) H2O
(iii) HI
(iv) H2S

Ans. (ii)
Solution.
HCl, HI and H2S do not from H-bonds. Only H20 forms hydrogen bonds. One H20 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_____.
(i) 3p6
(ii) 3p6, 4s2
(iii) 3p6, 3d2
(iv) 3d2, 4s
2
Ans. (d)
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 hybridisation?
(i) 90°
(ii) 120°
(iii) 180°
(iv) 109°
Ans. (ii)
Solution.
sp2 hybridisation gives three sp2 hybrid orbitals which are planar triangular forming an angle of 120° with each other.

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.

 A 1 s2 2s2 2p6

 B1 s2 2s2 2p6 3s2 3p3
 C1 s2 2s2  2p6 3s2 3p5

Q.14. Stable form of A may be represented by the formula :
(i) A
(ii) A2
(iii) A3
(iv) A4

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

Q.15. Stable form of C may be represented by the formula :
(i) C
(ii) C2
(iii) C3
(iv) C4
Ans. (ii)
Solution.
The electronic configuration of C represent 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
(i) BC
(ii) B2C
(iii) BC2
(iv) BC3
Ans. (iv)
Solution.
B represent P and C represents Cl. The stable compound is PCl3 i.e., BC3.

Q.17. The bond between B and C will be
(i) Ionic
(ii) Covalent
(iii) Hydrogen
(iv) Coordinate
Ans. (ii)
Solution.
Both B and C are non-metals and therefore, bond formed between them will be covalent.

Q.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)
Ans. (i)
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 view point of molecular orbital theory?
(i) Be2 is not a stable molecule.
(ii) He2 is not stable but He+2  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*2py) < σ*2pz
Ans. (iv)
Solution.
(i) 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
Here, 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, 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, dinitrogen (N2) molecule contains triple bond and no other molecule of second period have more than double bond. Hence, 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 : 
(i) O2 > O2 > O+2
(ii) O2 < O2 < O+2
(iii) O2 > O2 < O+2
(iv) O2 < O2 > O+2
Ans. (ii)
Solution.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Bond order, ONCERT 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 outer most shell of the most electronegative element is
(i) 2s22p5
(ii) 3s23p5
(iii) 4s24p5
(iv) 5s25p5
Ans. (i)
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 ionisation enthalpy is
(i) [Ne]3s23p1
(ii) [Ne]3s23p3
(iii) [Ne]3s23p2
(iv) [Ar]3d104s24p3
Ans. (ii)
Solution.
(i), (iii) and (iv) have exactly half-filled p-orbitals but (ii) is smaller in size than Hence, (ii) has 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?
(i) CN
(ii) NO+
(iii) O2
(iv) O2–2

Ans. (i,ii)
Solution.
CN (number of electrons = 6 + 7 + 1 = 14)
NO+ (number of electrons = 7 + 8 – 1 = 14)
02 (number of electrons = 8 + 8 + 1 = 17)
022- (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:
(i) BeCl2
(ii) NCO+
(iii) NO2
(iv) CS2

Ans. (i,iv)
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 C02.
N02 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
(i) NO+
(ii) N2
(iii) SnCl2
(iv) NO2
Ans. (i,ii)
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 N02 = 24

Q.26. Which of the following species have the same shape?
(i) CO2
(ii) CCl4
(iii) O3
(iv) NO2

Ans. (iii,iv)
Solution.
C02 →Linear, CCl4 → Tetrahedral, 03 →Angular (V-shaped), N02 →Angular (V-shaped)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.27. Which of the following statements are correct about CO2–3 ?
(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,iv)
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
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. Dimagnetic species are those which contain no unpaired electrons. Which among the following are dimagnetic?
(i) N2
(ii) N2–2
(iii) O2
(iv) O2–2
Ans. (i, iv)
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 02-2 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 same bond order are :
(i) N2
(ii) N2
(iii) F+2
(iv) O2
Ans. (iii,iv)
Solution.
The total number of electrons in F+2 and O-2 are same, i.e. 17 electrons.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

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


SHORT ANSWER TYPE QUESTIONS

Q.31. Explain the non linear shape of H2S and non planar shape of PCl3 using valence shell electron pair repulsion theory.
Ans. The Lewis structure of H2S is:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevS-atom is surrounded by four electron pairs (two bonded and two lone pairs). These four electron pairs adopt tetrahedral arrangement. The presence of two lone pairs brings distortion in the molecule on account of repulsion with bonded pairs of electrons. Thus, the shape of H2S molecule is V-shaped and not linear.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevThe Lewis structure of PCI3 is:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevP-atom is surrounded by four electron pairs (3 bonded and one lone pair). These four pairs adopt a tetrahedral geometry. Due to the presence of lone pair, PCl3 has a distorted tetrahedral geometry. Thus, it is pyramidal in shape and not non-planar shape.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Q.32. Using molecular orbital theory, compare the bond energy and magnetic character of O+2 and O2 species.
Ans. NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Bond energy of O+2 > O-2
Both are paramagnetic due to presence of unpaired electrons.

Q.33. Explain the shape of BrF5.

Ans. Br-atom has configuration:
1s2, 2s22p6 , 3s23p63d10, 4s24p5
To get pentavalency, two of the p-orbitals are unpaired and electrons are shifted to 4d-orbitals.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
In this excited state, sp3d2-hybridisation occurs giving octahedral structure. Five positions are occupied by F atoms forming sigma bonds with hybrid bonds and one position occupied by lone pair, i.e., the molecule as a square pyramidal shape.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.34. Structures of molecules of two compounds are given below :
(i) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(ii) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

(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 bond because N02 and OH groups are close together whereas it is not so in compound II. Compound II will have intermolecular hydrogen bonding. Bonding in both the cases is shown below:
(i)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(ii)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(b) As a large number of molecules can be linked together through intermo­lecular hydrogen bonding, compound II will show higher melting point.
(c) Due to intramolecular hydrogen bonding, compound I will not be able to form hydrogen bonds with H20 molecules. Hence, it will be less soluble in water. However, compound II can form hydrogen bonds with H20 molecules easily and hence it will be more soluble in water.

Q.35. Why does type of overlap given in the following figure not result in bond formation?
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevAns. In first figure, the ++ overlap is equal to +- overlap and therefore, these cancel out and net overlap is zero.
In second figure, no overlap is possible because the two orbitals are perpendicular to each other.

Q.36. Explain why PCl5 is trigonal bipyramidal whereas IF5 is square pyramidal.
Ans. PC15 – The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented below
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Q.37. In both water and dimethyl ether NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev, 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 has larger bond angle than water. This is because there is 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 group in ether is attached to three hydrogen atoms through c bonds and electron pairs of these bonds add to the electron charge density on carbon atom. Hence, repulsion between two CH3 groups will be more than that between two H atoms.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

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

HNO3,  NO2,   H2SO4
Ans. Formal charge on an atom in a Lewis structure
= [total number of valence electrons in free atom] – [total number of non-bonding (lone pairs) electrons]—1/2 [total number of bonding or shared electrons]
(i) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
F.C. on O (1) → 6 - 4 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = 0
F.C. on O (2) → 6 - 4 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = 0

F.C. on O (3) → 6 - 6 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = -1
F.C. on N → 5 - 0 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev= +1
F.C. on H  → 1 - 0 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev=0
(ii) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
F.C. on O (1) → 6 - 6 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = - 1
F.C. on O (2) → 6 - 4 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = 0
F.C. on N → 5 - 2 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = 0
(iii)NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
F.C. on O (1) and (4) → 6 - 4 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev= 0
F.C. on O (2) and (3) → 6 - 4 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = 0
F.C. on H → 1 - 0 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev= 0
F.C. on H → 6 - 0 - NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev = 0

Q.39. The energy of σ2pz 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, N+2 , N2 , N2+2
Ans. Sequence of energy levels
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
For N2 molecule the M.O. configuration is:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
B.O. = 1/2(10 - 4) = 3, diamagnetic
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
B.O. = 1/2(9 -4) = 2.5, paramagnetic
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
B.O. = 1/2(10 -5) = 2.5, paramagnetic
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
B.O. = 1/2(8 -4) = 2.5, diamagnetic
Stability order: N2 > N-2 > N+2 > N2+2
(N-2 has more bonding electrons as compared to N+2).

Q.40. What is the effect of the following processes on the bond order in N2 and O2?
(i) N2 → N+2 + e
(ii) O2 → O+2 + e

Ans.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.41. Give reasons for the following :
(i) Covalent bonds are directional bonds while ionic bonds are nondirectional.
(ii) Water molecule has bent structure whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.
Ans. (i) Since the covalent bond depends on the overlapping of orbitals between different orbitals, the geometry of the molecule is different. The orientation of overlap is different. The orientation of overlap is the factor responsible for their directional nature.
(ii) Due to presence of two lone pairs of electrons on oxygen atom in HiO the repulsion between Ip-lp is more. C02 undergoes sp hybridization resulting in linear shape (O = C = O) while H20 undergoes sp3 hybridisation resulting in distorted tetrahedral or bent structure.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(iii) In ethyne molecule carbon undergoes sp hybridization with two unhybridised orbitals. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C – C sigma bond while the other hybridized orbital of each carbon atom overlaps axially with S orbitals of hydrogen atoms forming σ bonds. Unhybridised orbitals form π bonds
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.42. What is an ionic bond? With two suitable examples explain the difference between an ionic and a covalent bond?
Ans. An ionic bond is formed as a result of the electrostatic attraction between the positive and negative ions formed by transfer of electrons from one atom to another.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Covalent bond is formed by sharing of electrons between atoms. Examples of covalent bonds:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevMultiple bonds are formed during covalent bonding while in ionic bond a number of bonds can be formed between different atoms.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.43. Arrange the following bonds in order of increasing ionic character giving reason.
N—H,  F—H,  C—H and O—H
Ans. Electronegativity difference
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Increasing order of electronegativity difference: C-H<N-H<0-H<F-H
Greater is the difference in electronegativity between two bonded atoms, greater is the ionic character.

Q.44. Explain why CO2–3 ion cannot be represented by a single Lewis structure. How can it be best represented?
Ans. A single Lewis structure of CO2-3  ion cannot explain all the properties of this ion. It can be represented as a resonance hybrid of the following structures:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
If it were represented only by one structure, there should be two types of bonds, i.e., C = O double bond and C – O single bonds but actually all bonds are found to be identical with same bond length and same bond strength.

Q.45. Predict the hybridisation of each carbon in the molecule of organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Ans.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Number of σ Bonds = 11
Number of π Bonds = 4

Q.46. Group the following as linear and non-linear molecules :
H2O, HOCl, BeCl2, Cl2O
Ans.
 NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.47. Elements X, Y and Z have 4, 5 and 7 valence electrons respectively. (i) Write the molecular formula of the compounds formed by these elements individually with hydrogen. (ii) Which of these compounds will have the highest dipole moment?
Ans. (i)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(ii) Z will be most electronegative element and hence HZ will have highest dipole moment.

Q.48. Draw the resonating structure of
(i) Ozone molecule
(ii) Nitrate ion
Ans. Resonating structures:
(i) Ozone molecule
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(ii) Nitrate ion
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.49. Predict the shapes of the following molecules on the basis of hybridisation.
BCl3, CH4, CO2, NH
3
Ans. BCl3 – sp2 hybridisation – Trigonal planar
CH4 – sp3 hybridisation – Tetrahedral .
NH3 – sp3 hybridisation – Distorted tetrahedral or Pyramidal

Q.50. All the C—O bonds in carbonate ion (CO2–3 ) are equal in length. Explain.
Ans.  Carbonate ion is represented by resonating structures as given below:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevActual structure will be the mixture of all resonating structures, also called resonance hybrid structure.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevQ.51. What is meant by the term average bond enthalpy? Why is there difference in bond enthalpy of O—H bond in ethanol (C2H5OH) and water?
Ans. All the similar bonds in a molecule do not have the same bond enthalpies, e.g., in CHNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevthe four C - H bonds do not have the same bond enthalpies because after breaking of the bonds one by one, the electronic environment around carbon changes. Hence, the actual bond enthalpy of C - H bond is taken as the average value.
O - H bond in ethanol NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevand that in water NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevdo not have similar electronic environment around oxygen atom. Hence, their O - H bond enthalpies are different.


MATCHING TYPE

Q.52.Match the species in Column I with the type of hybrid orbitals in Column II.

 Column I Column II
 (a) SF4 (a) sp3d2
 (b) IF5 (b) d2sp3
 (c) NO+2 (c) sp3d
 (d) NH+4 (d) sp3

 (e) sp

Ans. (i) → c; (ii) → a; (iii) → e; (iv) → d
(i) SF4 = number of bp (4) + number of lp (1) = sp3d hybridisation
(ii) IF5 = number of bp (5) + number of lp (1) = sp3d2 hybridisation
(iii) NO+= number of bp (2) + number of lp (0) = sp hybridisation
(iv) NH+4 = number of bp (4) + number of Ip (0) = sp3 hybridisation

Q.53. Match the species in Column I with the geometry/shape in Column II.

 Column I
 Column II
 (i) H3O+ (a) Linear
 (ii) HC ≡ CH (b) Angular
 (iii) ClO-2 (c) Tetrahedral
 (iv) NH+4 (d) Trigonal bipyramidal

 (e) Pyramidal

Ans. (i) → e; (ii) → a; (iii) → b; (iv) → c
(i) H3O+ → (e) Pyramidal
(ii) HC ≡ CH → (a) Linear
(iii) ClO-→ (b) Angular
(iv) NH+4 → (c) Tetrahedral


Q.54. Match the species in Column I with the bond order in Column II.

 Column I Column II
 (i) NO (a) 1.5
 (ii) CO (b) 2.0
 (iii) O-2 (c) 2.5
 (iv) O2 (d) 3.0

Ans. (i) → c; (ii) → d; (iii) → a; (iv) → b
NO → B.O. = 1/2(10 - 5) = 2.5
CO → B.O. = 1/2(10 4) = 3
O- B.O. = 1/2(10-7)= 1.5
O→ B.O. = 1/2(10-6) = 2

Q.55. Match the items given in Column I with examples given in Column II.

 Column I
 Column II
 (i) Hydrogen bond (a) C
 (ii) Resonance (b) LiF
 (iii) Ionic Solid (c) H2
 (iv) Covalent solid (d) HF

 (e) O3

Ans. (i) → d; (ii) → e: (iii) → b; (iv) → a
(i) H - F : hydrogen bond H - F ........... H - F
(ii) 03 : Resonance
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev(iii) LiF: Ionic solid
(iv) C: Covalent solid

Q.56. Match the shape of molecules in Column I with the type of hybridisation in Column II.

 Column I
 Column II
 (i) Tetrahedral (a) sp2
 (ii) Trigonal (b) sp
 (iii) Linear (c) sp3

Ans. (i) →c; (ii) → a; (iii) → b
sp3 hybridisation – Tetrahedralshape
sp2 hybridisation – Trigonal shape
sp hybridization – Linear shape


ASSERTION AND REASON TYPE

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q.57. Assertion (A) : Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R) : This is because sodium and chloride ions acquire octet in sodium chloride formation.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.

Ans. (i)
Solution.
Sodium chloride (Na+CL) is stable ionic compound because both Na+ and CL ions have complete octet in outermost shell.

Q.58. Assertion (A) : Though the central atom of both NH3 and H2O molecules are sp3 hybridised, yet H–N–H bond angle is greater than that of H–O–H.
Reason (R) : This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.

Ans. (i)
Solution.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
H20 has two lone pairs while NH3 has single lone pair, hence, H20 involves greater lone pair-bond pair repulsion.

Q.59. Assertion (A): Among the two O–H bonds in H2O molecule, the energy required to break the first O–H bond and the other O–H bond is the same.
Reason (R) : This is because the electronic environment around oxygen is the same even after breakage of one O–H bond.
(i) A and R both are correct, and R is correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.

(iv) A and R both are false.
Ans. (iv)
Solution.
Bond energy of two (-O – H) bonds in H20 will be different.


LONG ANSWER TYPE QUESTIONS

Q.60. (i) Discuss the significance/ applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in CO2 , NF3 and CHCl3.
Ans. (i) Dipole moment plays very important role in understanding the nature of chemical bond. A few applications are given below:
(a) Distinction between, polar and non-polar molecules. The measurement of dipole moment can help us to distinguish between polar and non-polar molecules. Non-polar molecules have zero dipole moment while polar molecules have some value of dipole moment.
(b) Degree of polarity in a molecule. Dipole moment measurement also gives an idea about the degree of polarity specially in a diatomic molecule. The greater the dipole moment, the greater is the polarity in such a molecule.
(c) Shape of molecules. In case of molecules containing more than two atoms, the dipole moment not only depends upon the individual dipole moments of the bonds but also on the arrangement of bonds. Thus, dipole moment is used to find the shapes of molecules.
(d) Ionic character in a molecule. Knowing the electronegativities of atoms involved in a molecule, it is possible to predict the nature of chemical bond formed. If the difference in electronegativities of two atoms is large, the bond will be highly polar. As an extreme case, when the electron is completely transferred from one atom to another, an ionic bond is formed. Therefore, the ionic bond is regarded as an extreme case of covalent bond. The greater the difference in electronegativities of the bonded atoms, the higher is the ionic character
(e) Distinguish between cis- and trans- isomers. Dipole moment measurements help to distinguish between cis- and trans- isomers because ds-isomer has usually higher dipole moment than trans isomer.
(f) Distinguish between ortho, meta and para isomers. Dipole moment measurements help to distinguish between o-, m- and p-isomers because the dipole moment of p-isomer is zero and that of o-isomers is more than that of m-isomer.
(ii)
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.61. Use t the molecular orbital energy level diagram to show that N2 would be expected to have a triple bond, F2, a single bond and Ne2, no bond.
Ans. Nitrogen molecule (N2): Electron configuration of N-atom (Z = 7) is Is2 2s2 2p1x 2p1y 2p1z. Total number o f electrons present in N2 molecule is 14, 7 from each N-atom. The electron configuration of N2 molecule will be
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevBond = 1/2 (10 - 4) = 3 (triple bond)
Fluorine molecule (F2): Electronic configuration of F-atom (Z = 9) is 1s2s2 2p2x 2p22p1z. Total number of electron in F2 molecule is 18, 9 from each F-atom. Thus, it has two electrons more than 02 molecule. These two additional electrons will enter into π*(2px) and π*(2py) orbitals, one in each of them so that pairing of electrons takes place in these orbitals. Thus, the electronic configuration of F2 will be
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevBond Order = 1/2 (10 - 8) = 1 (single bond)
Ne2:
Total number of electrons in Ne2 = 20.
The electron configuration of N2 molecule will be
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNo bond is formed between two Ne atoms or in other words, Ne2 does not exist.
Bond Order = 1/2(10 – 10) = 0 (No bond)

Q.62. Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Ans. The valence bond theory was put forward by Heitler and London in 1927. It was later improved and developed by L. Pauling and J.C. Slater in 1931. The valence bond theory is based on the knowledge of atomic orbitals and electronic configurations of elements, overlap criteria of atomic orbitals and stability of molecule.
The main points of valence bond theory are
(i) Atoms do not lose their identity even after the formation of the molecule.
(ii) The bond is formed due to the interaction of only the valence electrons as the two atoms come close to each other. The inner electrons do not participate in the bond formation.
(iii) During the formation of bond, only the valence electrons from each bonded atom lose their identity. The other electrons remain unaffected.
(iv) The stability of bond is accounted by the fact that the formation of bond is accompanied by release of energy. The molecule has minimum energy at a certain distance between the atoms known as intemuclear distance. Larger the decrease in energy, stronger will be the bond formed.
Valence bond Treatment of Hydrogen Molecule:
Consider two hydrogen atoms A and B approaching each other having nuclei Ha and HB and the corresponding electrons eA and eB respectively.
When atoms come closer to form molecules new forces begin to operate.
(a) The force of attraction between nucleus of atom and electron of another atom.
(b) The force of repulsion between two nuclei of the atom and electron of two atoms.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevFig. (a) Two hydrogen atoms at a large distance and hence, no interaction, (b) Two hydrogen atom closer to each other atomic orbitals begin to interact, (c) Attractive and repulsive forces in hydrogen atoms when interaction begins. In case of hydrogen: Figure ‘a’ shows that two hydrogen atoms are at farthest distances and their electron distribution is absolutely symmetrical.
(a) When two hydrogen atom start coming closer to each other, the electron cloud becomes distorted and new attractive and repulsive forces begin to operate as shown in figure ‘c’
(b) In figure ‘c’ dotted lines show attractive forces present in atom already and bold lines show the new attractive and repulsive forces.
(c) It has been found experimentally that the magnitude of net attractive forces is more than net repulsive forces. Thus stable hydrogen molecule is formed.
Potential energy diagram for formation of hydrogen molecules:
When two hydrogen atoms are at farther distance, there is no force operating between them, when they start coming closer to each other, force of attraction comes into play and their potential energy starts decreasing. As they come closer to each other potential goes on decreasing, but a point is reached, when potential energy acquires minimum value.
Note:
(a) This distance corresponding to this minimum energy value is called the distance of maximum possible approach, i.e. the point which corresponds to minimum energy and maximum stability.
(b) If atoms come further closer than this distance of maximum possible approach, then potential energy starts increasing and force of repulsion comes into play and molecules starts becoming unstable.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev


Q.63. Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer as compared to equatorial bonds in PCl5 whereas in SF6 both axial bonds and equatorial bonds have the same bond length. Explain.
Ans. Formation of PCI5 (sp3d hybridization): The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented below:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Three P—Cl bond lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial bonds. The remaining two P—Cl bonds—one lying above and the other lying below the equatorial plane, make an angle of 900 with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction than the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes PCl5 molecule more reactive.
Formation of SF6 (sp3d2 hybridization): Molecule has a regular octahedral geometry.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Q.64. (i) Discuss the concept of hybridisation. What are its different types in a carbon atom.
(ii) What is the type of hybridisation of carbon atoms marked with star.
(a) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(b) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(c) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(d) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
(e) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Ans. (i) Hybridisation: In order to explain the characteristic geometrical shapes of polyatomic molecules like CH4, NH3 and H2O etc., Pauling introduced the concept of hybridisation. According to him the atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals. Unlike pure orbitals, the hybrid orbitals are used in bond formation. The phenomenon is known as hybridisation which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.
For example, when one 2s and three 2p-orbitals of carbon hybridise, there is the formation of four new sp3 hybrid orbitals. The main features of hybridisation are as under:
1. The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.
2. The hybridised orbitals are always equivalent in energy and shape.
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
4. These hybrid orbitals are directed in space in some preferred direction to have minimum repulsion between electron pairs and thus a stable arrangement.
Therefore, the type of hybridisation indicates the geometry of the molecules.
Different types of hybridisation in carbon atom are:

  • In carbon compounds, if carbon is linked to carbon through triple bond, e.g. Alkynes, triple bonded carbon is sp hybridised.
  • In carbon compounds, if carbon is linked to carbon through double bond (C=C), e.g., Alkenes, double bonded carbon is sp2 hybridised.
  • In carbon compounds, if carbon is linked to carbon through single bond (C—C), e.g., Alkanes, single bonded carbon is sp3 hybridised.

(a) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Both the starred C atoms are sphybridised.
(b) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
The starred C atom is sp3 hybridised.
(c) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
The starred C atom is sp2 hybridised.
(d) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
The starred C atom is sp3 hybridised.
(e) NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
The starred C atom is sp2 hybridised.

Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti bonding molecular orbital (ABMO). Energy of anti bonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order:
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevand for oxygen and fluorine order of energy of molecular orbitals is given below :
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called ‘Sigma’, (σ)  and if the overlap is lateral, the molecular orbital is called ‘pi’, (π). The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.

Q.65. Which of the following statements is correct?
(i) In the formation of dioxygen from oxygen atoms 10 molecular orbitals will be formed.
(ii) All the molecular orbitals in the dioxygen will be completely filled.
(iii) Total number of bonding molecular orbitals will not be same as total number of anti bonding orbitals in dioxygen.
(iv) Number of filled bonding orbitals will be same as number of filled anti bonding orbitals.

Ans. (i)
Solution.
O2 = (8 + 8) = 16.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev

Q.66. Which of the following molecular orbitals has maximum number of nodal planes?
(i) σ*1 s
(ii) σ*2pz
(iii) π2px
(iv) π*2py
Ans. (ii)
Solution.
a*2pz has has maximum number of nodal planes.

Q.67. Which of the following pair is expected to have the same bond order?
(i) O2 , N2
(ii) O+2, N2
(iii) O2 , N+2
(iv) O2, N2
Ans. (ii) Bond order = (Nb — Na)/2
In O2, number of electrons in bonding orbitals are 10 and in antibonding are 5.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Bond order = (10-5)/2 =2.5
In N2, number of electrons in bonding orbitals are 10 and in anti- bonding are 5.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Bond order = (10- 5) / 2 = 2.5.

Q.68. In which of the following molecules, σ2pz molecular orbital is filled after π2px and π2py  molecular orbitals?
(i) O2
(ii) Ne2
(iii) N2
(iv) F2
Ans. (iii)
Solution.
NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
We have seen that 1s atomic orbitals on two atoms form two molecular orbitals designated as NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals on two atoms) give rise to the following eight molecular orbitals.
AntibondingNCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
Bonding NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
It has been observed experimentally that for molecules such as B2, C2, N2 etc, the increasing order of energies of various molecular orbitals is NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev
The important characteristic feature of this order is that the energy of NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevmolecular orbital is higher than that of NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRevand NCERT Exemplar - Chemical Bonding and Molecular Structure Notes | EduRev molecular orbitals.

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