Chemical Bonding-II

 Marked questions are recommended for Revision.

PART - I : SUBJECTIVE QUESTIONS

Section (A) : VSEPR theory
A-1. Why NO2+ and I3– are linear species ?
A-2. PCl5 has the shape of a trigonal bipyramidal where as IF5 has the shape of square pyramidal. Explain.
A-3. Write the geometry of XeF4 and OSF4 using VSEPR theory and clearly indicate the position of lone pair
of electrons.
A-4. Explain the structure of CIF3 on the basis of VSEPR theory.

Section (B) : Hybridisation

B-1. Explain hybridisation of central atom in :
(1) XeF2 (2) XeF4 (3) PCl3 (4) PCl5 (g)
(5) SF6 (6) IF3 (7) IF5 (8) IF7
(9) CH4 (10) CCl4 (11) SiCl4 (12) SiH4
(13) H2O (14) NH3 (15) PO43– (16) BrF5
(17) NO3– (18) CO32– (19) NH4+ (20) ClO3–
B-2. The order of size of the hybrid orbitals is as follows sp < sp2 < sp3. Explain.
B-3. Draw the structure of the following compounds. Clearly indicate the number of bond pairs and lone
pairs involved on central atom. Write (i) number of bond pairs and lone pairs on the central atom (ii) the
shape of the molecules (iii) hybridization of the central atom.
(a) SF4 (b) XeOF4
Section (C) : Bond angle, bond length comparison
C-1. Draw an electron dot structure for Br3–. Deduce an approximate value of the bond angle.
C-2. Which compound has the smallest bond angle in each series ?
(a) SbCl3 SbBr3 SbI3
(b) PI3 AsI3 SbI3
C-3. Compare the C–H bond strength in C2H6, C2H4 and C2H2.
C-4. The POCl3 molecule has the shape of an irregular tetrahedron with the P atom located centrally. The
Cl–P–Cl angle is found to be 103.5º. Give a qualitative explanation for the deviation of this structure
from a regular tetrahedron.
C-5. Which one has highest and least bond angle in the following ?
(1) CH4 PH3 AsH3 SbH3 (2) H2O H2S H2Te CO2
(3) PH3 H2O (4) Cl2O ClO2
(5) PF3 PH3 (6) BF3 NF3
(7) NH3 NF3 (8) PF3 PCl3
C-6. Write the Increasing order of Bond length of each :
(1) C–C, C=C, CC (2) C–N, C–O, C–F (3) H–Cl, H–Br, H–I, HF

Section (D) : Multicentered species

D-1. Find number of p–d bonds in
(a) Disulphate (b) triphosphate (c) trimetaphosphate
(d) trimer of SO3 (e) P4O10 (f) P4O6
D-2. In which of the following compounds, the p–d bonding take place ?
(a) P4O10 (b) HNO3 (c) N2O5 (d) HClO3

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D-3. Calculate individual and average oxidation number (if required) of the marked element and also draw
the structure of the following compounds or molecules.
(1) Na2 S2 O3 (2) Na2 S4O6 (3) H2SO5 (4) H2 S2O8 (5) H2S2O7 (6) S8
(7) HNO4 (8) C3O2 (9) OsO4 (10) PH3 (11) CrO42– (12) Cr2O72–
(13) Cr O2Cl2 (14) CrO5 (15) Na2 H PO4 (16) FeS2 (17) C6H12O6 (18) NH4 NO3

PART - II : ONLY ONE OPTION CORRECT TYPE

Section (A) : VSEPR theory
A-1. Which is the right structure of XeF4 ?
F
:

:
F F F F : F
F F
Xe Xe
(A) (B) Xe (C) Xe (D)
F F

:
F F F
F : F

:
A-2. Identify the correct match.
(i) XeF2 (a) Central atom has sp3 hybridisation and bent geometry.
(ii) N3– (b) Central atom has sp3d2 hybridisation and octahedral.
(iii) PCl6–(PCl5 (s) anion) (c) Central atom has sp hybridisation and linear geometry.
(iv) ICl2+ (I2Cl6 () cation) (d) Central atom has sp3d hybridisation and linear geometry.
(A) (i – a), (ii – b), (iii – c), (iv – d) (B) (i – d), (ii – b), (iii – d), (iv – c)
(C) (i – b), (ii – c), (iii – a), (iv – d) (D) (i – d), (ii – c), (iii – b), (iv – a)
A-3. Which of the following statement is true for IO2F2– ?
(A) The electrons are located at the corners of a trigonal bipyramidal but one of the equatorial pairs is
unshared.
(B) It has sp3d hybridisation and is T-shaped.
(C) Its structure is analogous to SF4.
(D) (A) and (C) both
A-4. Which reaction involves a change in the electron–pair geometry for the under lined element ?
(A) BF3 + F–  BF4– (B) NH3 + H+  NH4+
(C) 2SO2 + O2  2SO3 (D) H2O + H+  H3O+
A-5. In which of the following molecules number of lone paris and bond pairs on central atom are not equal ?
(A) H2O (B) I3– (C) O2F2 (D) SCl2
A-6. Which of the following species given below have shape similar to XeOF 4 ?
(A) XeO3 (B) OF4+ (C) PCl5 (D) XeF5

Section (B) : Hybridisation

4 3 2 1
B-1. The hybridization of carbon atoms in C2–C3 single bond of HC  C CH  CH2 is :
(A) sp3–sp3 (B) sp2–sp (C) sp–sp2 (D) sp3–sp
B-2. Specify the hybridisations of central atom in the following species respectively {N3–, NOCl, N2O}
(A) sp, sp2, sp (B) sp, sp, sp3 (C) sp2, sp, sp (D) sp2, sp2, sp.
B-3. In pent-3-en-1-yne the terminal carbon-atoms have following hybridisation
(A) sp & sp2 (B) sp2 & sp3 (C) sp2 & sp (D) sp & sp3
B-4. S1 : [XeF7]+ has sp3d3 hybridisation
S2 : [PCl4]+ has sp3d2 hybridisation
S3 : [SF6] has sp3d2 hybridisation
S4 : [PF4]+ has sp3 hybridisation
(A) T F F T (B) T T F T (C) T F T T (D) F F F T

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B-5. BF3 + F–  BF4–
What is the hybridiation state of B in BF3 and BF4– :
(A) sp2, sp3 (B) sp3, sp3 (C) sp2, sp2 (D) sp3, sp3d

Section (C) : Bond angle, bond length comparison

C-1. The ONO angle is maximum in :
(A) HNO3 (B) NO2+ (C) HNO2 (D) NO2
C-2. Which statement is correct for N3– ion.
(A) It is bent molecule (B) Bond angle is < 120°
(C) Central atom is sp2 hybridized (D) None of these
C-3. Consider the following molecules ; H2O H2S H2Se H2 Te
   V
Arrange these molecules in increasing order of bond angles.
(A) V (B) V (C) V (D) V
C-4. In which of the following bond angle is maximum
(A) NH3 (B) NH4+ (C) PCl3 (D) SCl2
C-5. In which of the following central atom is unhybridised?
(A) S(CH3)2 (B) SO2 (C) SiH4 (D) PCl3

Section (D) : Multicentered species

D-1. The no. of S-O-S bonds in the trimer of SO3 is
(A) 1 (B) 2 (C) 3 (D) None
D-2. Which of the following species do not contain S–S linkage?
(A) H2S2O5 (B) H2S2O7 (C) H2S2O3 (D) H2S4O6
D-3. Which statement is incorrect about pyrosilicate ion.
(A) sp3 hybridisation
(B) One oxygen atom is shared between two tetrahydron
(C) there are eight Si–O bond
(D) There is one Si-Si bond
D-4. Which is correct about the cyclic silicate [Si 6O18]n– :
(A) The value of n is 12
(B) each Si atom is bonded with three oxygen atoms
(C) each oxygen atom is bonded with two Si atoms
(D) all the above are correct.

PART - III : MATCH THE COLUMN

1. Match the following :
Column–I Column–II
(A) SF2 (p) sp3 and bent
(B) KrF4 (q) two lone pairs on central atom
(C) NOCl (r) bond angle < 109º28’
(D) NF3 (s) sp2 and bent
(t) sp3d2 and square planar
2. Match the compounds listed in column-I with characteristic(s) listed in column-II.
Column – I Column –II
(A) ClF2–, CIF2+ (p) Square pyramidal.
(B) IO2F2–, F2SeO (q) See–saw and pyramidal shaped respectively.
(C) IOF4–, XeOF2 (r) Linear and bent shaped respectively.
(D BrF5, XeOF4 (s) Square pyramidal and T-shaped respectively.
(t) Both sp3d2.

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 Chemical Bonding-II
3. Match the following :
Column-I Column-II
(A) H3P3O9 (p) S–O–S bond is present
(B) H2S2O7 (q) Di-basic acid
(C) H2S4O6 (r) P–O–P bond is present
(D) H4P2O5 (s) Central atom (S or P) in maximum oxidation state.

 Marked Questions may have for Revision Questions.

PART - I : ONLY ONE OPTION CORRECT TYPE

1. The hybridisation of P in phosphate ion (PO43–) is the same as :
(A)  in Cl4– (B) S in SO3 (C) N in NO3– (D) S in SO32–
2. Choose the molecules in which hybridisation occurs in the ground state ?
(a) BCl3 (b) NH3 (c) PCl3 (d) BeF2
The correct answer is -
(A) a, b, d (B) a, b, c (C) b, c (D) c, d
3. The bent or V–shape of the molecule can be resulted from which of the following hybridization.
(A) sp3 (B) sp2 (C) Both (A) and (B) (D) None of these
4. sp3d hybridization is considered to be a combination of two hybridization. They are
(A) p3 + sd (B) sp2 + pd (C) spd + p2 (D) none of these
5. If the equatorial plane is x–y plane in sp3d hybridisation then the orbital used in pd hybridisation are -
(A) pz and dz2 (B) px and dxy (C) py and dyz (D) none of these

6. A -bonded molecule MX3 is T-shaped. The number of lone pairs of electrons can be
(A) 0 (B) 2 (C) 1 (D) none of these
7. Which of the following should have pyramidal shape :
(A) [ClOF2]+ (B) ICl3 (C) [BrCl]– (D) All of these
8. Which of the following molecules has two lone pairs and bond angle (need not be all bond angles) <
109.5°?
(A) SF2 (B) KrF4 (C) Cl4– (D) All of these
9. The correct order of bond angle is :
(A) H2S < NH3 < BF3 < CH4 (B) NH3 < H2S < CH4 < BF3
(C) H2S < NH3 < CH4 < BF3 (D) H2S < CH4 < NH3 < BF3
10. In which of the following molecules are all the bonds not equal?
(A) NF3 (B) ClF3 (C) BF3 (D) AlF3
11. Which of the following is correct order of bond length ?
(A) BF4– < BF3 (B) NO2+ < NO2– (C) CCl4 < CF4 (D) +CH3 > CH4
12. Identify the correct statement :
(A) single N–N bond is stronger than single P–P bond
(B) single N–N bond is weaker than single P–P bond
(C) N N is weaker than PP
(D) None of these
13. In which of the following species peroxide group is not present :
(A) [B4O5(OH)4]2– (B) [S2O8]2– (C) CrO5 (D) HNO4

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 Chemical Bonding-II
14. Which of the following is correct ?
(A) S3O9 : contains no S–S linkage.
(B) S2O62– : contains –O–O– linkage.
(C) (HPO3)3 : contains P–P linkage
(D) S2O82– : contains S–S linkage
15. The percentage of s–character in the orbital forming P–S bonds in P4S3 is :
(A) 25 (B) 33 (C) 75 (D) 50
16. Indicate the incorrect statement :
(A) Number of hybrid orbitals formed is equal to no. of atomic orbitals involved.
(B) 2px and 2py - orbitals of carbon can be hybridized to yield two new more stable orbitals
(C) Effective hybridisation is not possible with orbitals of widely different energies
(D) The concept of hybridisation has a greater significance in the VB theory of localised orbitals than in
the MO theory.
17. In which of the following compounds B atoms are in sp2 and sp3 hybridisation states ?
(A) Borax (B) Diborane (C) Borazole (D) All

PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE

1. Find the number of planar species
(a) BF3 (b) BCl3 (c) CO32– (d) SO3
(e) NH3 (f) NCl3 (g) PCl3 (h) XeF4
2. Find the number of species having bond angle less than 109°28'.
(a) H2S (b) SO4– – (c) CCl4 (d) NH3
(e) PH3 (f) SiH4 (g) NH4+ (h) PF3
–
(i) NH2 (j) SO3 (k) H2O
3. Find out total number of  bond in following xenon compounds.
XeOF2, XeO2F4, XeO3, XeO4, XeO3F2, XeOF4, XeO2F2
4. P4O10 has two different types of P–O bonds. Find the no. of P–O bonds with shorter bond length.
5. Difference in the oxidation number of sulphur atom is in Na2S4O6 is x, that of H2S2O5 is y. Find value of
x × y is :
6. In a P4O6 molecule, the total number of P–O–P bonds is :

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

1. Which statement is correct about hybridization ?
(A) In hybridisation orbitals take part.
(B) In hybridisation electrons take part.
(C) In hybridisation fully filled, half filled or empty orbitals can take part.
(D) Hybridised orbitals only contains bond pair electron.
2. Which of the following represent the given mode of hybridisation sp2–sp2–sp–sp from left to right
(A) H2C=C=C=CH2 (B) HCC–CCH
(C) H2C=CH–CN (D) H2C=CH–CCH
3. Which is/are in linear shape ?
(A) NO2+ (B) XeF2 (C) 3– (D) 3+
4. Which is true about NH2–, NH3, NH4+ ?
(A) Hybridization of N is same. (B) No. of lone pair of electron on N are same.
(C) Molecular geometry (i.e. shape) is different. (D) Bond angle is same.
5. Which of the following molecule (s) has/have bond angle close to 90º ?
(A) NH3 (B) H2S (C) PH3 (D) ICl3

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 Chemical Bonding-II
6. Which of the following statements are true about borax :
(A) Boron atoms are present in 2 different oxidation state and it different by 1
(B) the average oxidation state of boron is same that in B2H6.
(C) Boron atoms are present in different hybridization
(D) 2 boron atoms are connected with 4 oxygen each and 2 boron atoms are connected with 3 oxygen
atom each.
7. Identify the correct statement
(A) H2S2O7 has peroxy linkage
(B) H2S2O6 has S–S linkage
(C) H2S2O8 has peroxy linkage
(D) H2SO3(Sulphurous acid) has S in +4 oxidation state

8. In which of the following compound(s) oxidation number of one central atom is/are  6 ?
(A) N2O6 (B) CrO5 (C) H3PO5 (D) H2S2O8

PART - IV : COMPREHENSION
Read the following passage carefully and answer the questions.
Comprehension # 1
VSEPR THEORY
The trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore
follows that the corners are not equivalent in ClF3 molecule. Lone pairs occupy two of the corners, and
F atoms occupy the other three corners. These different arrangements are theoretically possible, as
shown in figure.

(i) The most stable structure will be the one of lowest energy, that is the one with the minimum
repulsion between the five orbitals. The greatest repulsion occurs between two lone pairs. Lone pair
bond pair repulsions are next strongest, and bond pair-bond pair repulsions the weakest.

A rule of thumb can be theorised, that the position having maximum repulsion amongst them are
occupied at equatorial points. Therefore (3) structure is right.
(ii) Since double bond occupies more space compared to single bond therefore it will prefer equatorial
position.
(iii) More electronegative element will occupy axial position in case of trigonal bipyramidal geometry
(iv) In case of sp3d2 hybridisation lone pairs should be placed opposite to each other because all the
corners are identical.
1. Geometry (i.e. arrangement of electron pairs around central atom) of ClOF 3 is similar to the :
(A) XeF4 (B) SOCl2 (C) 3¯ (D) ClO4¯

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 Chemical Bonding-II
2. The shape of SF5– can be :
– – – –
F F F
F

••
F F F F F

••
S S S S
F F F F F F F F
•• F F F
   V
(A)  only (B)  and  only (C) V only (D) , , & 
3. Actual shape of the molecule BrF5 is similar to the molecule :
(A) PCl5 (B) XeF4 (C) PCl4+ (D) None of these
4.* Which of the following do not exist ?
(A) SH6 (B) HFO4 (C) SI6 (D) HClO3
Comprehension # 2
Answer Q.5, Q.6 and Q.7 by appropriately matching the information given in the three columns
of the following table.
Observe the three columns in which column-1 : compound, column-2 : shape while in column-3 :
Hybridisation are given.
Column-1 (Compound) Column-2 (Shape) Column-3 (Hybridisation)
(I) XeF4 (i) Tetrahedral (P) sp3
(II) ClF3 (ii) Square planar (Q) sp2
(III) SiF4 (iii) Bent (R) sp3d
(IV) CH3OCH3 (iv) T-shape (S) sp3d2
5. Which of the following combination is true for compound which have 2 lone pair of electrons on central
atom?
(A) (I), (ii), (S) (B) (I), (ii), (R) (C) (I), (i), (P) (D) (II), (iv), (Q)
6. Which combination is correct for the compound having bond angle > 109º28' ?
(A) (III), (i), (P) (B) (IV), (iii), (P) (C) (IV), (ii), (P) (D) (I), (ii), (S)
7. Which of the following is true for a planar compound ?
(A) (III), (i), (P) (B) (I), (iv), (P) (C) (II), (iv), (R) (D) (II), (i), (P)

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions may have more than one correct options.
1. The hybridization of atomic orbitals of nitrogen in NO2+, NO3– and NH4+ are : [JEE–2000(S), 1/135]
3 2 2 3
(A) sp, sp and sp respectively (B) sp, sp and sp respectively
(C) sp2, sp and sp3 respectively (D) sp2, sp3 and sp respectively

2. The number of P—O—P bonds in tricyclic metaphosphoric acid is : [JEE–2000(S), 1/135]

(A) zero (B) two (C) three (D) four

3. Draw the molecular structures of XeF2, XeF4 and XeO2F2, indicating the location of lone pair of
electrons. [JEE–2000(M), 3/135]

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4. The correct order of hybridisation of the central atom in the following species; NH 3, PCl5 and BCl3 is :
[JEE–2001(S), 1/135]
2 2 3 3 3 2 2 3 3
(A) dsp , sp , sp (B) sp , dsp , sp (C) dsp , sp , dsp (D) dsp , sp2, dsp3
2

5. The number of S–S bonds, in sulphur trioxide trimer (S3O9) is : [JEE–2001(S), 1/135]
(A) three (B) two (C) one (D) Zero

6. Which of the following are isoelectronic and isostructural ? [JEE–2003(S), 3/144]

NO3–, CO32–, ClO3–, SO3
–
(A) NO3 , CO32– (B) SO3, NO3– (C) ClO3–, CO32– (D) CO32–, SO3 .

7. Using VSEPR theory, draw the shape of PCl5 and BrF5. [JEE–2003(M), 2/144]

8. Use VSEPR model to draw the structures of OSF4 and XeF4 (indicate the lone pair(s) on central atom)
and specify their geometry. [JEE–2004(M), 2/144]

9. The percentage of p-character in the orbitals forming P–P bonds in P4 is : [JEE–2007, 3/162]
(A) 25 (B) 33 (C) 50 (D) 75

10.* The nitrogen oxide(s) that contain(s) N—N bond(s) is(are) : [JEE–2009, 4/160]
(A) N2O (B) N2O3 (C) N2O4 (D) N2O5

11. The species having pyramidal shape is : [IIT-JEE-2010, 5/163]

2–
(A) SO3 (B) BrF3 (C) SiO3 (D) OSF2

12. Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is : [JEE–2010, 3/163]

13. The shape of XeO2F2 molecule is [JEE–2012, 3/136]

(A) trigonal bipyramidal (B) square plannar
(C) tetrahedral (D) see-saw

14. The total number of lone pairs of electrons in N2O3 is : [JEE(Advanced) 2015, 4/168]

15. Among the triatomic molecules/ions, BeCl 2, N3–, N2O, NO2+, O3, SCl2, ICl2–, I3– and XeF2, the total
number of linear molecules(s)/ion(s) where the hybridization of the central atom does not have
contribution from the d-orbital(s) is [JEE(Advanced) 2015, 4/168]
[Atomic number : S = 16, Cl = 17,  = 53 and Xe = 54]

16.* The crystalline form of borax has [JEE Advanced 2016, 4/124]
(A) tetranuclear [B4O5(OH)4]2– unit
(B) all boron atoms in the same plane
(C) equal number of sp2 and sp3 hybridized boron atoms
(D) one terminal hydroxide per boron atom

17. The total number of compounds having at least one bridging oxo group among the molecules given
below is _______ . [JEE Advanced 2018, 3/120]
N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5

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PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS

1. The hybridisation of the underline atom changes in : [AIEEE-2002, 3/225]

(1) AlH3 changes to AlH4– (2) H2O changes to H3O+
+
(3) NH3 changes to NH4 (4) in all cases

2. Bond angle of 109º 28' is found in : [AIEEE-2002, 3/225]

(1) NH3 (2) H2O (3) CH3+ (4) NH4+

3. Which of the following compounds has the smallest bond angle in its molecule ? [AIEEE-2003, 3/225]
(1) SO2 (2) H2O (3) H2S (4) NH3

4. The pair of species having identical shapes for molecules of both species is : [AIEEE-2003, 3/225]
(1) CF4, SF4 (2) XeF2, CO2 (3) BF3, PCl3 (4) PF5, IF5.

5. The maximum number of 90º angles between bond pair–bond pair of electrons is observed in :
[AIEEE-2004, 3/225]
(1) dsp3 (2) sp3d (3) dsp2 (4) sp3d2

6. The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is : [AIEEE-2004, 3/225]
(1) H2S < SiH4 < NH3 < BF3 (2) NH3 < H2S < SiH4 < BF3
(3) H2S < NH3 < SiH4 < BF3 (4) H2S < NH3 < BF3 < SiH4

7. The molecular shapes of SF4, CF4 and XeF4 are : [AIEEE-2005, 3/225]
(1) the same with 2, 0 and 1 lone pairs of electrons on the central atom, respectively.
(2) the same with 1, 1 and 1 lone pair of electrons on the central atom, respectively.
(3) different with 0, 1 and 2 lone pairs of electrons on the central atom, respectively.
(4) different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively.

8. The hybridisation of orbitals of N atom in NO3–, NO2+ and NH4+ are respectively : [AIEEE-2011, 4/120]
(1) sp, sp2, sp3 (2) sp2, sp, sp3 (3) sp, sp3, sp2 (4) sp2, sp3, sp

9. The structure of IF7 is : [AIEEE-2011, 4/120]

(1) square pyramid (2) trigonal bipyramid (3) octahedral (4) pentagonal bipyramid

10. The molecule having smallest bond angle is : [AIEEE-2012, 4/120]

(1) NCl3 (2) AsCl3 (3) SbCl3 (4) PCl3

11. In which of the following pairs the two species are not isostructural ? [AIEEE-2012, 4/120]
(1) CO32– and NO3– (2) PCl4+ and SiCl4 (3) PF5 and BrF5 (4) AlF63– and SF6

12. The species in which the N atom is in a state of sp hybridization is : [JEE(Main)-2016, 4/120]
(1) NO2– (2) NO3– (3) NO2 (4) NO2

JEE(MAIN) ONLINE PROBLEMS

1. Which one of the following does not have a pyramidal shape?
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) (CH3)3N (2) (SiH3)3N (3) P(CH3)3 (4) P(SiH3)3

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 Chemical Bonding-II
2. The geometry of XeOF4 by by VSEPR theory is : [JEE(Main) 2015 Online (10-04-15), 4/120]
(1) pentagonal planar (2) octahedral (3) square pyramidal (4) trigonal bipyramidal

3. Which of the following compound has a P–P bond ? [JEE(Main) 2015 Online (11-04-15), 4/120]
(1) H4P2O5 (2) (HPO3)3 (3) H4P2O6 (4) H4P2O7

4. Choose the incorrect formula out of the four compounds for an element X below :
(1) X2O3 (2) X2Cl3 (3) X2(SO4)3 (4) XPO4

5. The group of molecules having identical shape is : [JEE(Main) 2016 Online (09-04-16), 4/120]
(1) PCl5, IF5, XeO2F2 (2) BF3, PCl3, XeO3
(3) ClF3, XeOF2, XeF3 (4) SF4, XeF4, CCl4

6. Assertion: Among the carbon allotropes, diamond is an insulator, Whereas, graphite is a good
conductor of electricity.
Reason: Hybridization of carbon in diamond and graphite are sp3 and sp2, respectively.
[JEE(Main) 2016 Online (10-04-16), 4/120]
(1) Both assertion and reason are correct, but the reason is not the correct explanation for the
assertion.
(2) Assertion is incorrect statement, but the reason is correct.
(3) Both assertion and reason are correct, and the reason is the correct explanation for the assertion.
(4) Both assertion and reason are incorrect.

7. The bond angle H–X–H is the greatest in the compound : [JEE(Main) 2016 Online (10-04-16), 4/120]
(1) NH3 (2) PH3 (3) CH4 (4) H2O

8. Identify the pair in which the geometry of the species is T-shape and square-pyramidal, respectively :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) ICl2– and ICl5 (2) IO3– and IO2F2–
(3) ClF3 and IO4– (4) XeOF2 and XeOF4

9. In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridization are
respectively : [JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 33 and 25 (2) 67 and 75 (3) 50 and 75 (4) 33 and 75

10. The decreasing order of bond angles in BF3, NH3, PF3 and I3 is :
(1) I3 > BF3 > NH3 > PF3 (2) BF3 > I3 > PF3 > NH3
(3) BF3 > NH3 > PF3 > I3 (4) I3 > NH3 > PF3 > BF3

11. The number of P–O bonds in P4O6 is : [JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 6 (2) 9 (3) 12 (4) 18

12. In XeO3F2, the number of bond pair(s), -bond(s) and lone pair(s) on Xe atom respectively are :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 5, 2, 0 (2) 4, 2, 2 (3) 5, 3, 0 (4) 4, 4, 0

13. Among the oxides of nitrogen : [JEE(Main) 2018 Online (16-04-18), 4/120]
N2O3, N2O4 and N2O5 ; the molecule(s) having nitrogen-nitrogen bond is/are :
(1) N2O3 and N2O4 (2) N2O4 and N2O5 (3) N2O3 and N2O5 (4) Only N2O5
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 Chemical Bonding-II
14. Which of the following conversions involves change in both shape and hybridisation ?
[JEE(Main) 2018 Online (16-04-18), 4/120]
(1) H2O  H3O+ (2) BF3  BF4– (3) CH4 C2H6 (4) NH3  NH4

15. The incorrect geometry is represented by : [JEE(Main) 2018 Online (16-04-18), 4/120]
(1) NF3 – trigonal planar (2) BF3 – trigonal planar
(3) AsF5 – trigonal bipyramidal (4) H2O – bent

16. The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF 4, respectively, are :
[JEE(Main) 2019 Online (10-01-19), 4/120]
3 2 3 2 3
(1) sp d and 1 (2) sp d and 2 (3) sp d and 1 (4) sp3d and 2

17. The pair that contains two P–H bonds in each of the oxoacid is:
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) H4P2O5 and H3PO3 (2) H4P2O5 and H4P2O6
(3) H3PO2 and H4P2O5 (4) H3PO3 and H3PO2

18. The element that shows greater ability to form p-p multiple bonds, is:
[JEE(Main) 2019 Online (12-01-19), 4/120]
(1) Ge (2) Sn (3) C (4) Si

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 Chemical Bonding-II

EXERCISE - 1
PART - I

A-1. 
In NO2+ the N has sp hybridisation; so it is linear O = N = O
In I3– there are 5 electron pairs around central iodine atom (3 lone pairs
and 2 bond pairs). The hybridisation of iodine is thus sp3d. To have
minimum repulsions between lp-lp and lp-bp it acquires linear shape as
shown below.

A-2. Cl
In PCl5 there are 5 electron pairs around central phosphorus atom and all are bond Cl
pairs. The hybridisation of phosphorus is thus sp3d. To have minimum repulsions P Cl
Cl
between bp-bp it acquires trigonal bipyramidal shape as shown below.
Cl 
In IF5 there are 6 electron pairs around central iodine atom. The hybridisation of iodine is thus sp3d2.
6 electron pairs contain 5 bond pairs and one lone pair so it will be square pyramidal to have minimum
repulsions between lp-bp and bp-bp.
..
F F

F F

A-3.

Square planar Trigonal bipyramidal

F F
..

87.5º
A-4. Cl F nearly 'T' shaped.
87.5º
..

F F

B-1. 1. 2. 
 sp3d2

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3. 
 sp3 4. 
 sp3d

5. 
 sp3d2 6. F3 : – 
 sp3d

7. F5 : – 
 sp3d2 8. F7 : – 
 sp3d3

9. CH4 : – 
 sp3 10. CCl4 : – 
 sp3

11. SiCl4 : – 
 sp3 12. SiH4 : – 
 sp3

..
13. H2O : – 
 sp3 14. N H3 :– 
 sp3

15. 16. BrF5 

 sp3d2

17. 
 sp2 18. 
 sp2

19. 
 sp3 20. 
 sp3

B-2. True, As the s character in hybrid orbital decrease, size of hybrid orbital increases.

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B-3. (a) SF4 4 bond pair & 1 lone pair, Hybridization = sp3d Shape : see saw

(b) XeOF4 5 bond pair & 1 lone pair, Hybridization = sp3d2 Shape : Square pyramidal

C-1. In Br3– there are 5 electron pairs around central bromine atom (3
lone pairs and 2 bond pairs). The hybridisation of bromine is thus linear
sp3d. To have minimum repulsions between lp-lp and lp-bp it
acquires linear shape as shown below.

C-2. (a) Cl, the most electronegative of the halogens in this

series, pulls shared electrons the most strongly away from
Sb, reducing electron density near Sb. The consequence
is that the lone pair exerts the strongest influence on
shape in SbCl3. 97.1º 98.2º 99º
(b) Phosphorus is the most electronegative of the central
atoms. Consequently, it exerts the strongest pull on
shared electrons, concentrating these electrons near P
and increasing bonding pair-bonding pair repulsions–
hence, the largest angle in PI3. Sb, the least
electronegative central atoms, has the opposite effect:
Shared electrons are attracted away from Sb, reducing 102º 100º 99º
repulsions between the Sb–I bonds.
The consequence is that the effect of the lone pair is greatest in SbI3, which has the smallest angle.
Atomic size arguments can also be used for these species. Larger outer atoms result in larger angles;
larger central atoms result in smallest angles.

C-3. Order of C–H bond strength is C2H2 > C2H4 > C2H6 as %s character decreases in the same order.

C-4. , double bond occupies large area and has large electron density. So there is intrinsic repulsion

between P=O and P–Cl bond pairs. To minimize this repulsion bond angle decrease from 109.5º to
103.5º.

C-5. Highest bond angle Lowest bond angle

(1) CH4 SbH3
(2) CO2 H2Te
(3) H2O PH3
(4) ClO2 Cl2O
(5) PF3 (sp3hybridisation) PH3 (nohybridisation)
(6) BF3 (sp2hybridisation) NF3 (sp3hybridisation)
(7) NH3 NF3
(8) PCl3 PF3

C-6. (1) CC < C=C < C–C (2) C–F < C–O < C–N (3) HF < H–Cl < H–Br < H–I

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 Chemical Bonding-II
D-1. (a) 4 (b) 3 (c) 3 (d) 6 (e) 4 (f) 0

D-2. (a) P4O10,

(b) and (c) nitrogen does not have empty d-orbital.

:

(d) HClO3, Cl p-d

O
OH O

D-3. (1) +2 (6, –2) (2) +5/2(5, 5, 0, 0) (3) +6 (4) +6 (+6, +6) (5) +6 (+6, +6)
(6) 0 (7) +5 (8) 4/3 (+ 2, +2, 0) (9) +8 (10) –3
(11) +6 (12) +6 (+6, +6) (13) +6 (14) +6 (15) +5
(16) +2 (17) 0 (18) –3, +5
Note: Inside the bracket ( ) answer for individual oxidation number is given.

PART - II
A-1. (C) A-2. (D) A-3. (D) A-4. (A) A-5. (B)

A-6. (D) B-1. (B) B-2. (A) B-3. (D) B-4. (C)

B-5. (A) C-1. (B) C-2. (D) C-3. (B) C-4. (B)

C-5. (A) D-1. (C) D-2. (B) D-3. (D) D-4. (A)

PART - III
1. (A – p, q, r) ; (B – q, r, t) ; (C – s) ; (D – r) 2. (A – r) ; (B – q) ; (C – s) ; (D – p, t)

3. (A – r,s); (B – p,q,s) ; (C – q); (D – q,r)

EXERCISE - 2
PART - I
1. (D) 2. (C) 3. (C) 4. (B) 5. (A)

6. (B) 7. (A) 8. (D) 9. (C) 10. (B)

11. (B) 12. (B) 13. (A) 14. (A) 15. (A)

16. (B) 17. (A)

PART - II
1. 5 (a, b, c, d, h) 2. 6 (a, d, e, h, i, k) 3. 16
4. 4 5. 10 6. 6

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PART - III
1. (AC) 2. (CD) 3. (ABC) 4. (AC) 5. (BCD)
6. (BCD) 7. (BCD) 8. (BD)

PART - IV
1. (C) 2. (D) 3. (D) 4.* (ABC) 5. (A)
6. (B) 7. (C)

EXERCISE - 3
PART - I
1. (B) 2. (C)
3. According to VSEPR theory
Number of electron pairs = 5, XeF2
Number of bond pairs = 2,
So, Number of lone pairs = 3.
(linear)
Thus XeF2 is linear with 3 lone pairs occupying 3 equatorial positions of
trigonal bipyramidal so as to minimize the repulsions.
Number of electron pairs = 6, XeF4
Number of bond pairs = 4,
So, Number of lone pairs = 2. (square planar)
Thus XeF4 is linear with 2 lone pairs occupying 2 axial positions of
octahedral pyramidal so as to minimize the repulsions.
Number of electron pairs (including super electron pairs) = 5,
XeO2F2
Number of bond pairs = 4,
So, Number of lone pairs = 1.
Thus XeO2F2 is see-saw with 1 lone pairs occupying one equatorial (see-saw)
position and two double bonds occupying other two equatorial positions
of trigonal bipyramidal so as to minimize the repulsions.

4. (B) 5. (D) 6. (A)

7. There are 5 electron pairs and all are bonds pairs in PCl5. So to have the minimum repulsions between
bond pairs it acquires trigonal bipyramidal shape. In BrF 5, there are 6 electrons pairs out of which one
lone pair and rest all are bond pairs. So to have the minimum repulsions between bond pairs and lone
pairs it acquires square pyramidal shape.

,
PCl5 (trigonal bipyramidal), BrF5 (square pyramidal)

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 Chemical Bonding-II
8.
According to VSEPR theory two lone pairs out of six electron pairs are
trans to each other to have minimum repulsion. The shape of XeF 4 is
square planar and geometry is octahedral with sp3d2 hybridisation. The
molecule looks like :

In OSF4, there are five electron pairs and all are bond pairs. So geometry
is trigonal bipyramidal. As double bond creates more repulsion than
singles bond, the double bond acquires one of equatorial position of
trigonal bipyramidal to have minimum repulsions.
The structure looks like:

9. (D) 10.* (ABC) 11. (D) 12. 0 or 8 13. (D)

14. 8 14. 4 16.* (ACD) 17. 5 or 6

PART - II

JEE(MAIN) OFFLINE PROBLEMS

1. (1) 2. (4) 3. (3) 4. (2) 5. (4)

6. (3) 7. (4) 8. (2) 9. (4) 10. (3)
11. (3) 12. (4)

JEE(MAIN) ONLINE PROBLEMS

1. (2) 2. (3) 3. (3) 4. (2) 5. (3)
6. (1) 7. (3) 8. (4) 9. (2) 10. (1)
11. (3) 12. (3) 13. (1) 14. (2) 15. (1)
16. (1) 17. (3) 18. (3)

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