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Best chemistry question for practice.. In chemical bonding chapter

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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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, Chemical Bonding-II
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 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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, Chemical Bonding-II
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.

Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Website : www.resonance.ac.in | E-mail :
ADVCBO-35
Toll Free : 1800 258 5555 | CIN: U80302RJ2007PLC024029

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