INORGANIC CHEMISTRY

(JEE-ADVANCED)

TOPIC : CHEMICAL BONDING DPP-4

1. Which of the following are correct statement(s).
(A) Bond angle order (X–S–X) is OSF2 < OSCl2 < OSBr2
(B) (O–O) bond length in O22– is more than O2F2
(C) dC–Cl bond length in CH3Cl > CF3Cl
(D) 4 P–P–P angles are equal to 90° in P4 molecule.

2. Choose the incorrect order of bond angle?

(A) H2S < H2O < NH3 (B) NO2 > NO2¯ > NO2
(C) OF2 < OH2 < O(CH3)2 (D) BF3 < BCl3 < BBr3

3. In compound PXn Y5–n ; X & Y are monovalent surrounding atoms and order of electronegativity
is X > Y then according to given information correct statement is / are:
(A) If n= 3 then bond angle between central atom and less electronegativity atom is greater than
120°.
(B) If n = 0, then all five bond lengths are not equal.
(C) if n =1, then axial plane can have two same substituent and equatorial plane can have three
same substituent.
(D) If n = 4 then axial plane can have three same substituent and equatorial plane can have two
same substituent.

4. Correct order regarding bond angle in the following species?

(A) NH4+ > NH3 > NH2– ( H N H bond angle)
(B) NO2+ > NO3– > NO2– ( O N O bond angle)
(C) OCl2 > OF2 > OH2 ( X O X ) (X = terminal atom)
(D) BCl3 > PCl3 > NCl3 ( Cl X Cl ) (X = central atom)

5. Consider skeleton of following two species :

H O
O z
3
1 x Cl
Cl y
2 O O O
O O
O

where 1, 2 and 3 are O – Ĉl – O bond angles and x, y and z are 'Cl – O' bond lengths then
which of the following order is incorrect.
(A) 2 > 1 (B) x > y (C) 2 < 3 (D) z > y

Download ATP STAR APP Page No. 1

And Practice For Free
6. If PF2Cl3 has zero dipole moment, the correct statement(s) is / are
(A) The two F atoms are at 180° with respect to each other.
(B) Maximum 4 atoms are present in the same plane.
(C) P–Cl bond is longer than P–F bond
(D) The d-orbital of 'P' involved in the formation of PF2Cl3 is d
z2

7. SO2 and SO3 are different in :

(A) Hybridisation (B) Compound polarity
(C) Number of p-p bond (D) Structure

8. Which of the following has non-zero dipole moment?

OH SH NH2 SH

(A) (B) (C) (D)

OH SH NF2 OH

9. Which of the following compounds are planar as well as non polar.

(A) XeF4 (B) XeF2 (C) XeF5– (D) XeF5+

10. Which of the following statements is incorrect regarding back bonding.

(A) Due to back bonding, bond angle always increases.
(B) Due to back bonding, bond length decreases.
(C) Due to back bonding, bond angle decreases.
(D) Due to back bonding hybridisation of central atom does not change always.
:

+
11. H3Si — N = N = N¯ :
(I) (II) (III)
If hybridisation of N (I) and N (II) nitrogen atoms are sp (s + px) and nodal plane of -bond present
between N(I) and N(II) lies in "xy" plane then which of the following overlapping is present in above
compound.
(A) py – dxy sideways overlapping (B) py – py sideways overlapping
(C) pz – pz sideways overlapping (D) sp3 – sp (s + px) ovrelapping

12. Which type of overlapping is / are present in H2Si NCNSiH3. (If skeleton of NCN group is present
on x-axis)
(A) py – py (B) pz – pz (C) pz – dxz (D) py – dxy

13. Correct statement(s) for B(OH)3 and B(OH)4¯ is/are

(A) Extent of back bonding : B(OH)3 > B(OH)4¯
(B) OBO: B(OH)3 > B(OH)4¯
(C) B(OH)4¯ does not form adduct with NH3
(D) Hybridization of O-atom in B(OH)4¯ : sp2

14. Which of the following is paramagentic as well as it has fractional bond?

(A) O2+ (B) B2 (C) O2 (D) N2

Download ATP STAR APP Page No. 2

And Practice For Free
15. In which of the following molecule(s) (s-p) mixing occur
(A) Li2 (B) B2 (C) C2 (D) H2

16. Which of the following species having bond order is equal to three?
(A) CO (B) N2 (C) O2 (D) S2

17. Which of the following species has center of symmetry (COS) in its HOMO
(A) B2 (B) C2 (C) F2 (D) NO

18. In O2 molecule, which of the following molecular orbitals are lower in energy than  2p orbital
z

* * *
(A)  2 px (B)  2 pz (C)  2 s (D)  2 py

19. Which of the following molecules are formed by the combination of orbitals having sp mixed
character
(A) NO (B) O2¯ (C) N2+ (D) N2¯

20. Which statement is correct

(A) Higher is the polarisation, higher will be relative solubility in non-polar solvent
(B) Higher is the polarisation, higher will be the intensity of colour
(C) Diamagnetic substances some times become coloured due to HOMO-LUMO transition
(D) Higher is the polarisation in metal oxide, higher will be the basic character

21. Choose the option(s) regarding correct order of acidic nature :

(A) MgO < ZnO < P2O5 < SO3 (B) MgO < ZnO < SO3 < P2O5
(C) Li2O < NO < CO2 < SO2 (D) Li2O < BeO < CO2 < NO

22. Polarisation may be called as the distortion of the shape of an anion by an adjacent cation. Which
of the following statements is/are not correct?
(A) Minimum polarization is brought about by a cation of low radius
(B) A large cation is likely to bring about a large degree of polarisation
(C) Maximum polarization is brought about by a cation of high charge
(D) A small anion is likely to undergo a large degree of polarisation

23. Correct order(s) of thermal stability is/are

(A) BeF2 > BaF2 (B) BaCO3 > CaCO3
(C) Cs2O2 > BaO2 (D) Be(NO3)2 > NaNO3

24. Consider the order of following metal hydroxides:

Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
Which of the given properties is following above order.
(A) Basic character (B) Thermal stability
(C) Solubility in water (D) All

25. Compound having lowest thermal stability is

(A) NaHCO3 (B) KHCO3 (C) RbHCO3 (D) CsHCO3

Download ATP STAR APP Page No. 3

And Practice For Free
 Paragraph for question nos. 26 to 27
All species tend to attain or adopt the electronic geometry in which overall repulsion is minimum.
If all the sites (positions) are equivalent (e.g. octahedral), lone pairs prefer to be trans (anti) to
each other, else lone pairs choose the largest site (e.g. equilateral position in trigonal bipyramidal).
In trigonal bipyramidal geometry, axial position (a) is more crowded than equatorial position (e)
which results in elongation of axial bond as compared to equatoraial bond, thereby increasing the
axial bond length. This holds valid if the substituents are same. Also bonding pairs to more
electronegative substituents occupy less space than those to more electropositive or less
electronegative substituents. Double bond occupy more space than single bond.

26. Choose the correct statement regarding ClF3 .

(A) Cl – Fa is shorter than Cl – Fe
(B) All Cl – F bonds are of equal length
(C) Fa – Cl – Fa is exactly equal to 180º
(D) Cl – Fa is longer than Cl – Fe

27. Consider the molecule given below,

F
 F
O S 
 F
F
Which of the following is correct order of bond angle?
(A)  >  >  (B)  >  >  (C)  >  >  (D)  >  > 

Paragraph for question nos. 28 to 30

The formal charge is the difference between the number of valence electrons in an isolated (i.e.
free) atoms and the number of electrons assigned to that atom in a Lewis structure.
For a molecule the net dipole moment is the vector addition of bond moment and lone pair moment.
28. In SnCl3– ion calculate the formal charge on Sn.
(A) +1 (B) –1 (C) zero (D) –2

29. In which of the following molecule has polar as well as non polar bond
(A) HCN (B) PCl5 (C) H2O2 (D) All are correct

30. Select the correct statement.

(A) NH3 has higher bond dipole than that of NF3
(B) CCl4 is polar molecule
(C) SF4 is polar molecule
(D) IF7 is polar molecule

Paragraph for question nos. 31 to 33

According to molecular orbital theory atomic orbitals of nearly same energy of different atoms are
combined and form molecule in homodiatomic molecule two atomic orbitals which have wave
function  A and  B are combined in two ways ( A   B ) . They have two type of electron
density, one is bonding [( A   B ) 2 ] where e– density increase between nucleus and another is

antibonding [( A   B ) 2 ] where e– density decreases between nucleus. These molecular orbitals
are filled according to Hund, Pauli, Afbau princple.

Download ATP STAR APP Page No. 4

And Practice For Free
31. Correct trend about bond energy is :
(A) N2– > N2+ (B) N2+ > N2– (C) N2+ = N2– (D) Not comparable

32. In which of the following molecular species s and p orbitals mixing does not occur?
(A) CN– (B) O22+ (C) N22– (D) B22–

33. Which of the following paramagnetic species have highest bond order with maximum number of
antibonding electrons?
(A) N2+ (B) O2+ (C) F2 (D) C2+

Paragraph for question nos. 34 to 36

The rule governing the transition from ionic to covalent bonding are called Fajan's rules. They are
based on the deformation of the interacting ions in the bond A+Br (Polarization of ions)

34. Select the correct statement.

(A) Solubility of hydroxides of group I A increases down the group while for group II A, it decreases
down the group.
(B) Solubility of hydroxides of s-block metal cations increases down the group.
(C) Solubility of carbonates of s-block metal cation decreases down the group.
(D) All are correct.

35. Select the correct statement.

(A) Solubility of bicarbonates of Na+, K+, Rb+, Cs+ decreases down the group
(B) Thermal stability of nitrates of group I A cations decreases down the group
(C) Solubility of chromates of group II A cation increases down the group
(D) Melting point of fluorides of Na+, K+, Rb+, Cs+ decreases down the group

36. In which of the following compound has highest thermal stability.

(A) NaCl (B) CsCl (C) KCl (D) LiCl

37. Match the orbital overlap figures shown in List-I with the description given in List-II and select the
correct answer using the code given below the lists.
List-I List-II

P. 1. p – d  antibonding

Q. 2. s – p  bonding

R. 3. d – d  bonding
S. 4. p – p  antibonding
Code:
P Q R S
(A) 3 1 4 2
(B) 3 1 2 4
(C) 1 3 2 4
(D) 1 3 4 2

Download ATP STAR APP Page No. 5

And Practice For Free
38. Match the ionization processes in List-I with the changes observed in List - II. For this, use the
codes given below:
List-I List-II
(P) N2  N2 + (1) bond order increases and magnetic property is changed
+
(Q) O2  O2 2+ (2) bond order decreases and magnetic property remains same.
(R) B2  B2 + (3) bond order increases and magnetic property remains same.
–
(S) NO  NO (4) bond order decreases and magnetic property is changed.
Code:
P Q R S
(A) 1 3 2 4
(B) 4 1 2 3
(C) 4 3 1 2
(D) 1 3 4 2

39. Column-I Column-II

(A) 2 lone pair (P) XeF5–
(B) Zero dipole moment (Q) NF3
(C) Planar (R) ICl3
(D) All adjacent bond angles are equal (S) XeF4

40. Column I Column II

(Number of electron pair (Probable geometry)
around the central atom)
(A) 2 (P) Bent
(B) 3 (Q) Linear
(C) 4 (R) Sea Saw
(D) 5 (S) Trigonal Pyramidal

41. Column I Column II

(Molecules) (Orbitals involved in overlapping)
(A) (CN)2 (P) sp – sp

(B) C2(CN)2 (Q) sp – p

(C) C2(CN)4 (R) p–p

(S) sp2 – sp2

42. Column-I Column-II

(A) PF3Cl2 (P) All ‘P-F’ bonds are identical (when P-F bonds are > 1)
(B) PF2Cl3 (Q) All ‘P-Cl’ bonds are identical (when P-Cl bonds are > 1)
(C) PFCl4 (R) dP–Cl(axial) > dp–Cl(equatorial)
(D) PF4Cl (S) dP–Cl(equatorial) > dP–F(axial)
(T) dP–F(axial) > dP–F(equatorial)

43. Which of the following has only  bonded molecule.

PCl5, PF5, POCl3, XeOF4, XeF2, XeF4, NCl3, IF7, SO2Cl2, SOCl2, SO2, SO3
[If your answer is 11 so write 0011.]

Download ATP STAR APP Page No. 6

And Practice For Free
44. How many compound which has p-d bonds -
CO32–, ClO4– , SO3, SO42–, P4O10, NO3–, HCN, [I(CN)2]–, PO43–

45. Find the number of non polar molecule having all polar bonds.
CCl4, PCl5, SF6, IF7, IF5, PCl4F, SF4 , NO2, BF3

46. The molecules / ions which are planar as well as polar :

SnCl2, I3+, NH2–, IF3, BF3, H2O, CO32–, I3–, SF2, XeF2, NO2–, XeF4, ICl4–, NH2+

47. Find total number of linkage having  character in F2BN(SiH3)2

48. How many total number of molecular orbitals which are known as gerade(g).
1s, 2py, 3s, *2py, *2pz, 2px, *2s, *2px

49. Which of the following is paramagnetic and has fractional bond order?
O2, O2+, O2–, C2+, B2+, H2+, He2+, N2 , N22+, NO+
[If your answer is 1 so write 0001.]

50. Which of following substance is having higher lattice energy than NaBr.
CaCl2, NaI , CsBr, LiF, MgO, Al2O3, TiO2

ANSWER KEY
1. (ABC) 2. (BD) 3. (ABCD) 4. (AB) 5. (C)
6. (ABCD) 7. (BD) 8. (ABCD) 9. (ABC) 10. (AC)
11. (ABCD) 12. (ABCD)
13. (ABC) 14. (ABC) 15. (ABC) 16. (AB) 17. (C)
18. (C) 19. (ACD) 20. (ABC) 21. (AC)
22. (ABD) 23. (ABC) 24. (D) 25. (A) 27. (D)
28. (B) 29. (C) 30. (C) 31. (B) 32. (B)
33. (B) 34. (B) 35. (D) 36. (D) 37. (B)
38. (B)
39. (A) P, R, S; (B) P, S; (C) P, R, S; (D) P, Q, R, S
40. (A) Q (B) P, Q (C) P,Q,S (D) P,Q,R
41. (A) PQR (B) PQR (C) QRS
42. (A)  QST; (B)  PQS; (C)  RS; (D)  S,T
43. 0006 44. 5 45. 0005 46. 8
47. 5 48. 4 49. 0006 50. 5

Download ATP STAR APP Page No. 7

And Practice For Free
 SOLUTION

1. P 60° P

12 P–P–P=60°

2. Bond angle of NO2+ > NO2 > NO2–

Bond angle of BF3– = BCl3 = BBr3 = BI3 Regular Geometry
3. Suppose
X & Y are F and Cl respectively
PXnY5–n
F Cl
2
F P 1
If n = 3 PF3Cl2 1 > 2
Cl
F
If n = 0 PCl5
Due to the presence of different bond angle it having different bond length.

4.
Cl
O
Cl
> H
O
H
> F
O
F
BCl3 > NCl3 > PCl3
sp 2 sp 3 sp 3

Fa Cl
e

Cle—P
6. For zero dipole moment, structure of PCl3F2 would be
Fa Cle

In sp3d, d is involved. As size of Cl > size of F, P–Cl bond is longer than P–F bond.
z2
O

7. S S sp2 hybridisation
O O O O
both change no. of p-p bond = 1

8. Due to non-linear structure of – OH, –SH groups. When groups attached to benzene ring are not
identical, dipole moment can not be zero.

10. (i) Due to back bonding bond angle does not always increases. If the empty orbital of central
atom participates in back bonding, then bond angle remains same provided all substituent
must be same.
(ii) Due to back bonding double bond character increases, hence angle increases (except in
symmetrical molecules having same substituents where angle remains same eg. BF3,
BeCl2)]
Download ATP STAR APP Page No. 8
And Practice For Free
14. (A) O2+  2.5  Paramagnetic
(B) B2  1  Paramagnetic
(C) O2  2  Diamagnetic
(D) N2  3  Diamagnetic

16. Apply molecular-orbital theory

20. Due to higher polarization of metal oxide, character shifts from basic towards acidic via amphoteric
stage.]
ch arg e
22. Polarizing power of cation 
size
Polarizibility of anion  charge  size

F
:

Cl F
26-27. (i)
:
F
In case of TBP
axial B.L. > Equatorial B.L.
 Cl–Fa > Cl – Fe
F
 F
O S 
(ii) F

F
 < 90° [(S = O) and (S – Fa) is more than (S – Fa) and (S– Fa) repulsion.]
In case of equatorial bond
OSF    120 ( double bond- single bond repulsion is greater than single - single bond
repulsioin)
  < 120° but > 90°
 >  > 

28. Sn (B) – 1 FC = 4

Cl Cl Cl

Cl H
polar
Cl
O
29. H – C  N Cl P all are polar bonds nonpolar
O
Cl H
Cl

Download ATP STAR APP Page No. 9

And Practice For Free
30. NH3 has more dipole moment than NF3 but bond moment (bond dipole) of NF3 i.e.N–F is more than

F
F

N–H S polar molecule.

F
F
34.
(A) Solubilities of hydroxides of A.M. and A.E.M. increase down ward. because in both groups
L.E. of hydroxides decrease at faster rate then their hydration energies.
(C)  Solubilities of carbonates of A.M. increase downward while those of A.E. Ms. decrease
downward.
Decreasing solubilities : BeCO3 > MgCO3 > CaCO3 > SrCO3 > BaCO3
of A.E.M. Carbonates : based on r+ << r¯
Increasing solubilities of A.M. carbonates
Li2CO3 < Na2CO3 < K2CO3 < Rb2CO3 < Cs2CO3 exception to r+ << r¯

35.
(A) Solubilities of bicarbonates : NaHCO3 < KHCO3 < RbHCO3 < CsHCO3 because extent of
association of HCO3– by intermolecular H.bonding decreases from NaHCO3 to CsHCO3.
(B) Thermal stability : LiNO3 < NaNO3 < KNO3 < RbNO3 < CsNO3 Thermal stability of an ionic
1
compound having polyatomic anion like NO3–  .
Polarization
(C) Solubilities of chromates of II A cations :
BeCrO4 > MgCrO4 > CaCrO4 > SrCrO4 > BaCrO4 based on r+ << r –
(D) M.P of fluorides of A.M.
NaF > KF > RbF > CsF decided by L.E.

36.
 Thermal stability of an ionic compound having monoatomic anion  L.E.
Hence LiCl is the most thermally stable ...... to highest L.E. among A.M chlorides.

37. P.  Waves are in same phase so they form BMO d-d -bonding

Q.  p-d -anti bonding (waves are in opposite phase)

R.  s-p -bonding (waves are in same phase)

S.  p – p  antibonding (waves are in opposite phase)

Download ATP STAR APP Page No. 10

And Practice For Free
 F F N
39. (i) F Xe¯ (ii) F F
F F F
NF3
XeF5¯

Cl
F F
< 90°
(iii) Cl I (iv) Xe
F F
Cl
ICl3 XeF4

43. O is divalent atom, so it bonded by double bond.

only  bonded molecules are PCl5, PF5, XeF2, XeF4, NCl3, IF7

44. Molecules Hybridisation -bonds

CO32– sp 2 p-p
ClO4– sp 3 p-d
SO 3 sp 2 p-p, p-d
SO42– sp 3 p-d
P4O10 sp 3 p-d
NO3– sp 2 p-p
HCN sp p-p
[I(CN)2]– C has sp p-p
PO43– sp 3 p-d

45. geometry Non-polar compund

Cl

C
µ = 0 , having all polar bonds
Cl Cl
Cl

Cl Cl

Cl P
µ = 0 , having all polar bonds
Cl
Cl

Download ATP STAR APP Page No. 11

And Practice For Free
 F
F F

S
µ = 0 , having all polar bonds
F F
F

F F

F F

µ = 0 , having all polar bonds

F F
F

B F µ = 0 , having all polar bonds

46. SnCl2, I3+, NH2–, IF3, H2O, SF2, NO2–, NH2+

48. Molecular orbitals Nodel plane g/Ug

1s 0 g
2p y 1 Ug
3s 0 g
*2p y 2 g
*2pz 1 Ug
2p x 1 Ug
*2s 1 Ug
*2p x 2 g

49. Species Bond order Magnetic behaviour

O2, 2 Paramagnetic
O2 + , 2.5 Paramagnetic 
O2–, 1.5 Paramagnetic 

C2+, 1.5 Paramagnetic  and having fractional bond order
B 2+ , 0.5 Paramagnetic 
H2+, 0.5 Paramagnetic 

He2+, 0.5 Paramagnetic 

N2 , 3 Diamagnetic
N22+, 2 Diamagnetic
NO+ 3 Diamagnetic

Download ATP STAR APP Page No. 12

And Practice For Free