Test: 35 Year JEE Previous Year Questions: Chemical Bonding & Molecular Structure

Both NH₃​ and (BF4​)^-1 have same hybridisation, sp³, we know in NH₃​, there is a lone pair, which makes the bond angle 109°28′. 

(BF4​)^-1 and NH₃​ has pyramidal structure. So they have same bond angle.

Bond order = 

Bond order = 

Bond order = (8 – 5) =  1.5

Bond order = (8 –6 ) =  1

NOTE : As we know that as the bond order decreases, stability also decreases and hence the bond strength also decreases. Hence the correct order of their increasing bond strength is

There is inter-molecular H binding in alcohols, thus giving them higher boiling point and lower volatility.

Hence C is correct.

Both NO₂ and O₃ have angular shape and hence will have net dipole moment.

In H₂S, due to low electronegativity of sulphur the L.P. - L. P. repulsion is more than B. P. - B. P. repulsion and hence the bond angle is minimum.

                      SO₂    H₂O   H₂S    NH₃
Bond angle 119.5° 104.5° 92.5° 106.5°

Both XeF₂ and CO₂ have a linear structure.
F — Xe — F O = C = O

The order of bond angles

Now sin ce bond order of NO⁺ given (3) is higher than that of NO (2.5). Thus bond length of NO⁺ will be shorter.

 XeF₄ (sp³d² square planar),
[ Ni(CN)₄ ]^2- (dsp² square planar ),  
BF₄^- (sp³ tetrahedral), SF₄ (sp³d see saw shaped)

The value of lattice energy depends on the charges present on the two ions and the distance between them.

The distribution of electrons in MOs is as follows :

N₂⁺(electrons 13) 

O₂ (electrons 16)

O₂^2– (electrons 18) 

B₂ (electrons 10)  

Only O₂^2– does not contain any unpaired electron.

In SF₄ the hybr idisation is sp³d and the shape of molecule is

The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.

NOTE : This can also be explained by the fact that as the size of central atom increases sp³ hybrid orbital becomes more distinct with increasing size of central atom i.e. pure p- orbitals are utilized in M–H bonding

Diamagnetic species have no unpaired electrons

Whereas paramagnetic species has one or more unpaired electrons as in

   unpaired electrons

unpaired electron

 unpaired electron

Smaller the size and higher the charge more will be polarising power of cation. Since the order of the size of cati on is K⁺ > Ca⁺⁺ > Mg ⁺⁺> Be⁺⁺ . So the correct order of polarising power is K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺

(a)N₂ : bond order 3, paramagnetic
N₂^– : bond order, 2.5, paramagnetic
(b) C₂ : bond order 2, diamagnetic
C₂⁺ : bond order 1.5, paramagnetic
(c) NO : bond order 2.5, paramagnetic
NO⁺ : bond order 3, diamagnetic
(d) O₂ : bond order 2, paramagnetic
O₂⁺ : bond order 2.5, paramagnetic

NOTE : Greater the difference between electronegativity of bonded atoms, stronger will be bond. Since F is most electronegative hence F – H ...... F is the strongest bond.

For any species to have same bond order we can expect them to have same number of electrons. Calculating the number of electrons in various species.
O₂^– (8 + 8 +1= 17) ; CN- (6 + 7 +1= 14)
NO⁺(7 + 8 – 1= 14); CN+ (6 + 7 –1 = 12)
We find CN^– and NO⁺ both have 14 electrons so they have same bond order. Correct answer is (a).

NOTE : The delocalised pπ -pπ bonding between filled p-orbital of F and vacant p-orbital of B leads to shortening of B–F bond length which results in higher bond dissociation energy of the B–F bond.

Bond order

Bond order in

Bond order in

Bond order in

Bond order in

Since Bond order

∴  Bond length is shortest in O₂²⁺ .

The proportion of covalent character in an ionic bond is decided by polarisability of the metal cation as well as the electrongativity of both elements involved in bonding. Polarisability is further decided by the density of positive charge on the metal cation. AlCl₃ is considered to show maximum covalent character among the given compounds. This is because Al³⁺ bears 3 unit of positive charge and shows strong tendency to distort the electron cloud, thus the covalent character in Al-Cl bond dramatically increases.

The formula to find the hybridisation of central atom is
Z =   [Number of valence electrons on central atom +No. of monovalent  atom altached to it + negative charge if any – positive charge if any]
For NO₃ – Z 

For 

For 

Pentagonal bipyramidal shape.

Compounds involved in chelation become non-polar.
Consequently such compounds are soluble in nonpolar solvents like ether, benzene etc. and are only sparingly soluble in water whereas meta and para isomers are more soluble in water & less soluble in non-polar solvents.

intra-molecular H-bonding

PF₅ trigonal bipyramidal

BrF₅ square pyramidal (distorted)

The molecular orbital structures of C₂ and N₂ are

both N₂ and C₂ have paired electrons in its molecular orbital hence are diamagnetic.

(a) 

(b) The central atom is sp² hybridized with one lone pair.

(c) It is a pale blue gase. At – 249.7°, it forms violet black crystals.

(d) It is diamagnic in nature due to presence of paired electrons.

bond order for

bond order for He₂ =

so both H₂²⁺ and He₂ does not exist

∴ Bond order = 

The bond order of Li₂⁺ and Li₂^– is same but Li₂⁺ is more stable than Li₂^– because Li₂⁺ is smaller in size and has 2 electrons in Anti bonding orbital whereas Li₂^– has 3 electrons in Anti bonding orbital. hence Li₂⁺ is more stable than Li₂^–

In both the molecules the bond moments are not canceling with each other and hence the molecules has a resultant dipole and hence the molecule is polar.

Hybridization (H) = [no. of valence electrons of central atom + no. of Monovalent atoms attached to it + (–ve charge if any) – (+ve charge if any)]

 = i.e. sp hybridisation

 i.e. sp² hybridisation

 i.e. sp² hybridisation

The lewis structure of NO₂ shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridization will be sp²