31 Year NEET Previous Questions: Chemical Bonding & Molecular Structure

The bond length of  O – O in O₂ is 1.21 Å, in H₂O₂ it is 1.48 Å and in O₃ it is 1.28 Å.
∴ correct order of bond length is H₂O₂ > O₃ > O₂.

In this configuration, there are four completely filled bonding molecular orbitals and one completely filled antibonding molecular orbital. So that N_b = 8 and
N_a = 2 .

∴ Bond order =

The bond represented by dots form the 3-centred electron pair bond. The idea of three centred electron pair bond B–H–B bridges is necessitated because diborane does not have sufficient electrons to  form normal covalent bonds. It has only 12 electrons instead of 14 required to give simple ethane like structure for diborane.

We know that due to polar nature, water molecules are held together by intermolecular hydrogen bonds. The structure of ice is open with large number of vacant spaces, therfore the density of ice is less than water.

In alkynes the hybridisation is sp i.e each carbon atom undergoes sp hybridisation to form two sp-hybrid orbitals. The two 2porbitals remain unhybridised. Hybrid orbitals form one sigma and two unhybridised orbitals form π-bonds.

Hence two π bond and one sigma bond between C — C lead to cylindrical shape.

For N2; 
σ1s² < σ*1s² < σ2s² < σ*2s² < П2px = 2py⁴ <σ2pz² < П*2px = 2py < σ*2pz
For O2;
σ1s²  < σ*1s² < σ2s² < σ*2s² <σ2pz2 < П2px = 2py⁴  <П*2px = 2py² < σ*2pz  
a) In N2+,the electron is removed from bonding molecular orbital.Hence bond order decreases and  N-N bond weakens.
b) In O²⁺ , the electron is removed from anti bonding molecular orbital.Hence O-O bond order increases.
c) In O²⁺, the no of unpaired electrons decreases and hence paramagnetism decreases.
d) In N²⁺, there is one unpaired electron.It is paramagnetic. This statement is wrong.

The electronic configuration of As is

sp³ d hybridisation

Thus the hybridisation involved in the AsF₅ molecule is trigonal bipyramidal. So, the hybrid orbitals used are s, p_x, p_y, p_z, d_z².

Total no. of electrons in O₂^2- = 16 + 2 = 18

Distribution of electrons in molecular orbital

Anti bonding electron = 8 (4 pairs)

Paramagnetic character is based upon presence of unpaired electron

Hence only Cl^– do not have unpaired electrons.

Bond order between P – O

or Formal charge on oxygen = 

(Formal charge = No. of electrons in valance

XeF₄ hybridisation is = 

hence V = 8 (no. of valence e^–) X = 4 (no of monovalent atom)

C = 0 charge on cation A = 0 (charge on anion). The shape is

  Square planar   shape.

As dipole moment = electric charge × bond length D. M. of AB molecules = 4.8 x 10 ^-10 x 2.82 x10 ^-8 = 13.53D
D.M. of CD molecules = 4.8 x10^-10 x 2.67 x10^-10 = 12.81D
now% ionic character

then % ionic character  in

% ionic character in

Bond order of NO⁺ =

Similarly, Bond order of NO =  

Bond order of NO^– 

Bon d order of  

By above calculation, we get Decreasing bond order NO⁺ > NO > NO^– > O^-₂

PF₃ has pyramidal shape
Phosphorus exist in sp³ hybridiation state hence it exist in tetrahedral shape. But due to presence of lone pair its shape is pyramidal.
F—H-----F—H-----F—H-----F

In the case of HCl we can see that hydrogen is bonded to chlorine atoms so it will not show hydrogen bonding.

HF: 
Here we can see that  HF show a linear polymeric structure due to hydrogen bonding.

Here we can see that though  H₂O show polymerisation due to hydrogen bonding it is not linear polymerisation.

NH₃

Here we can see that though  NH₃  show polymerisation due to hydrogen bonding it is not linear polymerisation.

Thus, the correct option is (B)  HF.

In BCl₃​ the valence electrons for boron are 6 (3 of boron and 3 of chlorine), so there are two electrons deficient for the octet configuration. Thus, electron deficient molecule is BCl₃​.

Dissociation energy of any molecules depends upon bond order. Bond order in N₂ molecule is 3 while bond order in N⁺₂ is 2.5.
Further we know that more the Bond order, more is the stability and more is the BDE.

We know that the electr ostatic force that binds the oppositely charged ions which are formed by transfer of electron from one atom to another is called ionic bond. We also know that cation and anion are oppositely charged particles therefore they form ionic bond in crystal.

In P–O bond, π bond is formed by the sidewise overlapping of d-orbital of P and p-orbital of oxygen. Hence it is formed by pπ and dπ overlapping.

We know that bond angles of  NH₃ = 107º, ₄⁺ = 109.5º,  PCl₃ = 100º. Therefore bond angle of NH₄⁺ is maximum.

In XeF₂ and IF₂^–. Both XeF₂ and IF₂^– are sp³d hybridized and have planar shape.

p_x and P_y orbitals do not have proper orientation to overlap and hence no bond is formed.

In X — H - - - Y, X an d Y both are electronegative elements (i.e attracts the electron pair) then electron density on X will increase and on H will decrease.

In SO₃^-2

S is sp³ hybridised, so

In 'S' unhybride d- orbit is present, which is involved in π bond formation

n oxygen two unpaired p- orbital is present in these one is involved in σ bond formation while other is used in π bond formation Thus in orbitals ar e involved for p_π - d_πbonding.

To form NO₃^- , nitrogen uses one p-electron for π-bond formation two p-electrons for σ-bond formation. 2s electrons are used for coordinate bond formation. Thus there is no lone pair on nitrogen but four bond pairs

As sigma bond is stronger than the π(pi) bond, so it must be having higher bond energy than π (pi) bond.

Shape (structure ) of a species can be ascertained on the basis of hybridisation of its central atom, which in turn can be known by determining the number of hybrid bonds

(H). Thus 

For SiF₄ : 

(Tetrahedral structure)

For SiF₄ : 

(See saw structure)

Hence the two compounds have different structures

Thus here  bond angles between X₄ - M - X₂ = 180°
X₁ - M - X₃ = 180°
X₅ - M - X₆ = 180°

In a linear symmetrical molecule like BeF₂, the bond angle between three atoms is 180°, hence the polarity due to one bond is cancelled by the equal polarity due to other bond, while it is not so in angular molecules, like H₂O