31 Year NEET Previous Questions: Chemical Bonding & Molecular Structure

Only those d or bital s whose  l obes are directed along X, Y and Z directions hybridise with s and p orbitals. In other three d or bitals n amely d_xy, d_yz and d_xz, the lobes are at an angle of 45° from both axis, hence the extent of their overlap with s and p orbitals is much lesser than and orbitals.

Bent T-shaped geometry in which both lone pairs occupy the equatorial positions of the trigonal bipyramid. Here (lp - lp) repulsion = 0, (lp - bp) repulsion = 4 and (bp - bp) repulsion = 2.

Molecules containing less than two valence electrons per bond are referred to as electron deficient.

B2H6B2H6 molecule has total 8 covalent bonds but only 12 electrons i.e. 6 pairs. Hence it has 2 electron pairs less than maximum required number for bonding, which makes it electron deficient.

B2H6B2H6 is called diborane. In this compound, the octet for B is incomplete which is why it dimerizes. So B2H6B2H6 is electron deficient.

As difference of electronegativity increases % ionic character increases and covalent character decreases i.e., negativity difference decreases covalent character increases.
Further greater the charge on the cation more will be its covalent character. Be has maximum (+2) charge.

Boron trifluoride BF₃ molecule has trigonal planar geometry.
The central B atom has 3 bond pairs and 0 lone pairs of electrons. It undergoes sp² hybridisation.
IF₃ has 'T' shaped geometry. PCl₃ and NH₃ have pyramidal geometry.

SF₄ has permanent dipole moment.
SF₄ has sp³d hybridization and see saw shape (irregular geometry).

Whereas XeF₄ shows squre planar geometry SiF₄ has tetrahedral shape and BF₃ has Trigonal planar shape. All these are symmetric molecules. Hence μ ≠ 0.

Statement (a), (c), (d) are correct. Statement (b) is incorrect statement.
AB₅ may have two structures as follows :

Electronic configuration of the molecule according to molecular orbital theory, is

Last two electrons are unpaired. So no. of unpaired electron is 2.

In BF₃, AlF₃ & NF₃ all fluoride atoms are symmetrically oriented with respect to central metal atom but in ClF₃ three fluorine atoms are arranged as follows :

Here two bonds are in equitorial plane & one bond is in axial plane.

In NH₃ the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite direction so in the former case they are added up whereas in the latter case net result is reduction of dipole moment. It has been shown in the following figure :

NO₂⁺ will have linear shape as it will have sp hybridisation.

In SiCl₄ there is sp³ hybridisation so the structure is tetrahedral. In SO₄^2–, PO₄^3–, NH₄⁺ the structure is tetrahedral because in all hybridisation is sp³. But in SCl₄, sp³d hybridisation present so shape is different which is see saw.

Hybridization of Br is BrO₃^–
Total valence elctrons = 7 + 3 × 6 = 25
Charge = –1 hence
Total = 25 + 1 = 26

hybridization = dsp³ hybridization of Xe in XeO₃ Total valence electrons

Hybridization = dsp³ In both cases, the structure is trigonal pyramidal.

All these structures exhibits resonance and can be represented by the following resonating structures.

(i) 

(ii) 

(iii)

More is the single bond character. More will be the bond length. Hence, the corret order is :

CO < CO₂ < CO₃^{– –}

Single bond contains = 1 σ bond
Double bond contains = 1 σ bond + 1 π bond
Triple bond contains = 1σ  bond + 2 π bond

The molecule of O₃ is bent with an angle of 116.8^o and equal O-O distance of 128 pm. The ozone molecule consists of 2 sigma bond and 1 pi bond.

From the structure of three species we can determine the number of lone pair electrons on central atom (i.e. N atom) and thus the bond angle.

We know that higher the number of lone pair of electron on central atom, greater is the lp – lp repulsion between Nitrogen and oxygen atoms. Thus smaller is bond angle..
The correct order of bond angle is

NO₂^- <NO₂<NO₂⁺  i .e. opti on (b) is correct.

Calculating the bond order of various species.

From these values we find the correct order of increasing bond order is

Due to intermolecular hydrogen bonding in methanol, it exist as assosiated molecule.

For neutral molecules: No. of electron pairs = No. of atoms bonded
to it + [Group number of central atom –Valency of the central atom].
For ions : No. of electron pairs = No. of atoms bonded
to it +[Group no. of central atom – valency of the central atom ± no. of electrons] On calcuting no. of electron pairs in given molecules. We find that in the given molecules hybridisation is

  3 means sp²

3 means sp²

  6 means sp³d²

I₃ , SF₄ , and PCl₅ ; all have sp³d hybridization.

Option B is correct.

• NO₃^-: The central nitrogen is bonded to three oxygen atoms and has three regions of electron density (steric number = 3). This leads to sp² hybridization and a trigonal planar molecular geometry (with resonance-delocalised N-O bonds).

• H₃O⁺: The central oxygen is bonded to three hydrogen atoms and has one lone pair, giving four regions of electron density (steric number = 4). This corresponds to sp³ hybridization and a trigonal pyramidal molecular geometry (electron-pair geometry is tetrahedral).

• Because the central atoms have different steric numbers, their hybridizations are dissimilar (sp² vs sp³), and their molecular geometries are also different (trigonal planar vs trigonal pyramidal). Therefore the correct statement is that they have dissimilar hybridization and different structures.

(a) NH⁺₄  : sp³ hybridisation
(b) CH₄: sp³ hybridisation
(c) SF₄: sp³ d hybridisation
(d) BF₄^– : sp³ hybridisation
∴ Correct choice : (c)

Hybridisation =  [No. of valenceelectrons of central atom + no. of monovalent atoms attached to it – Negative charge if any – positive charge if any]

SCN^–  = sp i.e., NO₂^– and NO₃^– have same hybridisation.

BF₄^- hybridisation sp³, tetrahedral  structure.
NH₄⁺ hybridisation sp³, tetrahedral structure.

(O₂) =

Bond order

Bond order 

Bond order

Bond order 

Both O₂^2– and B₂ has bond order equal to 1.

Bondorder = 

B₂ is known in the gas phase

Bond order  =