Test: Chemical Bonding and Molecular Structure (July 17) - NEET MCQ

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.

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.

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

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 :

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.

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.

Calculating the bond order of various species.

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

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

  6 means sp³d²

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

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