JEE Advanced (Fill in the Blanks): Chemical Bonding & Molecular Structure | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The angle between two covalen t bonds is maximum in ................. . (CH₄, H₂O, CO₂) (1981 - 1 Mark)

Ans. Sol. CO₂; Bond angle in CH₄ is 109°.28’, in H₂O it is 105 and in CO₂ it is 180°. So it is maximum in case of CO₂

Q.2. Pair of molecules which forms strongest intermolecular hydrogen bond is ................. .
 (SiH4 and SiF4, and CHCl₃,    and   (1981 - 1 Mark)  
Ans. Sol.   and     form strongest hydrogen bonds because of largest difference in electronegativities of bonded atoms.

Q.3. There are ............. π bonds in a nitrogen molecule. (1982 - 1 Mark) 

Ans. Sol.  2; N ≡ N (N₂) has 1σ and 2π bonds. (A triple bond consists of 1σ and 2π bonds)

Q.4. ............. hybrid orbitals of nitrogen atom are involved in the formation of ammonium ion. (1982 - 1 Mark) 

Ans. Sol. sp³; Hybridisation (H)=  [No. of valence electron in centralatom + No. of monovalent atoms – Charge on cation + Charge on anion]

For N in NH⁴⁺, hybridisation (H) = (5 + 4 – 1 + 0) = 4
∴ sp³ hybridisation.

Q.5. The shape of [CH₃]⁺ is ............... . (1990 - 1 Mark)

Ans. Sol.  Planar; ⁺CH₃  is a carbocation and such a species has aplanar shape.

Q.6. The two types of bonds present in B₂ H₆ are covalent and ........ (1994 - 1 Mark) 

Ans. Sol.  Three centred two electron bonds or banana bond;

NOTE : The formation of three centred two electron bond is due to one empty sp³ orbital of one of the B atom, 1s orbital of the bridge hydrogen atom and one of the sp³ (filled) orbital of the other B-atom. This forms a delocalized orbital covering the three nuclei giving the shape of a banana. Thus also known as banana bonds.

Q.7. When N₂ goes to N⁺₂ , the N–N bond distance ..., and when O₂ goes to O⁺₂ the  O–O bond distance.... (1996 - 1 Mark)

Ans. Sol.  Increases, decreases;
∴ Bond order in N₂ = 3 and  Bond order in N²⁺ = 2.5
Thus conversion of  N₂ to N⁺₂ decreases bond  order  (from 3 to 2.5) and hence increases  the N – N bond distance. [∵ Bond distance increases with decrease in B.O.)
Bond order in O₂ = 2 and Bond order in O⁺₂ = 2.5

NOTE : Thus conversion of O₂ to O⁺₂ increases bond order (from  2 to 2.5) hence decrease O – O bond distance.