Since Oxygen is more electronegative than Sulphur thus strong hydrogen bonding in water Answer: (d)

Hydrogen is not bonded with F in option 'd'Answer: (d)

Answer: (b)

Answer: (c)

Due to the presence of -OH groups glycerol undergoes extensive intermolecular H bonding and hence it has highest viscosity Answer: (a)

Answer:(d)

Due to intramolecular H bonding in o-nitrophenol it exists as a monomer while due to intermolecular H bonding in p-nitrophenol it exists as an associated molecule. As a result, vapour pressure of o-nitrophenol is higher than that of p-nitrophenol Answer: (a)

Strength depends on electro negativity Answer:(a)

Diethyl ether does not contain a H-atom attached to O and hence does not undergo H-bonding Answer: (d)

Due to absence of lone pairs of electrons on carbon, CH4 has regular tetrahedral geometry. Although CHCL3 also does not have loan pairs of electrons on carbon but due to the presence of three polar C-Cl bonds and one almost non polar C-H bond, its geometry is little distorted from regular shape. While SF4 has one lone pair and four bond pairs. therefore does not have regular geometry. Thus potion 'c' is correctAnswer: (c)

C2H6 is sp3 hybridized while other options are sp2 hybridizedAnswer: (d)

CH 3 in CH3CHO have sp3 hybridizationAnswer: (d)

The central atom carbon has four bonds which means 4×2 = 8e- Hence, in CO2 cotect is followed There are two bond pair and 2 lone pair and thus the oxygen atom has a total of eight e-. Thus its octet rule is satisfied. O2: In oxygen there exist two bond between each O atom and they have two lone pairs each, so there are four pair of electrons, which means 4×2=8e- on each atom. Thus, its octet rule is satisfied. In CO molecule, the carbon atom contains only 6 valence electrons, where as 8e- are require to fulfill its octet. Hence, CO molecule does not have an octet Answer: (d)

Total number of valence electrons in IF5 are=7 + 5 ×7=42 Now 42 ÷8=5(Q1 + 2(R1) R1 ÷ 2=2 ÷ 2=1(Q2) + 0 (R2) Now Q1+ Q2+R2=5+1+0=6 ∴ Hybridization of I=sp3d2 Since If5 has only five F atoms, therefore, sixth orbital will be occupied by a lone pair. So option 'a' is correctAnswer: (a)

Answer: (c)

Number of valence electrons in H2SO4 are =2 ×1 + 6 + 4×6=32 Now 32÷8=4 (Q)+ 0(R)=4sp3 hybridizationAnswer: (d)

Total valance electrons in BF3 is >=3 +3 ×7=24 Now 24 ÷ 8=3(Q) + 0(R) Thus Q+ R=3, sp>3 hybridization In H3O+ free electrons are=3 ×1 + + -1=8 Now 8 ÷2=4(Q) + 0(R)=4 Thus sp>3 hybridization in NH3 number of valence electrons=5 + 3 ×1=8 ∴ Hybridization is sp3 Total number of valence electrons in PCl3 =5+ 3 ×7=26 NOw 26÷8=3(q1 + 2 (R1) and R1 ÷ 2=1(Q2 + 0 (R2) By adding all quotients and final remainder=3+1+0=4 ∴ Hybridization in of P in PCl3 is sp3Answer: (a)

Answer:(d)

In option a: Coordnate bond is formed between lone pair of oxygen atom and empty s-orbital of H atom. Option b: Boron atom has empty 2p orbital after formation of BF3, So if can accept lone pair of fluorine atom and form BF4- In Option c: There is no coordinate bonding but very strong hydrogen bonding dye to high electronegativity of flourine. In Option d: Nitrogen has lone pair left after forming three covalent bonds with hydrogen, it forms coordinate bond by sharing that lone pair with H-atom Only HF2- has H bonding [ F-H ...F]- rest all the molecules have coordinate bonds Answer: (c)

SnCl2 is bent moleculeAnswer: (c)

Number of valence electrons in ICl2+ are=7 + 2 ×7 -1=20 Now 20 ÷ 8=2(Q1) + 4(R1)R1 ÷ 2=2(Q2 + 0 (R2Q1+Q2+R2=2+2+0=4sp3 hybridizationAnswer: (d)

Sn have two bond pairs and one loan pair hence structure 'b' is correct Answer:(b)

Higher the s- character of the hybrid orbital, more is its electro negativity. Thus sp-hybrid carbon is most electronegative Answer: (b)

The hybridization which leads to the formation of maximum bonds between two atoms is sp. therefore, at the maximum three bonds can be formed between two atoms example C-C ≡ N ; C-C = O Orbital geometry around carbons can not allow them to overlap in such a way that they form 4 bonds Answer: (c)

Valence electrons in POCl3 is 5 +6+3 ×7=32 Now 32÷8=4(Q) + 0(R)since Q+ R=4+0=4 sp3 hybridization Total number of valence electrons in PCl3 =5+ 3 ×7=26 NOw 26÷8=3(q1 + 2 (R1) and R1 ÷ 2=1(Q2 + 0 (R2) By adding all quotients and final remainder=3+1+0=4 ∴ Hybridization in of P in PCl3 is sp3Answer: (a)

In both C2H2 and CO2 c is sp2 hybridized Answer:(b)

Ch4 do not form hydrogen bond hence boiling point should be lowest in the series Answer: (d)

Answer: (a)

Answer:(a)

Total electrons in BeF3- =2 +3 ×7 +1=24Now 24÷8=3(Q)+ 0(R)=3 Thus Be in BeF3- is sp2 hybridized All other molecules, the hybridization of central atom is sp3Answer: (b)