Difference in electro negativity Answer: (b)

Element X of group VA ( or 15) has normal valency of three while element (Y) of group (VIA) ( 16) has a normal valency of 2. Therefore formula of the compound is X2Y3 Group VA → Group 15 → ns2 np3 either accept or losse 3e- (X3+/X3- Group VIA → Group 16 → ns2 np4 either loose 4e- or accept 2e- (Y4+/ Y2- In group -16 loosing 4e- is not easy. So it accept 2e- ∴ 2X3+ + 3Y2- X2 Y3 Answer: (a)

Answer: (a)

Answer:(c)

ICl = 7+7 = 14e- It is polar in nature due to the difference of its electronegativity. Br2 it is non polar in nature due to presence of large electronegativity element. OR ICl molecule are attracted to each other by dipole - dipole interaction as well as London dispersion forces, where as Br2 molecules are attracted to each other by London dispersion force Dipole interaction are stronger, so it take more energy to separate the molecules. Thus ICl is polar while Br2 is nonpolar Answer: (c)

Double bond in C2 has both π bonds between two carbon atoms because 4 electrons are present in π molecular orbitAnswer: (b)

there are four such resonating structures bond order=7/4=1.75Answer: (b)

Answer:(b)

Fajan's rule: 1) Larger anion + smaller cation ⇒ more covalent 2) higher chrge /oxidative state ⇒ more covalent Electronegativity difference decreases covalent character increases Answer: (b)

Coordinate bond means bond formed between two atom by sharing/ donation of e- from only one atom AlCl3, Al - 3 valency, cl - 6 valency Sharing of e- by two atom for bond formation. ∴ covalent. NaCl - Ionic compound Complete transfer of valence of e- from metal and non-metal Cl2 have covalent bond Answer:(a)

PCl3 number of valence electrons=5+21=26Now 26 ÷ 8=3(Q1) + 2(R1)2 ÷ 2=2(Q2 + 0 ( R2total of Q1 , Q2 and R2=3 +1=4 Thus sp3 hybridization, and have one lone pair, pyramidal shapeAnswer: (a)

O2 have double bondsO3 have resonance structure There is single bond between H and O and O and OBond length decreases with increase of bond multiplicityAnswer: (c)

More the number of bonds , shorter is the bond lengthIn C2H2, there is triple bond between two carbons Answer: (b)

C2H4 has sp2 hybridization thus trigonal planar Answer:(c)

CO2 number of valence electrons=4+ 12=16Now 12 ÷ 8=2 (Q) + 0(R) sp hybridization linear structureAnswer: (a)

Lattice energy is directly proportional to the charge on the ion and inversely proportional to the size of the ion. Answer: (d)

O2=σ1s2, σ*1s2, σ2s2 σ*2s2 σ2pz2,π2px2 π2py2 π*2px1, π*2py1 Answer: (d)

Carboxylic group have a strong Hydrogen bonding because of higher electro negativity of oxygenAnswer: (b)

According to Fajan's rule size smaller - cation and bigger size - Anion form covalent character. F-, Cl-, Br-, Br-, I- down the group size increase. ∴ AgI - Maximum covalent character AgF - Mainimum covalent character or we can say maximum ionic character Ionic bond or bonds with maximum ionic characternrequires more energy to break compared to covalent bond. smaller anions are polarised to lesser extent than larger anions. Therefore AgF will have highest melting pointAnswer: (b)

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)

NH3 > NF3 are pyramidal and have lone pairs. But NH3 have more dipole moment. BF3 have sp2 hybridization, and planer trigonal due to symmetry dipole moment is zeroAnswer: (d)

There are four pairs of electrons in Xe In XeF2 one pair is used 3 are loan pairs XeF4 two pairs is used 2 are loan pairs XeF6 three pairs are used one loan pairAnswer: (d)

Number of valence electrons=16 Now 16 ÷8=2(Q) + 0(R) CO2 , sp hybridization, linear molecule cancel dipole moment Answer:(d)

More the electronegative central atom more the bond angleAnswer: (a)

number of valence electron in SF6=48 Now 48÷8=6 hybridization is sp3d2 thus Octahedral and bond angle 90° Thus due to symmetry net dipole moment is zeroAnswer: (b)

If we require minimum ratio. S the size of anion should be minimum and the size of cation shold be maximum anion - minimum , cation - maximum Cl- is smaller than Br- and K+ is larger than Na+. HEnce ratio of Cl- : K+ is smallestAnswer: (c)

NH3 number of valence electron=8 now 8 ÷2=4(Q) + 0(R) ∴ hybridization sp2Answer: (b)

NaCl has ionic bond and HCl has polar covalent bond. Ionic between Cu and SO4Covalent between SO4Coordinate bonds between Cu and water Answer: (c)

H - C Ξ C - H 1σ and 2π bonds between carbon atoms Answer: (a)

ClO2- is sp2, Cl have loan pairAnswer: (c)