Chemistry MCQs for NEET — Practice Questions with Answers

(a) MO configuration of ${\mathrm{O}}_2^{+}$ (8+8-1=15)

 = $\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \stackrel{}{\mathrm{\sigma }}2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{y}}^2, \stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{x}}^1\approx \stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{y}}^0$

BO = (N_b -N_a)/2

(where, N_b = number of electrons in bonding molecular orbital

 N_a = number of electrons in antibonding molecular orbital)

   ... BO= (10-5)/2 = 2.5

Similarly,

(b) ${\mathrm{O}}_2^{-}$

for valance shell   (6-3)/2 = 1.5

c. (6-4)/2=1

d. (6 -2) /2 = 2

PCl3 contains three bond pairs formed by the three P-Cl bonds and one lone pair around the central phosphorus atom, giving it a trigonal pyramidal shape.

 sp hybridised bond angle is 180... so thy will maintain straight line.

(d) C-H: 0.109 nm

    C=C: 0.134 nm

    C-O: 0.143 nm

    C-C: 0.154 nm

.^.. Bond length order is

    C-H<C=C<C-O<C-C

As covalent character increase, melting point decreases, thus order of melting point is 

CaF₂ > CaCl₂ > CaBr₂ > Cal₂ 

Bond order of O₂⁺   = (10-5)/2=2.5

Bond order of O₂^-    = (10-7)/2=1.5

Bond order of O₂^2-  = (10-8)/2=1

Bond order of O₂     = (10-6)/2=2

Maximum bond order = minimum bond length.

Bond length is minimum for O₂⁺ 

(b) As the size of halogen atom increases, the acidic strength of boron halides increases. Thus, BF₃ is the weakest Lewis acid. This is because of the p$\mathrm{\pi }$ - p$\mathrm{\pi }$ back bonding between the fully-filled unutilised 2p orbitals of F and vacant 2p orbitals of boron which makes BF₃ less electron deficient. Such back donation is not possible in case of ₃ or BBr₃ due to larger energy difference between their orbitals. Thus, these are more electron deficient. Since on moving down the group the energy difference increases, the Lewis acid character also increases. Thus, the tendency to behave as Lewis acid follows the order

BBr₃> BCl₃ > BF₃

The dominant intermolecular force in liquid CH3OH (methanol) is hydrogen bonding due to the presence of the O-H bond. This hydrogen bonding must be overcome to convert liquid methanol into the gaseous state.