Chemical Bonding and Atomic Structure MCQs for NEET — Chemistry Questions with Answers

(c) Dipole-induced dipole interaction are present in the pair in which the first species is polar and the other is non-polar. H₂O and alcohol both are polar molecules, so there exists dipole-dipole interactions in between them. Cl₂ and CCl₄ both are non-polar so ther exists induced dipole-induced  dipole interactions in between them Similar is true for SiCl₄ and He atoms pair. HCl is a polar molecule, whereas He atoms are non-polar, so in between them dipole-induced dipole interactions.

(b) Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration. Molecular orbital configuration of the given species is given as

  CO (6+8=14) = $\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

(All the electrons are paired so it is diamagnetic)

 ${\mathrm{O}}_2^{-}$(8+8+1=17) = $\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2,\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2, \stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{y}}^1$

(b) All the molecules have O atom with lone pairs, but in the H₂O atom has no vacant orbital for $\mathrm{\pi }$ bonding. That's why it does not have any $\mathrm{\pi }$ bond. In all other given molecules, the central atom because of the presence of vacant orbitals is capable to farm $\mathrm{\pi }$ bonds.

(b) Symmetrical molecules are generally non-polar although they have polar bonds. This is because bond dipole of one bond is canceled by that of the other. BF₃, SiF₄ and XeF₄ being symmetrical as non-polar. SF₄ is unsymmetrical because of the presence of a lone pair of electrons. Due to which it is a polar molecule.

(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