Chemistry MCQs for NEET — Practice Questions with Answers

(a) Order of stability $\propto$ bond order

$\therefore$ Order of the stability of given species.

                     ${\mathrm{O}}_2^{+} > {\mathrm{O}}_2 > {\mathrm{O}}_2^{-} > {\mathrm{O}}_2^{2-}$

Bond order     2.5     2      1.5      1

Since, HCl is strong acid and dissociates completely. Hence, it conducts electricity best in its aqueous solution.

(c) The compounds with lower oxidation number and which cannot reduced by one another can exist together. Thus, FeCl₂ and SnCl₂ can exist together as Fe²⁺ cannot be reduced by Sn²⁺

Among the given options, NH3 (ammonia) has the maximum dipole moment. Dipole moment arises due to the asymmetric distribution of charge in a molecule. NH3 has a pyramidal shape with a lone pair of electrons, making it polar. CO2 and CH4 are non-polar due to their symmetric structure.

(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.