Chemistry MCQs for NEET — Practice Questions with Answers

Key Idea : As the number of lone pair of electrons increases, bond angle decreases.

NO₂⁺ ion is isoelectronic with CO₂ molecule. It is a linear ion and its central atom (N+) undergoes sp-hybridisation, hence bond angle is 180°.

In NO₂^- ion, N-atom undergoes sp²-hybridisation. The angle between hybrid orbital should be 120° but one lone pair of electrons is lying on N-atom, hence bond angle decreases to 115°.

In NO₂ molecule, N-atom has one unpaired electron in sp²-hybrid orbital. The bond angle should be 120° but actually it is 132°. It maybe due to one unpaired electron in sp²-hybrid orbital.

Therefore, the increasing order of bond angles is:

NO₂^-  < NO₂  < NO₂⁺

(115°) (132°) (180°)

Key Idea:

  -C-C- sp3

 -C=C- sp2

-C$\equiv$C- sp3

=C=C= sp

 sp³  sp²  sp² sp³  sp   sp

CH₃-CH=CH=CH₂-C$\equiv$CH

6      5     4    3      2    1

Hence, the state of hybridisation of carbons 1,3 and 5 are sp, sp³ and sp² respectively.

(b) The correct order regarding the electronegativity of hybrid orbitals of carbon is sp>sp²>sp³ because in sp, sp² and sp³ hybrid orbitals s-orbital character is 50%, 33.33% and 25% respectively and due to higher s-orbital character electron attraction tendency i.e, electronegativity increases.

(c) $\stackrel{+}{\mathrm{N}}{\mathrm{O}}_2$ has linear shape due to sp-hybridisation of N of $\stackrel{+}{\mathrm{N}}{\mathrm{O}}_2$

(c) In Al₂O₃, Sb₂O₃, Bi₂O₃ and SeO₂.Bi₂O₃ is most basic oxide due to higher reactivity with acid.

${\mathrm{Bi}}_2{\mathrm{O}}_3 + 6\mathrm{HCl} \stackrel{ }{\to }2{\mathrm{BiCl}}_3 + 3{\mathrm{H}}_2\mathrm{O}$

(c) Every AB₅ molecule does not in fact have square pyrimidal structure but AB₅ molecules have trigonal bipyramidal structures due to sp³d hybridisation.

SCl₄ is not isostructural of SiCl₄ because it shows square planar structure due to involvement of repulsion between lone pair and bond pair of electrons.

${\mathrm{SO}}_4^{2-}$ shows tetrahedral structure due to sp³ hybridisation.

${\mathrm{PO}}_4^{3-}$ shows tetrahedral structure due to sp³ hybridisation.

${\mathrm{NH}}_4^{+}$ shows tetrahedral structure due to sp³ hybridisation.

(c) stability order of carbon family hydride silane$<$German$>$stannane$>$Plumbane

(d)