NEET Practice Questions (MCQs) with Answers & Solutions

(a) No doubt $N{H}_3$ and $B{F}_3$ have $s{p}^3$ (pyramidal) and $s{p}^2$ (coplanar) hybridization respectively having one lone pair of electron on N atom which is responsible for pyramidal shape of $N{H}_3$ inspite of $s{p}^3$ hybridization. However, as soon as it is coordinated to $B{F}_3$,  both attain tetrahedral geometry and acquire $s{p}^3$- hybridization.  

 (a)

(a) :Incorrect : Because electronic transfer takes place and it is chemisorbtion.

(b) :correct: Adsorbtion is always exothermic.

(c) :correct: $O_2$ will accept the electron from metal into its $\mathrm{\pi }*2\mathrm{p}$ orbital.

(d) :correct: Since the electron enters into $\mathrm{\pi }*$ orbital bond order decrease  and bond length increases.

The BF3 molecule has a planar structure due to the absence of lone pairs on the central boron atom. In contrast, the NF3 molecule adopts a pyramidal structure because the nitrogen atom has one lone pair of electrons, causing repulsion and distorting the molecular geometry.

According to VSEPR theory, the IO2F2- ion has a trigonal bipyramidal structure with the two lone pairs and two I-O double bonds occupying the equatorial positions, which is analogous to the structure of SF4. Therefore, both options (A) and (C) are correct.

c. 

(A), (B) M.O. for C₂ = ${{\sigma }_1}^{s^2}<{{\sigma }_1}^{{*}_{s^2}}<{{\sigma }_2}^{s^2}<{{\sigma }_2}^{{*}_{s^2}}<\frac{{\mathrm{\pi }}_2{\mathrm{p}}^2\mathrm{y}={\mathrm{\pi }}_2{\mathrm{p}}^2\mathrm{z}}{HOMO}<\frac{{\sigma }_2{p}_x}{LUMO},$ two $\mathrm{\pi }$ molecular orbitals are involved in bonding .

(C) It is isoelectronic with $N_2$ and has one sigma and two pie-bonds.

(c) In both, all electrons are paired so diamagnetic.

(D) μ=q×d

$1.2\times {10}^{-18}=q\times {10}^{-8}cm$

$$\begin{gathered} q=\frac{1.2\times {10}^{-18}}{{10}^{-8}}=1.2\times {10}^{-{10}^{}} \\ \%charge =\frac{q}{e}\times =\frac{1.2\times {10}^{-10}}{4.8\times {10}^{-10}}\times 100=25\% \end{gathered}$$

All the given statements are correct. (CH3)3COH is less acidic than (CH3)3SiOH due to the poorer ability of silicon to stabilize the negative charge on the conjugate base. CO is a stable molecule, but its silicon analogue is not stable. In phosgene (COCl2), the C-O bond length is longer than expected due to the presence of resonance, while the C-Cl bond length is shorter.

(A) Strength of H-bonds on following factors.
      (i) Electronegativity of element covalently bonded to hydrogen atom.
      (ii) Size of electronegative element.
      (iii) Ease of donation of lone pair of electrons by electronegative element.

(C)  Number of polarisable electrons $\propto$ strength of vander waal’s force $\propto$ boiling point.
       boiling point $CHC{l}_3=61°C and CC{l}_4=73°C$.