Chemistry MCQs for NEET — Practice Questions with Answers

By Heisenberg's uncertainty principle

  $$\begin{gathered} ∆\mathrm{p} \mathrm{x} ∆x\mathit{\ge }h\mathit{/}\mathit{4}\mathrm{\pi } \\ \mathrm{or} ∆\mathrm{v} \mathrm{x} ∆x\mathit{\ge }h\mathit{/}\mathit{4}\mathrm{\pi m} \end{gathered}$$

$∆$p$\to$uncertainty in momentum

$∆x\to$uncertainty in position

$∆\mathrm{v}\to$uncertainty in velocity

m$\to$mass of particle

Given that,

$∆x = 0.1 \stackrel{°}{\mathrm{A}} = 0.1 \mathrm{x} {10}^{-10} \mathrm{m}$

m = 9.11 x 10^-31 kg

h = planck constant = 6.626 x 10^-34 Js

$\mathrm{\pi }$=3.14

In uncertain position $∆\mathrm{v} \mathrm{x} ∆x$ = h/4$\mathrm{\pi }$m

$∆\mathrm{v}$ x 0.1 x 10^-10 = (6.626 x 10^-34)/(4 x 3.14 x 9.11 x 10^-31) ms^-1

                  $∆$v = (6.626 x 10^-34)/(4 x 3.14 x 9.11 x 10-31 x 0.1 x 10^-10) ms^-1

                        =5.785 x 106 ms^-1

                         =5.79 x 106 ms^-1

The orientation of an atomic orbital is governed by magnetic quantum number.

The dissociation reaction is $I_2\to 2I$

Bond energy per single bond is

$\epsilon =\frac{240 kJ mo{l}^{-1}}{6.023\times {10}^3 mo{l}^{-1}}=3.985\times {10}^{-22} kJ =3.985\times {10}^{-19}J$

Energy of 1 photon of wavelength 4500 $\stackrel{\circ }{A}$ is

$$\begin{gathered} E=\frac{hc}{\lambda }=\frac{\left(6.626\times {10}^{-34} J s\right)\left(3\times {10}^8 m s^{-1}\right)}{\left(4500\times {10}^{-10 }m\right)} \\ =4.417\times {10}^{-19} J \end{gathered}$$

Of the energy $4.417\times {10}^{-19} J,$ the energy consumed to break I₂ molecules is $3.985\times {10}^{-19} J$. The remaining energy is used in imparting kinetic energy to the two iodine atoms. Hence, Kinetic energy per iodine atom

$=\frac{\left(4.417-3.985\right)\times {10}^{-19} J}{2}=2.16\times {10}^{-20} J.$

(a) Angular momentum of electron in d-orbital is

              $$\begin{gathered} =\sqrt{\mathrm{l}(\mathrm{l}+1)}\frac{\mathrm{h}}{2\mathrm{\pi }}; \mathrm{for} \mathrm{d}-\mathrm{orbital}, \mathrm{l}=2 \\ =\sqrt{2(2+!)} \Rightarrow \mathrm{h}=\sqrt{6} \mathrm{h} \left(\because \mathrm{h}=\frac{\mathrm{h}}{2\mathrm{\pi }}\right) \end{gathered}$$

2.

n-1 , for 6s , 6-1 = 5

5d , 5-1 = 4

4f, 4-1= 3

Radial Nodes=n-l-1

angular node  = l

 node = n-1 

5-1 = 4 for all.

No. of spherical nodes = n-l-1;

for s, p, d, f values of l=0, 1, 2, 3 respectively.

3.

$\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{x}}^2}+\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{y}}^2}+\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{z}}^2}+\frac{8{\mathrm{\pi }}^2\mathrm{m}}{{\mathrm{h}}^2}\left(\mathrm{E}-\mathrm{V}\right) \mathrm{\psi }=0$

The wave mechanical model of an atom depends on de Broglie's concept of the dual nature of electrons, the Heisenberg uncertainty principle, and the Schrodinger wave equation. All these principles are fundamental to understanding the behavior of electrons in atoms.