Atomic Structure MCQs for NEET — Chemistry Questions with Answers

Key Idea : According to Heisenberg’s uncertainly principle, it is impossible to determine simultaneously the position and momentum of a moving microscopic particle, i.e.,

Δx X Δp ≥ h/4$\mathrm{\pi }$

Δx X Δp ≥ h/4$\mathrm{\pi }$

Here Δx = Δp and Δp = m.Δv

Δv² = h/m²4$\mathrm{\pi }$

or Δ=1/2m $\sqrt{\frac{\mathrm{h}}{\mathrm{\pi }}}$

Note : The uncertainty principle in terms of energy and time is given as

ΔE.Δt≥h/4$\mathrm{\pi }$

Key Idea: $∆x \mathrm{x} ∆\mathrm{p}\ge \mathrm{h}/4\mathrm{\pi }$

where, $∆$p = uncertainty in momentum and $∆$v = uncertainty in velocity

Here, $∆$p = 1x10^-18 g cm s^-1

         m = 9x10^-28 g

        $∆$v =$∆$p/m = (1x10^-18)/(9x10^-28) = 1.1 x 10⁹ = 1.0 x10⁹ cm s^-1

(c) The value of l varies from 0 to n-l and the value of m varies from -l to +l through zero. The value of 's' $\pm$ 1/2 which signifies the spin of an electron. The correct sets of quantum number are the following:

         n              l            m             s

(ii)     2             1            1             +1/2

(iv)    1             0            0             -1/2

(v)    3              2            2             +1/2      

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.