Total number of electrons present in the penultimate shell of an element with atomic number $36$ is:

To draw the geometrical representation for the structure of the oxygen atom the following steps are given. Identify the correct sequence of steps.

$\left(1\right)$ The eight electrons present in the extra-nuclear part would be distributed in the first two orbits that is K and L. As per the rules, two electrons would occupy the K orbit and the remaining six electrons occupy the L orbit.

$\left(2\right)$ The atomic number of oxygen is $8$.

$\left(3\right)$ In the nucleus, $8$ protons and $8$ neutrons and present and in the extranuclear part, i.e., in the orbits, $8$ electrons are present.

$\left(4\right)$ Oxygen atom has $8$ electrons and $8$ protrons. The mass number is $16$, and hence, the number of neutrons is equal to 

Arrange the following statements in a sequence which involves the calculation of atomic number and mass number of an atom of an element with $15$ electrons and $16$ neutrons.

$\left(1\right)$ $\mathrm{A}=\mathrm{Number} \mathrm{of} \mathrm{protons} + \mathrm{Number} \mathrm{of} \mathrm{Neutrons}$

$$\begin{gathered} \mathrm{A}=\mathrm{Z} + \mathrm{Number} \mathrm{of} \mathrm{neutrons} \\ \mathrm{A}= 15 + 16 =31 \end{gathered}$$

$\left(2\right)$ Number of protons and number of electrons are equal in a neutral atom. Hence, the atomic number Z is equal to $15$.

$\left(3\right)$ Mass number is equal to the total number of protons and neutrons.

$\left(4\right)$ Atomic number is $15$ and mass number is $31$.

Rutherford's $\mathrm{\alpha }-$ray scattering experiment led to the discovery of the nucleus and to the conclusion that an atom consists of large empty space. Arrange the following steps in sequence which explains the experiment and also the above mentioned conclusions.
$\left(1\right)$ To make out the observations a spherical ZnS screen was placed surrounding the gold foil.
$\left(2\right)$ The substance which acts as a source of $\mathrm{\alpha }-$particles is taken in a lead container and made to pass through a slit between like charged positive plates.
$\left(3\right)$ It was observed that most of the particles passed straight through the gold foil, few were deflected through small angles and very few through large angles. Very few completely rebounded.
$\left(4\right)$ A narrow, condensed beam consisting of $\mathrm{\alpha } -$particles is made ot bombard on a thin a gold foil.