Mix Example - STRUCTURE OF THE ATOM Questions in English

(N/A) The three elementary particles that constitute an atom are $Electrons$,$Protons$,and $Neutrons$.
The mass of an atom is determined by the $Protons$ and $Neutrons$ present in the nucleus.
Justification:
Consider a Sodium $(Na)$ atom.
Number of $Protons = 11$
Number of $Neutrons = 12$
Number of $Electrons = 11$
Mass Number = (Number of $Protons$) + (Number of $Neutrons$)
Mass Number = $11 + 12 = 23$
Since the mass of $Electrons$ is negligible compared to $Protons$ and $Neutrons$,the atomic mass is primarily determined by the sum of $Protons$ and $Neutrons$.

(D) The neutron was discovered by James Chadwick in $1932$. He bombarded beryllium atoms with alpha particles,which resulted in the emission of neutral particles that he identified as neutrons.

(B) The electronic configuration of the atom is determined by the number of electrons in each shell.
According to the Bohr-Bury scheme,the maximum capacity of the $K$ shell is $2$ electrons and the $L$ shell is $8$ electrons.
Given that the $K$ and $L$ shells are full,they contain $2$ and $8$ electrons respectively.
The $M$ shell contains $5$ electrons.
Therefore,the total number of electrons = $2 + 8 + 5 = 15$.
Since the atomic number is equal to the number of electrons in a neutral atom,the atomic number is $15$.

(N/A) $(i)$ Similarity: Both ${ }_{6}^{14} C$ and ${ }_{6}^{12} C$ have the same atomic number,which is $6$.
$(ii)$ Difference: Both ${ }_{6}^{14} C$ and ${ }_{6}^{12} C$ have different atomic masses; ${ }_{6}^{14} C$ has a mass number of $14$,while ${ }_{6}^{12} C$ has a mass number of $12$.

(B) The atomic number is the fundamental property of an element that uniquely identifies it.
It is defined as the total number of protons present in the nucleus of an atom.
Since the number of protons is constant for all atoms of a specific element,it serves as the definitive characteristic for identification.

(11) The atomic number $(Z)$ is equal to the number of protons in the nucleus of an atom.
The mass number $(A)$ is the sum of the number of protons and the number of neutrons in the nucleus.
$A = \text{Number of Protons} + \text{Number of Neutrons}$
Given:
Mass number $(A)$ = $23$
Number of neutrons = $12$
Substituting the values:
$23 = \text{Number of Protons} + 12$
Number of Protons = $23 - 12 = 11$
Since the number of protons is equal to the atomic number $(Z)$, the atomic number is $11$.

(B) Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
In the given representations:
$A = { }_{17}^{35} A$ (Atomic number $Z = 17$,Mass number $A = 35$)
$B = { }_{20}^{40} B$ (Atomic number $Z = 20$,Mass number $A = 40$)
$C = { }_{17}^{37} C$ (Atomic number $Z = 17$,Mass number $A = 37$)
$D = { }_{19}^{38} D$ (Atomic number $Z = 19$,Mass number $A = 38$)
Since $A$ and $C$ both have the same atomic number $(Z = 17)$ but different mass numbers ($35$ and $37$),they are isotopes.

(N/A) The figure $9$ indicates the mass number of the element $X$,which is the sum of the number of protons and neutrons in the nucleus.
$(b)$ The number of neutrons is calculated as: $\text{Number of neutrons} = \text{Mass number} - \text{Atomic number}$.
Given: $\text{Mass number} = 9$,$\text{Atomic number} = 4$.
Therefore,$\text{Number of neutrons} = 9 - 4 = 5$.

(N/A) Isobars are atoms of different elements that have the same mass number but different atomic numbers.
For example,${ }_{18}^{40} Ar$ and ${ }_{19}^{40} K$ are isobars because both have a mass number of $40$ but different atomic numbers ($18$ and $19$ respectively).

(C) Noble gases are the elements belonging to Group $18$ of the periodic table. They are characterized by having a stable electronic configuration with a complete valence shell.
Two examples of noble gases are Helium $(He)$ and Argon $(Ar)$.

(N/A) $(i)$ The scientist who discovered neutrons is $J. Chadwick$.
$(ii)$ $A$ neutron has no electric charge (it is neutral) and its mass is approximately $1.67 \times 10^{-27} \text{ kg}$, which is nearly equal to the mass of a proton.
$(iii)$ Neutrons are located inside the nucleus of an atom along with protons.

(B) Element $Z$,having atomic number $16$,is chemically more reactive than element $X$ of atomic number $18$.
$1$. The electronic configuration of element $X$ (atomic number $18$) is $2, 8, 8$. Its outermost shell is completely filled,meaning it has a stable octet configuration. Therefore,it is chemically inert (a noble gas).
$2$. The electronic configuration of element $Z$ (atomic number $16$) is $2, 8, 6$. Its outermost shell contains $6$ electrons. To achieve a stable octet,it needs to gain $2$ electrons. Because its valence shell is incomplete,it is chemically reactive.

(A) An alpha particle is a helium nucleus,represented as ${ }_{2}^{4} He ^{2+}$.
$1$. The atomic number $(Z)$ of helium is $2$,which indicates the number of protons is $2$.
$2$. The mass number $(A)$ is $4$. The number of neutrons is calculated as $A - Z = 4 - 2 = 2$.
$3$. $A$ neutral helium atom has $2$ electrons. Since the alpha particle has a charge of $+2$,it has lost $2$ electrons. Therefore,the number of electrons is $2 - 2 = 0$.

(N/A) Ions are more stable than atoms because,except for the atoms of inert gases,the outermost shell of most atoms is incomplete. To achieve a stable electronic configuration (octet),atoms lose,gain,or share electrons to form ions.
$(b)$ The outermost shell of the atoms of noble gases is already complete (stable octet or duplet). Therefore,they have no tendency to lose,gain,or share electrons,resulting in low chemical reactivity.

(B) The electronic configuration of metal $A$ (atomic number $4$) is $2, 2$. Its valency is $2$.
The electronic configuration of metal $B$ (atomic number $11$) is $2, 8, 1$. Its valency is $1$.
The electronic configuration of metal $C$ (atomic number $13$) is $2, 8, 3$. Its valency is $3$.
Comparing the valencies: $B (1) < A (2) < C (3)$.
Therefore,the increasing order of their valency is $B < A < C$.

(N/A) The mass number of $C-14$ is $14$ and the atomic number is $6$. The number of neutrons is calculated as $\text{Mass number} - \text{Atomic number} = 14 - 6 = 8$ neutrons.
$(b)$ The atomic number of Helium $(He)$ is $2$,which represents the number of protons. The $He^{2+}$ ion is formed by the loss of electrons,but the number of protons remains unchanged. Thus,it has $2$ protons.
$(c)$ The maximum number of electrons in an orbit is given by the formula $2n^2$,where $n$ is the orbit number. For the $3^{rd}$ orbit,$n = 3$. Therefore,$2(3)^2 = 2 \times 9 = 18$ electrons.
$(d)$ $A$ neutral Helium atom has $2$ electrons. The $He^{2+}$ ion is formed by losing $2$ electrons. Therefore,$2 - 2 = 0$ electrons.

(C) The number of electrons in a neutral atom is equal to its atomic number $(Z)$. For element $A$,$Z = 20$,so it has $20$ electrons. For element $B$,$Z = 18$,so it has $18$ electrons.
$(b)$ The number of nucleons is equal to the mass number $(A)$. Both elements have a mass number of $40$,so both have $40$ nucleons.
$(c)$ The special term used to represent $A$ and $B$ is 'Isobars'.
Definition: Isobars are atoms of different chemical elements that have the same mass number but different atomic numbers.
Example: $_{18}^{40}Ar$ and $_{20}^{40}Ca$.

(N/A) Thomson's model of an atom is based on the following two postulates:
$(i)$ An atom consists of a positively charged sphere in which the electrons are embedded,similar to seeds in a watermelon or plums in a pudding.
$(ii)$ The negative and positive charges are equal in magnitude. Therefore,the atom as a whole is electrically neutral.

(N/A) $(i)$ Atomic number: It is defined as the total number of protons present in the nucleus of an atom and is denoted by the symbol $Z$. For example,the atomic number of Carbon $(C)$ is $6$,which means it has $6$ protons in its nucleus.
$(ii)$ Mass number: It is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom. It is denoted by the symbol $A$. For example,a Nitrogen $(N)$ atom has $7$ protons and $7$ neutrons,so its mass number is $7 + 7 = 14$.

(N/A) The electronic configuration is $2, 5$.
$(b)$ Its valency is $3$. This is because the atom has $5$ electrons in its outermost shell and requires $3$ more electrons to complete its octet (achieve a stable configuration of $8$ electrons).

(N/A)

Isobars	Isotopes
$1.$ Same mass number but different atomic number.	$1.$ Same atomic number but different mass number.
$2.$ Different chemical properties.	$2.$ Same chemical properties.

(A) The total number of electrons is the sum of electrons in each shell: $2 + 8 + 2 = 12$ electrons.
$(b)$ The atomic number of a neutral atom is equal to the total number of electrons. Therefore,the atomic number is $12$.
$(c)$ Since the element has $2$ electrons in its outermost shell,it tends to lose these electrons to achieve a stable octet configuration. Elements that lose electrons to form positive ions are classified as metals. Thus,$X$ is a metal.
$(d)$ The valency is determined by the number of electrons in the outermost shell. Since $X$ has $2$ valence electrons,it loses $2$ electrons to complete its octet. Therefore,the valency of $X$ is $+2$.

$(A)$ The ion $X^{2-}$ has $10$ electrons, which means it has gained $2$ electrons to form the ion.
Therefore, the number of protons in the neutral atom $X$ is $10 - 2 = 8$.
Atomic number is equal to the number of protons, so the atomic number of $X$ is $8$.
Mass number is the sum of protons and neutrons.
Mass number $= \text{Number of protons} + \text{Number of neutrons} = 8 + 8 = 16$.

(A) The ion $M^{3+}$ is formed by the loss of $3$ electrons from a neutral atom $M$.
Since the ion contains $10$ electrons, the number of electrons in the neutral atom $M$ is $10 + 3 = 13$.
In a neutral atom, the number of electrons is equal to the number of protons, which defines the atomic number.
Therefore, the atomic number of element $M$ is $13$.
The mass number is the sum of the number of protons and neutrons.
Mass number = $\text{Atomic number} + \text{Number of neutrons} = 13 + 14 = 27$.

(A) $Na^+$ has completely filled $K$ and $L$ shells.
$Na$ atom has an atomic number of $11$,with an electronic configuration of $(2, 8, 1)$.
When $Na$ loses one electron from its outermost $M$ shell to form $Na^+$,its electronic configuration becomes $(2, 8)$.
Here,the $K$ shell contains $2$ electrons and the $L$ shell contains $8$ electrons,both of which are completely filled.
In contrast,$He$ (atomic number $2$) has only the $K$ shell,which contains $2$ electrons,and it does not possess an $L$ shell.

(N/A) The atomic number of element $B$ is $16$. Its electronic configuration is $2, 8, 6$.
$(b)$ The atomic number of element $A$ is $9$. Its electronic configuration is $2, 7$. To complete its octet,it gains $1$ electron. Therefore,its valency is $1$.
$(c)$ The atomic number of an element is equal to the number of protons in its nucleus. For element $B$,the number of protons is $16$,so its atomic number is $16$.
$(d)$ The mass number of an element is the sum of the number of protons and neutrons. For element $D$,mass number $= 17 + 22 = 39$.

(N/A) Niels Bohr proposed the following postulates to address the limitations of Rutherford's atomic model:
$(a)$ Only certain special orbits,known as discrete orbits of electrons,are allowed inside the atom.
$(b)$ While revolving in these discrete orbits,the electrons do not radiate energy. These orbits or shells are referred to as energy levels.

(C) Rutherford performed the $\alpha$-particle scattering experiment by bombarding a thin gold foil with high-energy $\alpha$-particles.
He observed that while most $\alpha$-particles passed straight through the foil,a very small fraction was deflected at large angles,and some were even bounced back by $180^{\circ}$.
Since $\alpha$-particles are positively charged,their repulsion and deflection by the central part of the atom (the nucleus) indicated that the nucleus contains a concentrated positive charge.

(A) The average atomic mass is calculated by taking the weighted average of the masses of the isotopes based on their relative abundance.
Formula: $\text{Average Atomic Mass} = \frac{(M_1 \times P_1) + (M_2 \times P_2) + (M_3 \times P_3)}{100}$
Given:
Isotope $1$: Mass $= 54$,Abundance $= 5\%$
Isotope $2$: Mass $= 56$,Abundance $= 90\%$
Isotope $3$: Mass $= 57$,Abundance $= 5\%$
Calculation:
$\text{Average Atomic Mass} = \frac{(54 \times 5) + (56 \times 90) + (57 \times 5)}{100}$
$= \frac{270 + 5040 + 285}{100}$
$= \frac{5595}{100} = 55.95 \ u$

(N/A) $(a) (i)$ For $_{12}^{24} Mg$: Protons $(P) = 12$,Neutrons $(N) = 24 - 12 = 12$.
For $_{12}^{26} Mg$: Protons $(P) = 12$,Neutrons $(N) = 26 - 12 = 14$.
$(ii)$ Both isotopes have the same atomic number $(12)$,so their electronic configuration is $2, 8, 2$ and their valency is $2$.
$(b)$ Isotopes have different mass numbers because they possess a different number of neutrons in their nuclei,even though the number of protons remains the same.

(N/A) Canal rays are positively charged radiations produced in a gas discharge tube. They were discovered by $E. \text{ Goldstein}$. The charge of canal rays is positive,and their mass is approximately equal to the mass of a proton $(1 \text{ atomic mass unit})$.
$(b)$ Canal rays are positively charged,whereas electrons are negatively charged. The mass of an electron is extremely small,approximately $1/1837$ (or often approximated as $1/2000$) of the mass of a proton (canal ray).

(N/A)

Feature	Electron, Proton, and Neutron Differences
$1.$ Charge	Electron has a negative charge, Proton has a positive charge, and Neutron has no charge.
$2.$ Location	Electron is present in orbits around the nucleus, whereas both Proton and Neutron are present inside the nucleus.
$3.$ Mass	The mass of an electron is negligible ($\frac{1}{2000}$ times that of a proton), while the mass of both proton and neutron is approximately $1$ atomic mass unit $(amu)$.

(N/A) In the alpha particles scattering experiment,Rutherford bombarded a thin gold foil with fast moving alpha particles and observed that:
$(i)$ Most of the $\alpha$-particles passed straight through the gold foil.
$(ii)$ Some of the $\alpha$-particles were deflected by the foil through small angles.
$(iii)$ One out of every $12,000$ particles appeared to rebound.

(N/A) $(i)$ The maximum number of electrons present in a shell is given by the formula $2n^2$,where $n$ is the orbit number or energy level index $(1, 2, 3, ...)$.
$(ii)$ The outermost shell of an atom can accommodate a maximum of $8$ electrons.
$(iii)$ Electrons are not accommodated in a given shell unless the inner shells are completely filled; that is,the shells are filled in a stepwise manner.

(N/A) $(i)$ Some elements possess fractional atomic mass because they occur in nature in different isotopic forms. The atomic mass of an element is calculated as the weighted average of the masses of its naturally occurring isotopes.
$(ii)$ Isotopes of an element have similar chemical properties because they possess the same atomic number and the same number of valence electrons,which determine the chemical behavior of an atom.
$(iii)$ Noble gases are inert because the outermost shell of their atoms is completely filled with electrons (stable octet/duplet configuration). Consequently,they do not show affinity towards chemical reactions.

(N/A) Bohr$-$Bury Rules:
$(i)$ The maximum number of electrons present in a shell is given by the formula $2n^2$,where $n$ is the orbit number or shell number.
$(ii)$ The maximum number of electrons that can be accommodated in the outermost orbit is $8$.
$(iii)$ Electrons are not accommodated in a given shell unless the inner shells are filled. The shells are filled in a step-wise manner.
$(b)$ The atomic number of sulphur $(S)$ is $16$. Its electronic configuration is $2, 8, 6$.
Schematic diagram of the sulphur atom:
The atom has a nucleus at the center,with $2$ electrons in the $K$-shell,$8$ electrons in the $L$-shell,and $6$ electrons in the $M$-shell.

(N/A) Applications of any three isotopes are:
$(a)$ An isotope of Uranium $(U-235)$ is used as a fuel in nuclear reactors for generating electricity.
$(b)$ An isotope of Cobalt $(Co-60)$ is used in the treatment of cancer through radiotherapy.
$(c)$ An isotope of Iodine $(I-131)$ is used in the treatment of goiter,a disease related to the thyroid gland.

(N/A) Features of Bohr's Model of an Atom:
$(i)$ Electrons revolve in discrete orbits,which are associated with a fixed amount of energy. These are called energy levels or shells,designated as $K, L, M, N, ...$
$(ii)$ While revolving in these discrete orbits,electrons do not radiate energy.
$(iii)$ The energy of an orbit increases as its distance from the nucleus increases. The orbit closest to the nucleus has the lowest energy.
$(iv)$ When energy is supplied to an electron,it jumps to a higher energy level. When it returns to a lower energy level,it emits energy in the form of radiation.
$(b)$ $(i)$ $_{26}^{58} A$ and $_{28}^{58} B$ are isobars because they have the same mass number $(58)$ but different atomic numbers ($26$ and $28$).
$(ii)$ $_{35}^{79} X$ and $_{35}^{80} Y$ are isotopes because they have the same atomic number $(35)$ but different mass numbers ($79$ and $80$).
$(c)$ Element $B$ (atomic number $16$) is more reactive than element $A$ (atomic number $18$). The electronic configuration of $A$ is $2, 8, 8$,which is a stable octet (noble gas configuration). The electronic configuration of $B$ is $2, 8, 6$,which requires two more electrons to complete its octet,making it chemically reactive.

(N/A) Rutherford performed an experiment by bombarding a very thin gold foil with high-energy $\alpha$-particles. The observations were as follows:
$(i)$ Most of the fast-moving $\alpha$-particles passed straight through the gold foil without any deflection.
$(ii)$ Some of the $\alpha$-particles were deflected by the foil by small angles.
$(iii)$ Out of every $12000$ particles,one appeared to rebound (deflected by $180^{\circ}$).
Based on these observations,the following conclusions were drawn:
$(i)$ There is a positively charged center in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
$(ii)$ The electrons revolve around the nucleus in well-defined circular orbits.
$(iii)$ The size of the nucleus is very small compared to the total size of the atom.

(N/A) $(i)$ There is a positively charged center in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
$(ii)$ The electrons revolve around the nucleus in well-defined orbits.
$(iii)$ The size of the nucleus is very small as compared to the size of the atom.

(N/A) The three isotopes of hydrogen are:
$1.$ ${ }_{1}^{1} H$ - Protium
$2.$ ${ }_{1}^{2} H$ - Deuterium
$3.$ ${ }_{1}^{3} H$ - Tritium
$(b)$ An isotope of uranium,${ }_{92}^{235} U$,is used as a fuel in nuclear reactors to generate electricity.
$(c)$ Isotopes of an element have similar chemical properties because they have the same atomic number and the same number of valence electrons,which determines their chemical behavior.

(A) Let the percentage composition of $_{17}^{37}y$ be $x$.
Then the percentage composition of $_{17}^{35}y$ will be $100-x$.
Now,the average atomic mass is calculated as:
$\frac{x}{100} \times 37 + \frac{100-x}{100} \times 35 = 35.5$
$\frac{37x + 3500 - 35x}{100} = 35.5$
$2x + 3500 = 3550$
$2x = 50$
$x = 25\%$
Therefore,the percentage of $_{17}^{37}y$ is $25\%$ and the percentage of $_{17}^{35}y$ is $100 - 25 = 75\%$.

(N/A) $(i)$ The postulates of the Bohr model of an atom are as follows:
- Only certain special orbits,known as discrete orbits of electrons,are allowed inside the atom.
- While revolving in these discrete orbits,the electrons do not radiate energy.
$(ii)$ For an atom with atomic number $15$ (Phosphorus),the electronic configuration is $2, 8, 5$. The sketch shows the nucleus at the center,surrounded by three shells $(K, L, M)$: the $K$-shell contains $2$ electrons,the $L$-shell contains $8$ electrons,and the $M$-shell contains $5$ electrons.

(N/A) Bohr$-$Bury Rules:
$(i)$ The maximum number of electrons present in a shell is given by the formula $2n^2$,where $n$ is the shell number.
$(ii)$ The maximum number of electrons that can be accommodated in the outermost orbit is $8$.
$(iii)$ Electrons are not accommodated in a given shell unless the inner shells are filled.
$(b)$ Electronic structure:
$(i)$ For element $X$ (Atomic number $17$): Electronic configuration is $2, 8, 7$. It has $2$ electrons in the $K$ shell,$8$ in the $L$ shell,and $7$ in the $M$ shell.
$(ii)$ For element $Y$ (Atomic number $16$): Electronic configuration is $2, 8, 6$. It has $2$ electrons in the $K$ shell,$8$ in the $L$ shell,and $6$ in the $M$ shell.

(N/A) Rutherford selected gold for his experiment because it is highly malleable and can be beaten into a very thin foil.
$(b)$ He used $\alpha$-particles,which are doubly charged helium ions $(He^{2+})$.
$(c)$ The major drawback of Rutherford's model was that it could not explain the stability of the atom. According to classical electromagnetic theory,an accelerating charged particle should emit energy; thus,electrons revolving around the nucleus should lose energy and eventually fall into the nucleus,making the atom unstable.

(N/A) The electronic configuration is determined by the atomic number $(Z)$:
Sulphur $(S)$: Atomic number $Z = 16$. Electronic configuration: $2, 8, 6$.
Oxygen $(O)$: Atomic number $Z = 8$. Electronic configuration: $2, 6$.
$(b)$ Valency is the combining capacity of an element.
For Oxygen $(O)$: It needs $2$ electrons to complete its octet,so its valency is $2$.
For Sulphur $(S)$: It needs $2$ electrons to complete its octet,so its valency is $2$.
$(c)$ No,they are not isotopes. Isotopes are atoms of the same element having the same atomic number but different mass numbers. Since sulphur and oxygen have different atomic numbers ($16$ and $8$ respectively),they are different elements.

(D) The three subatomic particles of an atom are electrons,protons,and neutrons.
$(b) (i)$ The $K$ shell is the first shell and can hold $2$ electrons. The $L$ shell is the second shell and has $5$ electrons. Therefore,the total number of electrons is $2 + 5 = 7$. Since the atom is neutral,the atomic number is equal to the number of electrons,which is $7$.
$(ii)$ The valency is determined by the number of electrons required to complete the octet. Since there are $5$ electrons in the outermost shell ($L$ shell),it needs $3$ more electrons to complete the octet. Thus,its valency is $3$ (or $-3$ as it gains $3$ electrons).
$(iii)$ The element with atomic number $7$ is Nitrogen $(N)$.

(N/A) $(i)$ Isotopes are atoms of the same element having the same atomic number but different mass numbers. Example: Hydrogen has three isotopes: Protium $(_{1}^{1}H)$,Deuterium $(_{1}^{2}H)$,and Tritium $(_{1}^{3}H)$.
$(ii)$ Isobars are atoms of different elements having the same mass number but different atomic numbers. Example: Argon $(_{18}^{40}Ar)$ and Calcium $(_{20}^{40}Ca)$ are isobars as both have a mass number of $40$.
$(b)$ $(i)$ The isotope of Uranium $(U-235)$ is used as a fuel in nuclear reactors.
$(ii)$ The isotope of Cobalt $(Co-60)$ is used in the treatment of cancer.

(N/A) For an element with atomic number $Z = 20$ and mass number $A = 40$:
Number of protons $= Z = 20$
Number of electrons $= Z = 20$
Number of neutrons $= A - Z = 40 - 20 = 20$
Electronic configuration: $K=2, L=8, M=8, N=2$.
The structure of the atom consists of a nucleus containing $20$ protons and $20$ neutrons,surrounded by four shells $(K, L, M, N)$ containing $2, 8, 8,$ and $2$ electrons respectively.
$(b)$ An atom with complete $K$ and $L$ shells has an electronic configuration of $2, 8$. The total number of electrons is $10$,which corresponds to the element Neon $(Ne)$. Since its outermost shell ($L$-shell) is completely filled (octet is complete),it is chemically inert and is classified as a noble gas.

(A) Electronic configurations $Y: 3, 8, 2$ and $Z: 2, 8, 9$ are not possible. According to the Bohr-Bury scheme,the first shell ($K$-shell) can hold a maximum of $2$ electrons,and the outermost shell of an atom can hold a maximum of $8$ electrons.
$(b)$ The electronic configurations and valencies are as follows:

Element	Electronic Configuration $(K, L, M)$	Valency
Fluorine $(9)$	$2, 7$	$1$
Aluminium $(13)$	$2, 8, 3$	$3$
Argon $(18)$	$2, 8, 8$	$0$