Atomic Number, Mass Number, Isotopes and Isobars Questions in English

(B) The number of neutrons in an atom is calculated by subtracting the number of protons from the atomic mass.
Number of neutrons $=$ Atomic mass $-$ Number of protons
Given that the atomic mass is $4 \, u$ and the number of protons is $2$.
Therefore,the number of neutrons $= 4 - 2 = 2$.

(A) $(i)$ The atomic number of an element is defined as the total number of protons present in the nucleus of its atom. Since the number of protons is $8$,the atomic number is $8$.
$(ii)$ The charge on an atom is determined by the net difference between the number of protons (positive charge) and electrons (negative charge). Since the number of protons $(8)$ equals the number of electrons $(8)$,the net charge is $8 - 8 = 0$. Thus,the atom is electrically neutral.

(A) The mass number of an atom is defined as the sum of the number of protons and the number of neutrons in its nucleus.
For Oxygen $(O)$:
Atomic number $= 8$ (protons $= 8$)
Number of neutrons $= 8$
Mass number $= 8 + 8 = 16$
For Sulphur $(S)$:
Atomic number $= 16$ (protons $= 16$)
Number of neutrons $= 16$
Mass number $= 16 + 16 = 32$
Therefore,the mass numbers are $16$ and $32$ respectively.

(N/A) The isotopes of hydrogen are Protium $(H)$,Deuterium $(D)$,and Tritium $(T)$. All isotopes of an element have the same number of protons and electrons,but differ in the number of neutrons. The sub-atomic particles for each are as follows:

Symbol	Protons,Neutrons,Electrons
$H$ (Protium)	$1$ Proton,$0$ Neutron,$1$ Electron
$D$ (Deuterium)	$1$ Proton,$1$ Neutron,$1$ Electron
$T$ (Tritium)	$1$ Proton,$2$ Neutrons,$1$ Electron

(N/A) Isotopes are atoms of the same element having the same atomic number but different mass numbers. For example,Carbon isotopes: ${}^{12}C_{6}$ and ${}^{14}C_{6}$. Both have an atomic number of $6$,so their electronic configuration is $(2, 4)$.
Isobars are atoms of different elements having the same mass number but different atomic numbers. For example,${}^{22}Ne_{10}$ and ${}^{22}Na_{11}$.
Electronic configuration of ${}^{22}Ne_{10}$ $(Z=10)$: $(2, 8)$.
Electronic configuration of ${}^{22}Na_{11}$ $(Z=11)$: $(2, 8, 1)$.

(A) $(i)$ Atomic number: The atomic number of an element is defined as the total number of protons present in the nucleus of an atom of that element. It is denoted by $Z$. For example,a nitrogen atom contains $7$ protons,so its atomic number is $7$.
$(ii)$ Mass number: The mass number of an element is the sum of the total number of protons and neutrons present in the nucleus of an atom of that element. It is denoted by $A$. For example,a boron atom has $5$ protons and $6$ neutrons,so its mass number is $5 + 6 = 11$.
Uses of isotopes:
$1$. An isotope of uranium is used as a fuel in nuclear reactors.
$2$. An isotope of iodine is used in the treatment of goitre.

(N/A) $(i)$ Isotopes: These are atoms of the same element having the same atomic number,but different mass numbers. For example,chlorine has two isotopes with atomic number $17$ but mass numbers $35$ and $37$ represented by $_{17}^{35}Cl$ and $_{17}^{37}Cl$.
$(ii)$ Isobars: These are atoms of different elements having the same mass number,but different atomic numbers. For example,Neon $(Ne)$ has atomic number $10$ and Sodium $(Na)$ has atomic number $11$,but both have a mass number of $22$,represented by $_{10}^{22}Ne$ and $_{11}^{22}Na$.
Two uses of isotopes:
$\to$ One isotope of Uranium is used as a fuel in nuclear reactors.
$\to$ One isotope of Cobalt is used in the treatment of cancer.

(C) Let the percentage of isotope $_8^{18}X$ be $y \%$. Then,the percentage of isotope $_8^{16}X$ is $(100 - y) \%$.
The average atomic mass is given by the formula:
$\text{Average atomic mass} = \frac{(16 \times (100 - y)) + (18 \times y)}{100} = 16.2$
Multiplying by $100$ on both sides:
$1600 - 16y + 18y = 1620$
Simplifying the equation:
$2y = 1620 - 1600$
$2y = 20$
$y = 10$
Thus,the percentage of isotope $_8^{18}X$ is $10 \%$ and the percentage of isotope $_8^{16}X$ is $(100 - 10) \% = 90 \%$.

(D) Mass number of $X = \text{Number of protons} + \text{Number of neutrons} = 6 + 6 = 12$.
Mass number of $Y = \text{Number of protons} + \text{Number of neutrons} = 6 + 8 = 14$.
These two atomic species $X$ and $Y$ have the same atomic number (number of protons $= 6$),but different mass numbers ($12$ and $14$). Hence,they are isotopes.

(D) Isotopes are defined as atoms of the same element that have the same atomic number but different mass numbers. Since the atomic number is the same (number of protons),the number of protons is identical. However,because the mass number is the sum of protons and neutrons,a difference in mass number implies a different number of neutrons in the nucleus. Therefore,isotopes of an element have a different number of neutrons.

(C) The atomic number $(Z)$ of an element is equal to the number of protons, which is equal to the number of electrons in a neutral atom. Here, the number of electrons is $15$, so $Z = 15$.
The mass number $(A)$ is the sum of the number of protons and neutrons. $A = Z + \text{number of neutrons} = 15 + 16 = 31$.
The standard representation of an element is written as $_{Z}^{A}X$, where $Z$ is the atomic number (subscript) and $A$ is the mass number (superscript).
Therefore, the correct representation is $_{15}^{31}X$.

(B) For a neutral atom,the atomic number is defined as the number of protons present in the nucleus,which is also equal to the number of electrons orbiting the nucleus. Thus,statement $(iv)$ is correct.
The mass number of an element is defined as the sum of the number of protons and the number of neutrons present in the nucleus of an atom. Thus,statement $(ii)$ is correct.
Statement $(i)$ is incorrect because the atomic number is not the sum of protons and electrons.
Statement $(iii)$ is incorrect because atomic mass is not equal to the number of protons or neutrons.
Therefore,statements $(ii)$ and $(iv)$ are true.

(A) The mass number $(A)$ is the sum of protons $(p)$ and neutrons $(n)$. Given $A = 27$ and $n = 14$.
Therefore,the number of protons (atomic number) is $p = A - n = 27 - 14 = 13$.
In a neutral atom,the number of electrons equals the number of protons,so the neutral atom has $13$ electrons.
The ion has $3$ positive charges,which means it has lost $3$ electrons.
Thus,the number of electrons in the ion $= 13 - 3 = 10$.

(C) Isotopes are atoms of the same element that have the same atomic number (number of protons and electrons) but different mass numbers.
The difference in mass number arises because isotopes have a different number of neutrons in their nuclei.
Since oxygen atoms in a sample can exist as different isotopes (e.g.,$^{16}O$,$^{17}O$,or $^{18}O$),they will have the same number of electrons but a different number of neutrons.

(N/A) Yes,it is possible. The atom of the element hydrogen $(H)$ consists of one proton and one electron,but it contains no neutrons. It is represented as $_1^1H$.

(A) The fact that atoms are divisible is supported by the following observations:
$1$. Discovery of subatomic particles: The discovery of electrons by $J$.$J$. Thomson and protons by $E$. Goldstein proved that atoms are composed of even smaller particles.
$2$. Existence of isotopes and isobars: The existence of different forms of the same element (isotopes) and different elements with the same mass number (isobars) suggests that the internal structure of the atom is complex and divisible.

(B) The atoms $^{35}Cl$ and $^{37}Cl$ are isotopes of chlorine.
Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
Since they have the same atomic number $(Z = 17)$,they possess the same number of electrons ($17$ electrons) and the same electronic configuration $(2, 8, 7)$.
Valency is determined by the number of valence electrons in the outermost shell.
Since both isotopes have the same number of valence electrons,their valencies are identical.

(N/A) The student's statement is incorrect.
$1$. In a neutral atom,the number of protons is always equal to the number of electrons.
$2$. The number of protons is not necessarily greater than the number of neutrons. In many stable isotopes,the number of neutrons is equal to or greater than the number of protons.
$3$. Therefore,the proposed relationship (protons > neutrons > electrons) is scientifically invalid.

(A) The mass number $(A)$ of an element is the sum of the number of protons and neutrons in its nucleus.
Given the representation $^{31}_{15}X$,the mass number $A = 31$ and the atomic number $Z$ (number of protons) $= 15$.
The formula for the number of neutrons $(n)$ is: $n = A - Z$.
Substituting the values: $n = 31 - 15 = 16$.
Therefore,the number of neutrons present in the nucleus is $16$.

(B) Elements that have the same mass number but different atomic numbers are known as isobars.
In this case,both calcium $(Ca)$ and argon $(Ar)$ have a mass number of $40$,but their atomic numbers are $20$ and $18$ respectively.
Therefore,they are classified as isobars.

(N/A) To find the number of protons $(n_{(p)})$ and neutrons $(n_{(n)})$, we use the notation ${ }_{Z}^{A} X$, where $Z$ is the atomic number (number of protons) and $A$ is the mass number (protons + neutrons).
$1$. For ${ }_{17}^{35} Cl$: $n_{(p)} = Z = 17$. $n_{(n)} = A - Z = 35 - 17 = 18$.
$2$. For ${ }_{6}^{12} C$: $n_{(p)} = Z = 6$. $n_{(n)} = A - Z = 12 - 6 = 6$.
$3$. For ${ }_{35}^{81} Br$: $n_{(p)} = Z = 35$. $n_{(n)} = A - Z = 81 - 35 = 46$.

Element	$n_{(p)}$	$n_{(n)}$
$Cl$	$17$	$18$
$C$	$6$	$6$
$Br$	$35$	$46$

(N/A) To complete the table,we use the following relationships:
$1$. $\text{Atomic Number} (Z) = \text{Number of Protons} = \text{Number of Electrons (in a neutral atom)}$.
$2$. $\text{Mass Number} (A) = \text{Number of Protons} + \text{Number of Neutrons}$.

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
$9$	$19$	$10$	$9$	$9$	Fluorine
$16$	$32$	$16$	$16$	$16$	Sulphur
$12$	$24$	$12$	$12$	$12$	Magnesium
$1$	$2$	$1$	$1$	$1$	Deuterium
$1$	$1$	$0$	$1$	$0$	Hydrogen ion

(B) The mass of an atom is primarily determined by the sum of the number of protons and neutrons present in the nucleus.
These particles are collectively known as nucleons.
Electrons have negligible mass compared to protons and neutrons,so they do not contribute significantly to the atomic mass.

(N/A) Elements exist in nature as a mixture of different isotopes,each with a specific relative abundance. Because these isotopes have different atomic masses,a single atomic mass value cannot represent the element accurately. Therefore,the average atomic mass is calculated by taking the weighted average of the masses of all naturally occurring isotopes based on their percentage abundance.

(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 $Neutrons$.

(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).

(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) 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)

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 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 $= \text{Number of protons} + \text{Number of neutrons} = 8 + 8 = 16$.

(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 $= 17 + 22 = 39$.

(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) $(i)$ 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) 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) 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) $(i)$ 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.

(A) $(i)$ Element $A$ (Chlorine) has an atomic number of $17$,and its electronic configuration is $(2, 8, 7)$. It needs to gain $1$ electron to complete its octet.
$(ii)$ Element $B$ (Sulfur) has an atomic number of $16$,and its electronic configuration is $(2, 8, 6)$. It needs to gain $2$ electrons to complete its octet.
$(iii)$ Reactivity in non-metals is determined by the ease of gaining electrons. Since element $A$ requires only $1$ electron to achieve a stable noble gas configuration,it is more chemically reactive than element $B$,which requires $2$ electrons.

(D) The species $K$ (Potassium atom) is electrically neutral.
An atom is considered electrically neutral when the number of positively charged protons in its nucleus is exactly equal to the number of negatively charged electrons orbiting the nucleus.
In the case of $K^{+}$,$Cl^{-}$,$S^{2-}$,and $P^{3-}$,these are ions that have either lost or gained electrons,resulting in a net electrical charge.
Therefore,$K$ is the only neutral species among the given options.

(N/A) In a chlorine atom,the number of electrons is equal to the number of protons,which is $17$; hence,it is electrically neutral.
In a chloride ion,one electron is gained in the valence shell to attain a stable octet configuration. Consequently,the number of protons remains $17$,but the number of electrons becomes $18$ $(17 + 1)$. Since there is one excess electron compared to the number of protons,it carries a net charge of $-1$ and is thus negatively charged.

(D) Atoms of different elements with different atomic numbers but the same mass number are called isobars.
$(b)$ $(i)$ The electronic configuration of $Y$ $(Z=17)$ is $2, 8, 7$.
$(ii)$ The number of valence electrons in $Y$ is $7$.
$(iii)$ To complete the octet,$Y$ needs $8 - 7 = 1$ electron.
$(iv)$ Since $Y$ has $7$ valence electrons,it tends to gain $1$ electron to complete its octet,making it a non-metal.
$(c)$ The electronic configuration of $Na$ $(Z=11)$ is $2, 8, 1$. To achieve a stable noble gas configuration,it is energetically easier for $Na$ to lose its $1$ valence electron than to gain $7$ electrons. Therefore,its valency is $1$ and not $7$.

(D) $(i)$ The electronic configuration is $K=2, L=8, M=2$. The total number of electrons is $2 + 8 + 2 = 12$. Therefore, the atomic number is $12$.
$(ii)$ The element with atomic number $12$ is Magnesium $(Mg)$.
$(iii)$ Since it has $2$ electrons in its outermost shell, it loses $2$ electrons to achieve a stable octet. Thus, its valency is $2$.
$(iv)$ The number of neutrons is calculated as: $\text{Number of neutrons} = \text{Mass number} - \text{Atomic number} = 24 - 12 = 12$.

(N/A) Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
For uranium $(U)$,the atomic number is $92$.
The two common isotopes of uranium are:
$1$. Uranium-$238$: represented as ${ }_{92}^{238} U$.
$2$. Uranium-$235$: represented as ${ }_{92}^{235} U$.