Atomic number, Mass number, Atomic species Questions in English

(C) An atom is electrically neutral because the total positive charge of protons is balanced by the total negative charge of electrons.
An atom contains an equal number of protons and electrons,not necessarily protons and neutrons.
Therefore,the Assertion is correct,but the Reason is incorrect.

(D) Isobars are defined as atoms of different elements that have the same mass number $(A)$ but different atomic numbers $(Z)$.
The mass number $(A)$ is defined as the sum of the number of protons and neutrons in the nucleus.
Since isobars have the same mass number,the sum of protons and neutrons is the same for them,not different.
Therefore,the Assertion is incorrect,and the Reason is correct.

(D) Atoms are electrically neutral because the number of electrons (negatively charged) and protons (positively charged) are equal in a neutral atom. Therefore,both the Assertion and the Reason are incorrect.

(A) For the neutral atom $_{35}^{80}Br$:
$Z = 35$ (Atomic number)
$A = 80$ (Mass number)
Number of protons $= Z = 35$
Number of electrons $= Z = 35$ (since the atom is neutral)
Number of neutrons $= A - Z = 80 - 35 = 45$
Therefore,the number of protons,neutrons,and electrons are $35, 45,$ and $35$ respectively.

(A) The atomic number $(Z)$ is equal to the number of protons $= 16$. The element with atomic number $16$ is sulphur $(S)$.
Atomic mass number $(A) =$ number of protons $+$ number of neutrons $= 16 + 16 = 32$.
The species is not neutral because the number of protons $(16)$ is not equal to the number of electrons $(18)$.
Since the number of electrons is greater than the number of protons,it is an anion (negatively charged) with a charge equal to the excess electrons $= 18 - 16 = 2$.
Therefore,the symbol is $_{16}^{32} S^{2-}$.

For any nucleus $_{Z}^{A}X$,the number of protons is equal to the atomic number $(Z)$,and the number of neutrons is equal to the mass number $(A)$ minus the atomic number $(Z)$: $n = A - Z$.
$1. _{6}^{13}C$: Protons = $6$,Neutrons = $13 - 6 = 7$.
$2. _{8}^{16}O$: Protons = $8$,Neutrons = $16 - 8 = 8$.
$3. _{12}^{24}Mg$: Protons = $12$,Neutrons = $24 - 12 = 12$.
$4. _{26}^{56}Fe$: Protons = $26$,Neutrons = $56 - 26 = 30$.
$5. _{38}^{88}Sr$: Protons = $38$,Neutrons = $88 - 38 = 50$.

(N/A) The general representation of an atom is given by $_{Z}^{A}X$,where $X$ is the chemical symbol of the element,$Z$ is the atomic number,and $A$ is the atomic mass.
$(i)$ For $Z=17$,the element is Chlorine $(Cl)$. Thus,the symbol is $_{17}^{35}Cl$.
$(ii)$ For $Z=92$,the element is Uranium $(U)$. Thus,the symbol is $_{92}^{233}U$.
$(iii)$ For $Z=4$,the element is Beryllium $(Be)$. Thus,the symbol is $_{4}^{9}Be$.

(N/A) The general convention of representing an element along with its atomic mass $(A)$ and atomic number $(Z)$ is $_{Z}^{A} X$.
Hence,$_{35}^{79} Br$ is acceptable because $Z = 35$ and $A = 79$,whereas $_{79}^{35} Br$ is not acceptable as the atomic number of Bromine is $35$,not $79$.
$^{79} Br$ is acceptable as it specifies the isotope,but $^{35} Br$ is not acceptable because the atomic number $35$ is already fixed for Bromine,and the notation should represent the mass number to distinguish isotopes.

(D) Let the number of protons in the element be $x$.
$\therefore$ Number of neutrons in the element
$= x + 31.7\% \text{ of } x$
$= x + 0.317x = 1.317x$
According to the question,the mass number is $81$.
$\therefore \text{Number of protons} + \text{Number of neutrons} = 81$
$x + 1.317x = 81$
$2.317x = 81$
$x = \frac{81}{2.317} \approx 34.95 \approx 35$
Since the atomic number is equal to the number of protons,the atomic number is $35$.
The element with atomic number $35$ is Bromine $(Br)$.
$\therefore$ The atomic symbol is $_{35}^{81}Br$.

(A) Let the number of electrons in the ion be $x$.
Since the ion has one unit of negative charge,the number of electrons is one more than the number of protons $(Z)$.
So,$Z = x - 1$.
Number of neutrons $(n)$ $= x + 0.111 \, x = 1.111 \, x$.
Mass number $(A)$ $= Z + n = 37$.
Substituting the values: $(x - 1) + 1.111 \, x = 37$.
$2.111 \, x = 38$.
$x \approx 18$.
Therefore,$Z = 18 - 1 = 17$.
The element with atomic number $17$ is Chlorine $(Cl)$.
Thus,the symbol of the ion is $_{17}^{37}Cl^{-}$.

(A) Let the number of electrons in the ion $A^{3+}$ be $x$.
Number of neutrons $= x + 0.304x = 1.304x$.
For a tripositive ion $A^{3+}$,the number of protons $= x + 3$.
Mass number $= \text{Protons} + \text{Neutrons} = 56$.
$(x + 3) + 1.304x = 56$.
$2.304x = 53$.
$x = 23$.
Number of protons $= 23 + 3 = 26$.
The element with atomic number $26$ is Iron $(Fe)$.
Thus,the symbol of the ion is $_{26}^{56} Fe^{3+}$.

(N/A) Atomic Number $(Z)$: The number of protons present in the nucleus of an atom is called the atomic number.
Nucleons: Protons and neutrons present in the nucleus are collectively known as nucleons.
Mass Number $(A)$: The total number of protons and neutrons present in the nucleus is called the mass number.
Relation: $A = Z + N$,where $N$ is the number of neutrons.

(N/A) The atomic number $(Z)$ is defined as the number of protons present in the nucleus of an atom.
For a neutral atom,the number of electrons is equal to the number of protons,which is equal to the atomic number $(Z)$.
For hydrogen: Number of protons $= 1$. Therefore,atomic number $(Z) = 1$ and number of electrons $= 1$.
For sodium: Number of protons $= 11$. Therefore,atomic number $(Z) = 11$ and number of electrons $= 11$.

(N/A) An atom of an element $X$ is represented by placing the mass number $A$ as a left superscript and the atomic number $Z$ as a left subscript.
The notation is written as ${ }_{Z}^{A}X$.
Here,$X$ is the element symbol,$A$ is the mass number,and $Z$ is the atomic number.
Example: For sodium $(Na)$,the atomic number $(Z)$ is $11$ and the mass number $(A)$ is $23$.
Therefore,the notation is ${ }_{11}^{23}Na$.

(N/A) Isobars: Isobars are atoms of different elements that have the same mass number but different atomic numbers.
Example: ${ }_{6}^{14}C$ and ${ }_{7}^{14}N$ both have a mass number of $14$.
Isotopes: Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
Example: Hydrogen has three isotopes: Protium $({ }_{1}^{1}H)$,Deuterium $({ }_{1}^{2}H)$,and Tritium $({ }_{1}^{3}H)$.

(N/A)

Atom	Isotope	Proton	Electron	Neutron
Hydrogen	$Protium \ ({ }_{1}^{1}H)$	$1$	$1$	$0$
Hydrogen	$Deuterium \ ({ }_{1}^{2}D)$	$1$	$1$	$1$
Hydrogen	$Tritium \ ({ }_{1}^{3}T)$	$1$	$1$	$2$
Carbon	${ }_{6}^{12}C$	$6$	$6$	$6$
Carbon	${ }_{6}^{13}C$	$6$	$6$	$7$
Carbon	${ }_{6}^{14}C$	$6$	$6$	$8$
Chlorine	${ }_{17}^{35}Cl$	$17$	$17$	$18$
Chlorine	${ }_{17}^{37}Cl$	$17$	$17$	$20$

(N/A) The number of protons is equal to the atomic number $(Z)$.
The number of neutrons is equal to the mass number $(A)$ minus the atomic number $(Z)$,i.e.,$n = A - Z$.
The number of electrons is equal to the atomic number $(Z)$ minus the charge on the species.

Species	Protons $(p)$,Electrons $(e^-)$,Neutrons $(n)$
$(i) \ _{15}^{31}P$	$p=15, e^-=15, n=16$
$(ii) \ _{11}^{23}Na^{+}$	$p=11, e^-=10, n=12$
$(iii) \ _{16}^{32}S^{2-}$	$p=16, e^-=18, n=16$
$(iv) \ _{101}^{235}X$	$p=101, e^-=101, n=134$
$(v) \ _{20}^{40}Ca^{2+}$	$p=20, e^-=18, n=20$
$(vi) \ _9^{19}F^{-}$	$p=9, e^-=10, n=10$

(N/A) $(i)$ The species has $8$ protons,so it is Oxygen $(O)$. With $10$ electrons and $8$ protons,it is $O^{2-}$. Mass number $= 8 + 8 = 16$. Symbol: ${ }_{8}^{16} O^{2-}$.
$(ii)$ Calcium $(Ca)$ with $Z = 20$ and mass number $40$. Symbol: ${ }_{20}^{40} Ca$.
$(iii)$ Magnesium $(Mg)$ with $Z = 12$,mass number $25$,and $10$ electrons. Symbol: ${ }_{12}^{25} Mg^{2+}$.
$(iv)$ Bromine $(Br)$ with $Z = 35$,$36$ electrons,and $45$ neutrons. Mass number $= 35 + 45 = 80$. Symbol: ${ }_{35}^{80} Br^{-}$.
$(v)$ $Al^{3+}$ has $13 - 3 = 10$ electrons. $A$ $+2$ ion with $10$ electrons must have $12$ protons $(Mg^{2+})$. $A$ $-2$ ion with $10$ electrons must have $8$ protons $(O^{2-})$. Symbols: ${ }_{12}^{25} Mg^{2+}$ and ${ }_{8}^{16} O^{2-}$.

(N/A) $(i)$ Isobars: Species with the same mass number $(A=40)$ but different atomic numbers.
$(ii)$ Isotopes: Species with the same atomic number $(Z=17)$ but different mass numbers $(A=35, 37)$.
$(iii)$ Isobars: All three species have the same mass number $(A=40)$ but different atomic numbers.
$(iv)$ Isotopes: Species with the same atomic number $(Z=38)$ but different mass numbers $(A=87, 90)$.

(N/A) The number of nucleons is equal to the mass number $(A)$ of the atom.
$(i)$ For $^{35}_{17}Cl$,the mass number is $35$. Therefore,the number of nucleons is $35$.
$(ii)$ For $^{40}_{19}K$,the mass number is $40$. Therefore,the number of nucleons is $40$.

(D) For $Cl$ $(Z=17, A=35)$: Protons $(p)$ = $Z = 17$. Electrons $(e^{-})$ = $Z = 17$. Neutrons $(n)$ = $A - Z = 35 - 17 = 18$.
For $Cl^{-}$: Protons $(p)$ = $17$. Electrons $(e^{-})$ = $17 + 1 = 18$. Neutrons $(n)$ = $18$.
For $Cl_{2}$: Protons $(p)$ = $2 \times 17 = 34$. Electrons $(e^{-})$ = $2 \times 17 = 34$. Neutrons $(n)$ = $2 \times 18 = 36$.

(N/A) For a neutral atom: Atomic number $= Z$,number of electrons $= e$,number of protons $= p$,number of neutrons $= n$,mass number $A = (p + n)$.

Symbol	Element Name	$Z$	$A$	$p$	$e$	$n$
${ }_{1}^{1} H$	Hydrogen	$1$	$1$	$1$	$1$	$0$
${ }_{2}^{4} He$	Helium	$2$	$4$	$2$	$2$	$2$
${ }_{3}^{7} Li$	Lithium	$3$	$7$	$3$	$3$	$4$
${ }_{4}^{9} Be$	Beryllium	$4$	$9$	$4$	$4$	$5$
${ }_{5}^{11} B$	Boron	$5$	$11$	$5$	$5$	$6$
${ }_{6}^{12} C$	Carbon	$6$	$12$	$6$	$6$	$6$
${ }_{7}^{14} N$	Nitrogen	$7$	$14$	$7$	$7$	$7$
${ }_{8}^{16} O$	Oxygen	$8$	$16$	$8$	$8$	$8$
${ }_{9}^{19} F$	Fluorine	$9$	$19$	$9$	$9$	$10$
${ }_{10}^{20} Ne$	Neon	$10$	$20$	$10$	$10$	$10$

(N/A) Isoelectronic species are atoms or ions that contain the same number of electrons.
For example,$O^{2-}, F^{-}, Na^{+}, Mg^{2+}$ all have $10$ electrons and are isoelectronic.

Ion and number of electrons	Isoelectronic species
$(i)$ $F^{-} (10)$	$O^{2-}$
$(ii)$ $Ar (18)$	$Cl^{-}$
$(iii)$ $Mg^{2+} (10)$	$Na^{+}$
$(iv)$ $Rb^{+} (36)$	$Kr$

(B) Mass number $(A)$ is the sum of the number of protons $(p)$ and neutrons $(n)$ present in the nucleus.
$A = n + p$
Given: $A = 13$,$n = 7$.
$p = A - n = 13 - 7 = 6$.
The atomic number $(Z)$ is equal to the number of protons $(p)$ in a neutral atom.
Therefore,the atomic number is $6$.

(A) Isobars are atoms with the same mass number but different atomic numbers.
Here,$_6^{14}C$ and $_7^{14}N$ have the same mass number $(14)$,so they are isobars.
Isotopes are atoms of the same element with the same atomic number but different mass numbers.
Here,$_6^{14}C, _6^{13}C,$ and $_6^{12}C$ have the same atomic number $(6)$,so they are isotopes.

(N/A) The atomic number of carbon is $Z = 6$.
The symbols of its three isotopes are ${ }_{6}^{12}C$,${ }_{6}^{13}C$,and ${ }_{6}^{14}C$.

(A) $_{17}^{35}Cl$ and $_{17}^{37}Cl$ are known as isotopes.
Isotopes are atoms of the same element having the same atomic number $(Z = 17)$ but different mass numbers ($A = 35$ and $A = 37$).
This difference in mass number is due to the presence of a different number of neutrons in their nuclei ($35 - 17 = 18$ neutrons in $_{17}^{35}Cl$ and $37 - 17 = 20$ neutrons in $_{17}^{37}Cl$).

(N/A) The isotopes of chlorine are ${ }_{17}^{35} Cl$ and ${ }_{17}^{37} Cl$.
Isotopes: ${ }_{17}^{35} Cl$ and ${ }_{17}^{37} Cl$.
Mass number: $35$ and $37$.
Number of neutrons: $18$ and $20$.
They differ in their mass numbers due to the difference in the number of neutrons present in their nuclei.

(A) species with $1$ electron is hydrogen-like.
$H$ (Atomic number $Z=1$) has $1$ electron.
$He^+$ (Atomic number $Z=2$) has $2-1 = 1$ electron.
$Li^{2+}$ (Atomic number $Z=3$) has $3-2 = 1$ electron.
$Be^{3+}$ (Atomic number $Z=4$) has $4-3 = 1$ electron.
Thus,the correct set is $H, He^+, Li^{2+}, Be^{3+}$.

(A, B) Isoelectronic species are those that contain the same number of electrons.
$1$. For $Na^{+}$ $(Z=11)$,number of electrons $= 11 - 1 = 10$.
$2$. For $Mg^{2+}$ $(Z=12)$,number of electrons $= 12 - 2 = 10$.
$3$. For $K^{+}$ $(Z=19)$,number of electrons $= 19 - 1 = 18$.
$4$. For $Ca^{2+}$ $(Z=20)$,number of electrons $= 20 - 2 = 18$.
$5$. For $S^{2-}$ $(Z=16)$,number of electrons $= 16 + 2 = 18$.
$6$. For $Ar$ $(Z=18)$,number of electrons $= 18$.
Thus,$Na^{+}$ and $Mg^{2+}$ are isoelectronic ($10$ electrons),and $K^{+}, Ca^{2+}, S^{2-},$ and $Ar$ are isoelectronic ($18$ electrons).

(N/A) $(i)$ True: Isobars are atoms of different elements with the same mass number but different atomic numbers.
$(ii)$ True: Isotones have the same number of neutrons. For $_{14}^{30}\text{Si}$,$n = 30 - 14 = 16$. For $_{16}^{32}\text{S}$,$n = 32 - 16 = 16$. Since both have $16$ neutrons,they are isotones.
$(iii)$ False: $J$.$J$. Thomson was the first to suggest that the atom is spherical.
$(iv)$ True: Rutherford's alpha-particle scattering experiment led to the discovery of the nucleus,establishing the nuclear model of the atom.

(N/A) $Mg$ is a neutral atom,while $Mg^{2+}$ is a positively charged cation.
The radius of $Mg$ is greater than the radius of $Mg^{2+}$ $(r_{Mg} > r_{Mg^{2+}})$.
The number of electrons in $Mg$ and $Mg^{2+}$ is different. $Mg^{2+}$ has two fewer electrons than $Mg$.
$Mg$ is the parent element,and $Mg^{2+}$ is the cation derived from it.
Similarity: Both $Mg$ and $Mg^{2+}$ have the same nuclear charge $(Z = 12)$. However,the effective nuclear charge experienced by the outermost electrons is higher in $Mg^{2+}$.

(A) $(i)$ For $Na$ and $Na^+$,the atomic number $Z = 11$,which is identical.
$(ii)$ For $F$ and $F^-$,the atomic number $Z = 9$,which is identical.

(B) Isoelectronic species are those that have the same number of electrons.
For $F^-$,the number of electrons is $9 + 1 = 10$.
For $Na^+$,the number of electrons is $11 - 1 = 10$.
Since both $F^-$ and $Na^+$ have $10$ electrons,they are isoelectronic.

(B) For the neutral atom ${ }_{71}^{175} Lu$:
The atomic number $(Z)$ is $71$,which represents the number of protons $(p^+)$.
Since the atom is neutral,the number of electrons $(e^-)$ is equal to the number of protons,so $e^- = 71$.
The mass number $(A)$ is $175$. The number of neutrons $(n^0)$ is calculated as $A - Z = 175 - 71 = 104$.
Therefore,the number of protons,neutrons,and electrons are $71, 104$,and $71$ respectively.

(D) Iso-electronic species are those that have the same number of electrons.
$O^{2-}$ has $8 + 2 = 10$ electrons; $F^{-}$ has $9 + 1 = 10$ electrons. (Iso-electronic)
$Na^{+}$ has $11 - 1 = 10$ electrons; $Mg^{2+}$ has $12 - 2 = 10$ electrons. (Iso-electronic)
$Mn^{2+}$ has $25 - 2 = 23$ electrons; $Fe^{3+}$ has $26 - 3 = 23$ electrons. (Iso-electronic)
$Fe^{2+}$ has $26 - 2 = 24$ electrons; $Mn^{2+}$ has $25 - 2 = 23$ electrons. (Not iso-electronic)
Therefore,the pair $Fe^{2+}, Mn^{2+}$ is not iso-electronic.

(D) Isoelectronic species are those that have the same number of electrons.
$Al^{3+}$ has $13 - 3 = 10$ electrons.
$O^{2-}$ has $8 + 2 = 10$ electrons.
$Mg^{2+}$ has $12 - 2 = 10$ electrons.
Since $O^{2-}$ and $Mg^{2+}$ both have $10$ electrons,they are isoelectronic with $Al^{3+}$.

(A) For the ion ${}_{22}^{48}X^{3-}$:
Atomic number $(Z)$ = $22$.
Mass number $(A)$ = $48$.
Number of neutrons = $A - Z = 48 - 22 = 26$.
Number of electrons = $Z + 3 = 22 + 3 = 25$.
We are given that the nucleus contains '$a$'$\%$ more neutrons than the number of electrons.
Percentage difference = $\frac{\text{Number of neutrons} - \text{Number of electrons}}{\text{Number of electrons}} \times 100$.
Percentage difference = $\frac{26 - 25}{25} \times 100 = \frac{1}{25} \times 100 = 4$.
Thus,the value of '$a$' is $4$.

(B) Isoelectronic species are those species which have the same number of electrons.
Total number of electrons in $O^{2-} = 8 + 2 = 10$.
Total number of electrons in $Zn^{2+} = 30 - 2 = 28$.
Total number of electrons in $Mg^{2+} = 12 - 2 = 10$.
Total number of electrons in $K^{+} = 19 - 1 = 18$.
Total number of electrons in $Ni^{2+} = 28 - 2 = 26$.
Thus,$O^{2-}$ is isoelectronic with $Mg^{2+}$.

(C) Isoelectronic species are those that have the same number of electrons.
$Na^{+}$ ($11-1 = 10$ electrons)
$Mg^{2+}$ ($12-2 = 10$ electrons)
$F^{-}$ ($9+1 = 10$ electrons)
$O^{2-}$ ($8+2 = 10$ electrons)
Since $Na^{+}, Mg^{2+}, F^{-},$ and $O^{2-}$ all contain $10$ electrons,they are isoelectronic.
Therefore,option $(c)$ is correct.

(B)
Total number of electrons in $S^{2-} = 16 + 2 = 18$.
Total number of electrons present in the ions given in options are as follows:
$(I)$ $Na^{+}$: Total electrons = $11 - 1 = 10$.
$(II)$ $Ca^{2+}$: Total electrons = $20 - 2 = 18$.
$(III)$ $Mg^{2+}$: Total electrons = $12 - 2 = 10$.
$(IV)$ $Sr^{2+}$: Total electrons = $38 - 2 = 36$.
Thus,$S^{2-}$ and $Ca^{2+}$ have the same number of electrons.

(C) The number of neutrons in an atom is calculated as $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
$A)$ $^{12}_{6}C$: $12 - 6 = 6$ neutrons; $^{24}_{12}Mg$: $24 - 12 = 12$ neutrons.
$B)$ $^{23}_{11}Na$: $23 - 11 = 12$ neutrons; $^{19}_{9}F$: $19 - 9 = 10$ neutrons.
$C)$ $^{23}_{11}Na$: $23 - 11 = 12$ neutrons; $^{24}_{12}Mg$: $24 - 12 = 12$ neutrons.
$D)$ $^{23}_{11}Na$: $23 - 11 = 12$ neutrons; $^{39}_{19}K$: $39 - 19 = 20$ neutrons.
Since $^{23}_{11}Na$ and $^{24}_{12}Mg$ both have $12$ neutrons,option $C$ is correct.

(C) The correct option is $(C)$.
For the ion ${}_{19}^{40} K^{+}$:
Atomic number $(Z) = 19$,Mass number $(A) = 40$.
Number of electrons in neutral $K$ atom $= 19$. Since it is a $K^{+}$ ion,it has lost $1$ electron,so number of electrons $= 19 - 1 = 18$.
Number of neutrons $= A - Z = 40 - 19 = 21$.
Sum of electrons and neutrons $= 18 + 21 = 39$.

(D) Isoelectronic species are those that have the same number of electrons.
Number of electrons in:
$K^{+} = 19 - 1 = 18$
$Cl^{-} = 17 + 1 = 18$
$Ca^{2+} = 20 - 2 = 18$
$Sc^{3+} = 21 - 3 = 18$
Since all these ions have $18$ electrons,they represent a collection of isoelectronic species.

(A) Isoelectronic species are those that have the same number of electrons.
$O^{2-} = 8 + 2 = 10 \text{ electrons}$
$F^{-} = 9 + 1 = 10 \text{ electrons}$
$Mg^{2+} = 12 - 2 = 10 \text{ electrons}$
$Na^{+} = 11 - 1 = 10 \text{ electrons}$
$Al^{3+} = 13 - 3 = 10 \text{ electrons}$
Other species:
$Al = 13 \text{ electrons}$
$O^{+} = 7 \text{ electrons}$
$Mg = 12 \text{ electrons}$
$F = 9 \text{ electrons}$
The isoelectronic species are $O^{2-}, F^{-}, Mg^{2+}, Na^{+}, \text{ and } Al^{3+}$.
Total count = $5$.

(B) $(i)$ Same atomic number $(Z)$,different mass number $(A)$ $\rightarrow$ Isotopes.
$(ii)$ Same mass number $(A)$,different atomic number $(Z)$ $\rightarrow$ Isobars.
$(iii)$ Same number of neutrons $(n = A - Z)$ $\rightarrow$ Isotones.
$(iv)$ Same difference between neutrons and protons $(n - p)$ $\rightarrow$ Isodiaphers.
$(v)$ Same atomic number $(Z)$ $\rightarrow$ Isotopes.

(A) Isoelectronic species are those that have the same number of electrons.
$1$. For $Ne$ $(Z=10)$: Electrons = $10$. For $O^{2-}$ $(Z=8)$: Electrons = $8 + 2 = 10$. Both have $10$ electrons,so they are isoelectronic.
$2$. For $Cl^{-}$ $(Z=17)$: Electrons = $17 + 1 = 18$. For $Ca$ $(Z=20)$: Electrons = $20$. Not isoelectronic.
$3$. For $Ar$ $(Z=18)$: Electrons = $18$. For $F^{-}$ $(Z=9)$: Electrons = $9 + 1 = 10$. Not isoelectronic.
$4$. For $K^{+}$ $(Z=19)$: Electrons = $19 - 1 = 18$. For $Al^{3+}$ $(Z=13)$: Electrons = $13 - 3 = 10$. Not isoelectronic.
Therefore,the correct pair is $Ne$ and $O^{2-}$.

(A) To find the number of electrons in each species:
$(a)$ $O^{2-}$: Oxygen has atomic number $8$. $O^{2-}$ has $8 + 2 = 10$ electrons. Thus,$(a)-(iii)$.
$(b)$ $Li^{2+}$: Lithium has atomic number $3$. $Li^{2+}$ has $3 - 2 = 1$ electron. Thus,$(b)-(iv)$.
$(c)$ $He$: Helium has atomic number $2$. It has $2$ electrons. Thus,$(c)-(ii)$.
$(d)$ $Ca^{2+}$: Calcium has atomic number $20$. $Ca^{2+}$ has $20 - 2 = 18$ electrons. Thus,$(d)-(i)$.
Therefore,the correct match is $a-iii, b-iv, c-ii, d-i$.

(C) The number of electrons in each species is as follows:
$Ne$ (Atomic number $10$): $10$ electrons.
$O^{2-}$ (Atomic number $8$): $8 + 2 = 10$ electrons.
$Na^{+}$ (Atomic number $11$): $11 - 1 = 10$ electrons.
$Na$ (Atomic number $11$): $11$ electrons.
Since $Ne$,$O^{2-}$,and $Na^{+}$ are isoelectronic (all have $10$ electrons),$Na$ is the species that does not have the same number of electrons.

(D) Isotopes are defined as atoms of the same element having the same atomic number $(Z)$ but different mass numbers $(A)$.
Since they have the same atomic number,they have the same number of protons and electrons,which leads to similar chemical properties.
They occupy the same position in the periodic table because the periodic table is arranged based on atomic number.
However,isotopes have different numbers of neutrons because their mass numbers $(A = Z + N)$ differ while their atomic numbers $(Z)$ remain the same.
Therefore,the statement that they have an equal number of neutrons is false.