Atomic number, Mass number, Atomic species Questions in English

(B) The number of protons $(Z)$ is calculated by subtracting the number of neutrons $(N)$ from the mass number $(A)$:
$Z = A - N = 40 - 21 = 19$.
Since the atomic number is $19$,the element is potassium $(K)$.
The standard representation of an element is ${ }_{Z}^{A} X$,where $A$ is the mass number and $Z$ is the atomic number.
Substituting the values,we get ${ }_{19}^{40} X$.
Therefore,the correct option is $(B)$.

(B) Isobars are defined as atoms of different elements that have the same mass number but different atomic numbers.
For example,${ }_{6}^{14}C$ and ${ }_{7}^{14}N$ both have a mass number of $14$ but different atomic numbers ($6$ and $7$ respectively).

(B) To determine the number of electrons in each species,we look at their atomic numbers $(Z)$:
$1$. For $K^{+}$: Atomic number of $K$ is $19$. Since it is a cation with a $+1$ charge,it has $19 - 1 = 18$ electrons.
$2$. For $Ca$: Atomic number of $Ca$ is $20$. As a neutral atom,it has $20$ electrons.
$3$. For $Mg$: Atomic number of $Mg$ is $12$. As a neutral atom,it has $12$ electrons.
$4$. For $Cl$: Atomic number of $Cl$ is $17$. As a neutral atom,it has $17$ electrons.
Therefore,the species containing $20$ electrons is $Ca$.

(B) Atoms and ions having the same number of electrons are isoelectronic.
Neon $(Ne)$ has an atomic number of $10$,so it has $10$ electrons.

Species	Number of electrons
Neon $(Ne)$	$10$
$O^{2-}$	$8 + 2 = 10$
$Na$	$11$
$Mg^{2+}$	$12 - 2 = 10$
$Al^{3+}$	$13 - 3 = 10$

Since $Na$ has $11$ electrons,it is not isoelectronic with neon.

(D) $Na^{+}$ has $10$ electrons $(11 - 1 = 10)$.
Among the given options,$Ne$ (Neon) has an atomic number of $10$,meaning it also has $10$ electrons.
Therefore,$Na^{+}$ and $Ne$ are isoelectronic.

(A) Isoelectronic species are atoms or ions that have the same number of electrons.
For $O^{2-}$: Atomic number of $O$ is $8$,so $O^{2-}$ has $8 + 2 = 10$ electrons.
For $Na^{+}$: Atomic number of $Na$ is $11$,so $Na^{+}$ has $11 - 1 = 10$ electrons.
Since both $O^{2-}$ and $Na^{+}$ contain $10$ electrons each,they are isoelectronic species.

(D) In a neutral atom,the number of electrons is equal to the number of protons.
Given that the atom has $29$ electrons,it must have $29$ protons.
The number of nucleons is the sum of the number of protons and the number of neutrons.
$\text{Number of nucleons} = \text{Number of protons} + \text{Number of neutrons} = 29 + 34 = 63$.

(A) Isobars are atoms of different elements that have the same mass number but different atomic numbers.
For ${ }_{20} Ca^{40}$,the mass number is $40$ and the atomic number is $20$.
Among the options,${ }_{18} Ar^{40}$ has the same mass number $(40)$ but a different atomic number $(18)$.
Therefore,${ }_{18} Ar^{40}$ is an isobar of ${ }_{20} Ca^{40}$.

(C) hydrogen-like species is an atom or ion that contains only one electron.
To determine this,we check the number of electrons in each species:
$A) \ Li^{2+}$: Lithium has $3$ electrons. $Li^{2+}$ has $3 - 2 = 1$ electron. It is hydrogen-like.
$B) \ Be^{3+}$: Beryllium has $4$ electrons. $Be^{3+}$ has $4 - 3 = 1$ electron. It is hydrogen-like.
$C) \ Li^{+}$: Lithium has $3$ electrons. $Li^{+}$ has $3 - 1 = 2$ electrons. It is $NOT$ hydrogen-like.
$D) \ He^{+}$: Helium has $2$ electrons. $He^{+}$ has $2 - 1 = 1$ electron. It is hydrogen-like.
Therefore,$Li^{+}$ is the correct answer.

(A) hydrogen-like species is an ion that contains only one electron.
$He$ (Helium) has an atomic number of $2$ and its electronic configuration is $1s^2$,meaning it has $2$ electrons.
$He^{+}$ has $1$ electron.
$Li^{2+}$ has $1$ electron.
$Be^{3+}$ has $1$ electron.
Therefore,$He$ is not a hydrogen-like species.

(B) For the ion ${ }_{16}^{32} S^{2-}$:
Number of protons $= Z = 16$.
Number of neutrons $= A - Z = 32 - 16 = 16$.
Number of electrons $= Z + (\text{charge magnitude}) = 16 + 2 = 18$.
Therefore,the number of protons,neutrons,and electrons are $16, 16, 18$ respectively.

(C) Isoelectronic species are those that have the same number of electrons.
$Mg$ has an atomic number of $12$,so $Mg^{2+}$ has $12 - 2 = 10$ electrons.
$Na$ has an atomic number of $11$,so $Na^{+}$ has $11 - 1 = 10$ electrons.
Since both $Mg^{2+}$ and $Na^{+}$ have $10$ electrons,they are isoelectronic.

(B) Naturally occurring carbon consists of two stable isotopes,which are $^{12}C$ and $^{13}C$.
Among these,$^{12}C$ is the most abundant isotope.

(A) For a neutral atom,the number of protons and electrons is equal to the atomic number $(Z)$.
For ${}^{13}_{6}C$,the atomic number $Z = 6$,so the number of protons = $6$ and the number of electrons = $6$.
The number of neutrons is calculated by subtracting the atomic number from the mass number $(A)$:
Number of neutrons = $A - Z = 13 - 6 = 7$.
Therefore,the number of protons,neutrons,and electrons are $6, 7, 6$ respectively.

(B) Given: $p = 17$,$e = 18$,$n = 18$.
Atomic number $(Z)$ is equal to the number of protons,so $Z = 17$,which corresponds to the element Chlorine $(Cl)$.
Since the number of electrons $(18)$ is greater than the number of protons $(17)$,the species carries a negative charge of $-1$,forming the chloride ion $(Cl^{-})$.
Mass number $(A)$ is the sum of protons and neutrons: $A = p + n = 17 + 18 = 35$.
Therefore,the proper symbol is $_{17}^{35} Cl^{-}$.

(A) An element is represented as ${}_Z^A X$,where:
$Z$ is the atomic number (number of protons).
$A$ is the mass number (sum of protons and neutrons).
For ${}_{16}^{32} S^{2-}$:
Number of protons $= Z = 16$.
Number of neutrons $= A - Z = 32 - 16 = 16$.
Since the ion has a charge of $-2$,it has gained $2$ electrons.
Number of electrons $= Z + 2 = 16 + 2 = 18$.
Thus,the number of protons,neutrons,and electrons are $16, 16, 18$ respectively.
The correct option is $A$ (or $D$ as they are identical).

(C) Isoelectronic species are atoms or ions that have the same number of electrons.
For $O^{2-}$: $8 + 2 = 10$ electrons.
For $F^{-}$: $9 + 1 = 10$ electrons.
For $Na^{+}$: $11 - 1 = 10$ electrons.
For $Mg^{2+}$: $12 - 2 = 10$ electrons.
Since all these species have $10$ electrons,they are isoelectronic.

(C) Isobars are defined as atoms of different chemical elements that have the same mass number but different atomic numbers.
For example,$^{40}_{18}Ar$ and $^{40}_{20}Ca$ are isobars because both have a mass number of $40$ but different atomic numbers ($18$ and $20$ respectively).

(A) Let $a$ be the number of protons in the element.
The number of neutrons is $a + 0.32a = 1.32a$.
The mass number is the sum of protons and neutrons,which is $181$.
$a + 1.32a = 181$
$2.32a = 181$
$a = \frac{181}{2.32} \approx 78$.
The element with atomic number $78$ is Platinum $(Pt)$.

(A) For element $X$: Mass number $A = 52$. Let the number of protons be $p$. Number of neutrons $n = p + 0.166p = 1.166p$. Since $A = p + n$,we have $52 = p + 1.166p = 2.166p$. Thus,$p = 52 / 2.166 \approx 24$. The element with atomic number $24$ is $Cr$.
For element $Z$: Mass number $A = 75$. Let the number of protons be $p'$. Number of neutrons $n' = p' + 0.273p' = 1.273p'$. Since $A = p' + n'$,we have $75 = p' + 1.273p' = 2.273p'$. Thus,$p' = 75 / 2.273 \approx 33$. The element with atomic number $33$ is $As$.

(D) The atomic number $(Z)$ of $Uuq$ is $114$,so the number of protons $(p)$ is $114$.
Let the number of neutrons be $n$.
According to the problem,the number of neutrons is $12.8 \%$ more than the number of protons:
$n = p + 0.128 \times p = 1.128 \times p$
$n = 1.128 \times 114 = 128.592 \approx 129$.
The mass number $(A)$ is the sum of protons and neutrons:
$A = p + n = 114 + 129 = 243$.
Therefore,the approximate mass number is $243$.

(A) The spectrum of an atom or ion depends on the number of electrons present in it.
Helium $(He)$ has $2$ electrons.
Among the given options,$Li^{+}$ (Lithium ion) has an atomic number of $3$,so $Li^{+}$ has $3 - 1 = 2$ electrons.
Since both $He$ and $Li^{+}$ have the same number of electrons $(2)$,their spectra are expected to be similar.

(D) The number of electrons in a species is calculated by subtracting the charge from the atomic number of the neutral atom.
$1$. For $Be^{2+}$: Atomic number of $Be$ is $4$. Electrons = $4 - 2 = 2$. Thus,$(i-c)$.
$2$. For $H^{+}$: Atomic number of $H$ is $1$. Electrons = $1 - 1 = 0$. Thus,$(ii-a)$.
$3$. For $Na^{+}$: Atomic number of $Na$ is $11$. Electrons = $11 - 1 = 10$. Thus,$(iii-b)$.
$4$. For $Mg^{+}$: Atomic number of $Mg$ is $12$. Electrons = $12 - 1 = 11$. Thus,$(iv-d)$.
Therefore,the correct match is $(i-c), (ii-a), (iii-b), (iv-d)$.

(D) Isoelectronic species are those that have the same number of electrons.
$O^{2-}: 8 + 2 = 10 \ e^-$
$F^{-}: 9 + 1 = 10 \ e^-$
$Na^{+}: 11 - 1 = 10 \ e^-$
$Mg^{2+}: 12 - 2 = 10 \ e^-$
Since all the given species have $10$ electrons,they are all isoelectronic.

(C) Isoelectronic species are atoms or ions having the same number of electrons.
In option $(C)$:
$N^{3-}: 7 + 3 = 10$ electrons
$Mg^{2+}: 12 - 2 = 10$ electrons
$F^{-}: 9 + 1 = 10$ electrons
$O^{2-}: 8 + 2 = 10$ electrons
All species have $10$ electrons,therefore they are isoelectronic.

(A) Given,${ }_{Z_1} A^{M_1}$ and ${ }_{Z_2} B^{M_2}$.
Here,$M_1 \neq M_2$ and $Z_1 \neq Z_2$ $(M = \text{atomic weight}, Z = \text{atomic number})$.
But,$M_1 - Z_1 = M_2 - Z_2$.
We know that $\text{atomic weight} = \text{number of protons} + \text{number of neutrons}$ and $\text{atomic number} = \text{number of protons}$.
Therefore,$\text{number of neutrons} = \text{atomic weight} - \text{atomic number}$.
Since $M_1 - Z_1$ and $M_2 - Z_2$ represent the number of neutrons in elements $A$ and $B$ respectively,and they are equal,these elements have the same number of neutrons.
Species having the same number of neutrons are known as isotones.
Thus,these elements are isotonic.

(D) Isoelectronic species are atoms or ions that have the same number of electrons.
For $(I)$: $O^{2-} (8+2=10)$,$F^{-} (9+1=10)$,$Na^{+} (11-1=10)$,$Mg^{2+} (12-2=10)$. All have $10$ electrons,so $(I)$ is isoelectronic.
For $(II)$: $Na^{+} (10)$,$Mg^{+} (11)$,$Al^{3+} (10)$,$F^{-} (10)$. Since $Mg^{+}$ has $11$ electrons,$(II)$ is not isoelectronic.
For $(III)$: $N^{3-} (7+3=10)$,$O^{2-} (8+2=10)$,$F^{-} (9+1=10)$,$Ne (10)$. All have $10$ electrons,so $(III)$ is isoelectronic.
Therefore,$(I)$ and $(III)$ are isoelectronic species.

(C) Isotones are species that have an equal number of neutrons.
Number of neutrons = Mass number $(A)$ - Atomic number $(Z)$.
For $^{39}_{19} K$: Neutrons = $39 - 19 = 20$.
For $^{40}_{20} Ca$: Neutrons = $40 - 20 = 20$.
Since both have $20$ neutrons,they are isotones.

(C) Iso-electronic species are those that have the same number of electrons.
$1$. For $Mg^{2+}$ ($Z=12$,$12-2=10$ electrons) and $C^{4-}$ ($Z=6$,$6+4=10$ electrons),both have $10$ electrons. Thus,it is an iso-electronic pair.
$2$. For $N^{3-}$ ($Z=7$,$7+3=10$ electrons) and $O^{2-}$ ($Z=8$,$8+2=10$ electrons),both have $10$ electrons. Thus,it is an iso-electronic pair.
$3$. For $N^{2-}$ ($Z=7$,$7+2=9$ electrons) and $O^{2-}$ ($Z=8$,$8+2=10$ electrons),they have different numbers of electrons ($9$ and $10$). Thus,it is not an iso-electronic pair.
$4$. For $F^{-}$ ($Z=9$,$9+1=10$ electrons) and $Al^{3+}$ ($Z=13$,$13-3=10$ electrons),both have $10$ electrons. Thus,it is an iso-electronic pair.
Therefore,the correct option is $C$.

(A) For the ion ${}^{14}_{7}N^{3-}$,the atomic number $(Z)$ is $7$,which represents the number of protons. Therefore,protons = $7$.
The mass number $(A)$ is $14$. The number of neutrons is calculated as $A - Z = 14 - 7 = 7$.
Since the ion has a charge of $-3$,it means the nitrogen atom has gained $3$ electrons. The number of electrons = (Atomic number) + (Magnitude of negative charge) = $7 + 3 = 10$.
Thus,the number of protons,neutrons,and electrons are $7, 7, 10$ respectively.

(B) $({}_{32}Ge^{76}, {}_{34}Se^{76})$ have the same mass number $(A = 76)$ but different atomic numbers ($Z = 32$ and $Z = 34$),therefore they are isobars.
Number of neutrons in ${}_{14}Si^{30} = 30 - 14 = 16$.
Number of neutrons in ${}_{16}S^{32} = 32 - 16 = 16$.
Since both have the same number of neutrons,${}_{14}Si^{30}$ and ${}_{16}S^{32}$ are isotones.

(D) The number of protons $(p)$ determines the atomic number $(Z)$.
Here,$Z = 16$,which corresponds to the element Sulfur $(S)$.
The number of electrons $(e^-)$ is $18$.
The net charge is calculated as: $\text{Charge} = \text{Number of protons} - \text{Number of electrons} = 16 - 18 = -2$.
Therefore,the species is $S^{2-}$.

(A) $1$. The anion $A^-$ has $36$ electrons. Since it is a monoatomic anion with a $-1$ charge, the neutral atom $A$ must have $36 - 1 = 35$ electrons.
$2$. The atomic number $(Z)$ of the element is equal to the number of protons, which is $35$ for a neutral atom. This element is Bromine $(Br)$.
$3$. The mass number $(A_{mass})$ is the sum of protons and neutrons: $A_{mass} = Z + n = 35 + 45 = 80$.
$4$. Bromine belongs to Group $17$ (Halogens) of the periodic table.
$5$. Bromine is one of the few elements that exists as a liquid at room temperature.
$6$. Therefore, the atomic mass is $80$, the group is $17$, and the physical state is liquid.