Atomic number, Mass number, Atomic species Questions in English

(C) Atomic weight is the mass of a single atom of an element,which can be fractional due to the existence of isotopes.
Equivalent weight is the mass of an element that reacts with or displaces $1 \ g$ of hydrogen or $8 \ g$ of oxygen,which can also be fractional.
Atomic volume is the volume occupied by one mole of an atom,which is not necessarily a whole number.
Atomic number is defined as the number of protons present in the nucleus of an atom. Since protons are discrete particles,the count must be an integer. Therefore,the atomic number is always a whole number.

(C) Positive ions are formed from the neutral atom by the loss of electrons.

(A) The atomic number of carbon is $6$,which represents the number of protons.
The mass number of ${}^{12}C$ is $12$.
The number of neutrons is calculated as: $\text{Neutrons} = \text{Mass number} - \text{Protons}$.
$\text{Neutrons} = 12 - 6 = 6$.

(C) The nucleus of a helium atom is represented as $_2He^4$.
This notation indicates that the atomic number $(Z)$ is $2$,which corresponds to the number of protons.
The mass number $(A)$ is $4$.
The number of neutrons $(n)$ is calculated as $A - Z = 4 - 2 = 2$.
Therefore,the nucleus contains $2$ protons and $2$ neutrons.

(B) In a neutral atom,the number of electrons is equal to the number of protons,which is defined as the $Atomic \ number$ $(Z)$.

(A) The number of protons is equal to the atomic number,which is $25$.
The number of neutrons is calculated as: $\text{Mass number} - \text{Atomic number} = 55 - 25 = 30$.
Therefore,the nucleus contains $25$ protons and $30$ neutrons.

(B) The number of neutrons is calculated as the mass number minus the number of protons.
Since $W$ is the atomic weight (mass number) and $N$ is the atomic number (number of protons),the number of neutrons is $W - N$.
Therefore,the number of $_{0}n^{1} = W - N$.

(B) The atomic number of zinc $(Zn)$ is $30$.
For the isotope with mass number $A = 70$,the number of neutrons is calculated as $N = A - Z$.
$N = 70 - 30 = 40$.
The formation of a dipositive ion $(Zn^{2+})$ involves the loss of electrons from the valence shell,which does not affect the number of protons or neutrons in the nucleus.
Therefore,the number of neutrons remains $40$.

(A) Isoelectronic species are those that contain the same number of electrons.
$Na^{+}$ has $11 - 1 = 10$ electrons.
$Ne$ has $10$ electrons.
Since both contain $10$ electrons,$Na^{+}$ and $Ne$ are isoelectronic.
Therefore,the correct option is $A$.

(C) The chlorine atom $(Cl)$ has $17$ protons and $17$ electrons.
The chloride ion $(Cl^-)$ is formed by the gain of one electron,so it has $17$ protons and $18$ electrons.
Therefore,they differ in the number of electrons.

(B) Atomic number is defined as the number of protons in the nucleus.

(B) For a neutral atom,the number of electrons $(E)$ is equal to the number of protons $(P)$,which is the atomic number $(Z)$.
Given: $E = 26$,so $Z = 26$.
The atomic mass number $(A)$ is given as $56$.
The relationship between mass number,protons,and neutrons is $A = P + N$.
Therefore,the number of neutrons $(N)$ is calculated as $N = A - Z$.
$N = 56 - 26 = 30$.

(C) The atomic number of $Na$ is $11$,so $Na^{+}$ has $11 - 1 = 10$ electrons.
$O^{2-}$ has $8 + 2 = 10$ electrons.
$F^{-}$ has $9 + 1 = 10$ electrons.
$Li^{+}$ has $3 - 1 = 2$ electrons.
$Al^{3+}$ has $13 - 3 = 10$ electrons.
Therefore,$Na^{+}$ has a different number of electrons from $Li^{+}$.

(C) An atom is electrically neutral because the number of protons (positive charge) equals the number of electrons (negative charge).
When an atom loses one electron,the number of protons exceeds the number of electrons by $1$.
Therefore,the atom acquires a net positive charge of $+1$ and becomes a cation.

(A) The atomic number $(Z)$ is equal to the total number of electrons in a neutral atom.
Summing the electrons: $2 + 2 + 6 + 2 + 6 + 10 + 2 + 5 = 35$.
Thus,the atomic number $(Z)$ is $35$.
Given the atomic mass $(A)$ is $80$.
The number of neutrons $(N)$ is calculated as $N = A - Z$.
$N = 80 - 35 = 45$.
Therefore,the atomic number is $35$ and the number of neutrons is $45$.

(C) To determine the number of electrons and neutrons,we use the atomic number $(Z)$ and mass number $(A)$:
$1$. For $C$ $(Z=6, A=12)$: $p=6, e=6, n=6$.
$2$. For $F^{-}$ $(Z=9, A=19)$: $p=9, e=10, n=10$.
$3$. For $O^{-2}$ $(Z=8, A=16)$: $p=8, e=10, n=8$.
$4$. For $Al^{+3}$ $(Z=13, A=27)$: $p=13, e=10, n=14$.
In $O^{-2}$,the number of electrons $(10)$ is greater than the number of neutrons $(8)$.

(A) The two atoms are isotopes of the same element because they have the same atomic number $(Z = 6)$ but different mass numbers ($A = 12$ and $A = 13$).
For the atom with mass number $12$: Number of protons $(p)$ = $6$,number of electrons $(e)$ = $6$,and number of neutrons $(n)$ = $12 - 6 = 6$.
For the atom with mass number $13$: Number of protons $(p)$ = $6$,number of electrons $(e)$ = $6$,and number of neutrons $(n)$ = $13 - 6 = 7$.
Therefore,the atom with atomic weight $13$ contains one more neutron than the atom with atomic weight $12$.

(C) The atomic number $(Z)$ of Calcium $(Ca)$ is $20$,which represents the number of protons $(P)$.
The mass number $(A)$ is $40$.
The number of neutrons $(N)$ is calculated as $N = A - Z = 40 - 20 = 20$.
Therefore,the nucleus contains $20$ protons and $20$ neutrons.

(B) $Na^{+}$ ion has $11 - 1 = 10$ electrons.
$Mg^{2+}$ ion has $12 - 2 = 10$ electrons.
Since both have the same number of electrons,they are isoelectronic.

(C) For $Ca$ atom $(_{20}Ca^{40})$:
Atomic number $(Z)$ = $20$ (Number of protons = Number of electrons = $20$).
Mass number $(A)$ = $40$ (Number of nucleons = $40$).
Number of neutrons $(n)$ = $A - Z = 40 - 20 = 20$.
Evaluating the options:
$(A)$ Number of electrons $(20)$ = Number of neutrons $(20)$. (Correct)
$(B)$ Number of nucleons $(40)$ = $2 \times$ Number of electrons $(20)$. (Correct)
$(C)$ Number of protons $(20)$ is half of the number of neutrons $(20)$. (Incorrect,as $20$ is equal to $20$,not half).
$(D)$ Number of nucleons $(40)$ = $2 \times$ Atomic number $(20)$. (Correct)
Therefore,the incorrect statement is $(C)$.

(A) Isoelectronic species are atoms or ions that have the same number of electrons.
For option $A$:
$K^{+}$ has $19 - 1 = 18$ electrons.
$Cl^{-}$ has $17 + 1 = 18$ electrons.
Since both have $18$ electrons,they are isoelectronic.
Electronic configuration of $K^{+}$: $1s^2 2s^2 2p^6 3s^2 3p^6$ ($18$ electrons).
Electronic configuration of $Cl^{-}$: $1s^2 2s^2 2p^6 3s^2 3p^6$ ($18$ electrons).

(C) The lightest element is hydrogen,which has an atomic weight of approximately $1$.
Given that the atomic weight of the element is $23$ times that of hydrogen,its mass number $(A)$ is $23$.
The number of protons $(Z)$ is given as $11$.
Since the mass number is the sum of protons and neutrons $(A = Z + N)$,we have $23 = 11 + N$,which gives $N = 12$ neutrons.
For a neutral atom,the number of electrons is equal to the number of protons,so there are $11$ electrons.
Thus,the element contains $11$ protons,$12$ neutrons,and $11$ electrons.

(C) The number of neutrons is calculated as $N = A - Z$,where $A$ is the mass number and $Z$ is the atomic number.
For carbon $(_{6}^{12}C)$: $N = 12 - 6 = 6$.
For silicon $(_{14}^{28}Si)$: $N = 28 - 14 = 14$.
The ratio of neutrons in $C$ to $Si$ is $6 : 14$,which simplifies to $3 : 7$.

(D) The atomic number $(Z)$ of an element is defined as the number of protons present in the nucleus of its atom.
Since each proton carries a unit positive charge $(+1)$ and neutrons are electrically neutral,the total electrical charge of the nucleus is directly proportional to the number of protons.
Therefore,the atomic number is equal to the electrical charge of the nucleus.

(D) The atomic number of carbon $(C)$ is $6$,so it has $6$ electrons.
Isoelectronic species are those that have the same number of electrons.
For $N^{+}$,the atomic number of nitrogen $(N)$ is $7$. Since it has a $+1$ charge,it has $7 - 1 = 6$ electrons.
Thus,$N^{+}$ is isoelectronic with $C$ as both have $6$ electrons.
Therefore,the correct option is $(D)$.

(C) The total number of electrons is $2 + 8 + 18 + 1 = 29$.
Since the element is neutral,the atomic number $(Z)$ is equal to the number of electrons,which is $29$.
The atomic mass $(A)$ is given as $63$.
The number of neutrons $(N)$ is calculated using the formula: $N = A - Z$.
$N = 63 - 29 = 34$.

(B) For an element represented as $_{Z}E^{A}$,the atomic number $Z$ represents the number of protons.
Here,$Z = 21$,so there are $21$ protons.
The mass number $A$ is the sum of protons and neutrons $(A = Z + N)$.
Therefore,the number of neutrons $N = A - Z = 45 - 21 = 24$.
Thus,the nucleus contains $21$ protons and $24$ neutrons,and the element is Scandium $(Sc)$.

(B) The mass number $(A)$ is defined as the sum of the number of protons $(Z)$ and the number of neutrons $(n)$: $A = Z + n$.
Given the mass number $A = 14$.
For the anion $X^{3-}$,the number of electrons is $10$. Since the anion is formed by gaining $3$ electrons,the number of protons $(Z)$ in the neutral atom $X$ is $Z = 10 - 3 = 7$.
Using the mass number formula: $14 = 7 + n$.
Therefore,the number of neutrons $n = 14 - 7 = 7$.

(C) The charge on an atom or ion is calculated by the difference between the number of protons and the number of electrons.
Charge = (Number of protons) - (Number of electrons)
Charge = $17 - 18 = -1$
Therefore,the species is a chloride ion $(Cl^-)$ which has $17$ protons,$18$ neutrons,and $18$ electrons.

(C) In a neutral atom,the number of negatively charged electrons $(e^-)$ is equal to the number of positively charged protons $(p^+)$ to maintain electrical neutrality.
Therefore,the correct option is $(C)$.

(C) Isoelectronic species are those that have the same number of electrons.
The oxygen anion ${O^{2-}}$ has $8 + 2 = 10$ electrons.
Checking the options:
$1$. ${N^{3-}}$ has $7 + 3 = 10$ electrons.
$2$. ${F^{-}}$ has $9 + 1 = 10$ electrons.
$3$. ${Na^{+}}$ has $11 - 1 = 10$ electrons.
$4$. ${Tl^{+}}$ (Thallium ion) has an atomic number of $81$,so ${Tl^{+}}$ has $81 - 1 = 80$ electrons.
Therefore,${Tl^{+}}$ is not isoelectronic with ${O^{2-}}$.

(D) The atomic number of potassium $(K)$ is $19$,which represents the number of protons in a neutral atom.
For the anion $[_{19}^{40}K]^{-1}$,the charge of $-1$ indicates that the atom has gained one electron.
Therefore,the number of electrons $= \text{Atomic number} + \text{Charge magnitude} = 19 + 1 = 20$.

(A) For a neutral atom,the number of protons is equal to the number of electrons.
Given,number of electrons $= 18$,so number of protons $(P) = 18$.
Given,number of neutrons $(N) = 20$.
The mass number $(A)$ is defined as the sum of the number of protons and neutrons.
$A = P + N = 18 + 20 = 38$.

(C) For an element represented as $_{Z}^{A}Y$,the number of protons is equal to the atomic number $(Z)$.
In the given element $_{89}^{231}Y$,the atomic number $Z = 89$,so the number of protons is $89$.
Since the atom is neutral,the number of electrons is equal to the number of protons,which is $89$.
The number of neutrons is calculated as $A - Z = 231 - 89 = 142$.
Therefore,the number of protons,neutrons,and electrons are $89, 142, 89$ respectively.

(C) $Be^{2+}$ has atomic number $4$,so it has $4 - 2 = 2$ electrons.
$Li^{+}$ has atomic number $3$,so it has $3 - 1 = 2$ electrons.
Since both have $2$ electrons,they are isoelectronic.
The electronic configuration for both is $1s^{2}$,which is the same as $He$.

(A) The atomic number of nitrogen $(N)$ is $7$,which means it has $7$ protons.
For a neutral nitrogen atom,the number of electrons is also $7$.
The nitride ion $(N^{3-})$ is formed by the gain of $3$ electrons by a nitrogen atom.
Therefore,the number of electrons in $N^{3-}$ is $7 + 3 = 10$.
Thus,the nitride ion consists of $7$ protons and $10$ electrons.

(D) $Na^{+}$ has $10 \ e^-$.
$Mg^{2+}$ has $10 \ e^-$.
$O^{2-}$ has $10 \ e^-$.
$Cl^{-}$ has $18 \ e^-$.
Since $Cl^{-}$ has $18 \ e^-$ while the others have $10 \ e^-$,$Cl^{-}$ is not isoelectronic with the others.

(C) The charge on any ion is given by the formula $q = n \times e$,where $n$ is the net charge number and $e$ is the elementary charge $(1.6 \times 10^{-19} \ C)$.
For a $Li^{+}$ ion,the net charge is $+1$.
Therefore,the charge on the $Li^{+}$ ion is $1 \times (+1.6 \times 10^{-19} \ C) = +1.6 \times 10^{-19} \ C$.

(A) Iso-electronic species are those that have the same number of electrons.
For $F^{-}$: Atomic number of $F$ is $9$. Number of electrons = $9 + 1 = 10$.
For $O^{2-}$: Atomic number of $O$ is $8$. Number of electrons = $8 + 2 = 10$.
Since both have $10$ electrons,they are iso-electronic.

(A) The electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ corresponds to a total of $2 + 2 + 6 + 2 + 6 = 18$ electrons.
For a neutral atom,the atomic number $(Z)$ is equal to the number of electrons,so $Z = 18$.
The atomic mass $(A)$ is given as $40$.
The number of neutrons $(n)$ is calculated using the formula: $n = A - Z$.
$n = 40 - 18 = 22$.
Therefore,the atomic number is $18$ and the number of neutrons is $22$.

(B) Isoelectronic species are those that have the same number of electrons.
For $N^{3-}$,the number of electrons is $7 + 3 = 10$.
For $F^{-}$,the number of electrons is $9 + 1 = 10$.
For $Na^{+}$,the number of electrons is $11 - 1 = 10$.
Since all three species have $10$ electrons,they are isoelectronic.

(C) The atomic number of chlorine $(Cl)$ is $17$,which means a neutral chlorine atom has $17$ electrons.
In a $Cl^{-}$ ion,one extra electron is gained.
Therefore,the total number of electrons in $Cl^{-}$ is $17 + 1 = 18$.

(C) The number of protons in an atom is equal to its atomic number.
Since the atomic number of $Carbon$ $(C)$ is $6$,it contains $6$ protons in its nucleus.

(A) The atomic number $(Z)$ of nitrogen is $7$,which represents the number of protons.
In a neutral nitrogen atom,the number of electrons is equal to the number of protons,which is $7$.
When a nitrogen atom forms a nitride ion $(N^{3-})$,it gains $3$ electrons.
Therefore,the number of electrons in $N^{3-} = 7 + 3 = 10$.
The number of protons remains unchanged at $7$.
Thus,the nitride ion has $7$ protons and $10$ electrons.

(D) Atomic number represents the number of protons in the nucleus of an atom,which must be an integer. Therefore,it is always a whole number.

(B) The spectrum of an atom or ion depends on the number of electrons present in it.
$He$ (Helium) has $2$ electrons.
$Li^{+}$ (Lithium ion) also has $3 - 1 = 2$ electrons.
Since both species are isoelectronic (contain the same number of electrons),their spectra are expected to be similar.

(D) The specific charge is defined as the ratio of charge to mass,$\frac{q}{m}$.
For a proton $(p)$,the charge is $e$ and the mass is $m_p$.
For an $\alpha$-particle,the charge is $2e$ and the mass is $4m_p$.
Therefore,$(\frac{q}{m})_{\alpha} = \frac{2e}{4m_p} = \frac{1}{2} (\frac{e}{m_p}) = \frac{1}{2} (\frac{q}{m})_p$.
Given $(\frac{q}{m})_p = 9.6 \times 10^7 \ C \ kg^{-1}$,the specific charge for an $\alpha$-particle is $\frac{1}{2} \times 9.6 \times 10^7 \ C \ kg^{-1} = 4.8 \times 10^7 \ C \ kg^{-1}$.

(C) The statement that the number of protons is equal to the number of neutrons is false.
In a neutral atom,the number of electrons is equal to the number of protons.
Key points regarding atomic structure:
$1$. The number of protons in the nucleus is equal to the atomic number $(Z)$.
$2$. In a neutral atom,the number of electrons equals the number of protons.
$3$. The mass number $(A)$ is the sum of the number of protons and neutrons.
$4$. The number of neutrons is calculated as $(A - Z)$.
Therefore,the number of protons and neutrons is not necessarily equal.

(A) For a neutral atom,the number of electrons is equal to the number of protons.
Since the atom has $20$ protons,it must also have $20$ electrons to maintain electrical neutrality.
Therefore,the correct option is $A$.