
${{\text{H}}^{\text{ + }}}{\text{,}}{{\text{D}}^{\text{ + }}}$ and ${{\text{T}}^{\text{ + }}}$ differ in all except in:
(A)- Number of electrons
(B)- Number of neutrons
(C)- Ionic mass
(D)- All are correct
Answer
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Hint: ${{\text{H}}^{\text{ + }}}{\text{,}}{{\text{D}}^{\text{ + }}}$ and ${{\text{T}}^{\text{ + }}}$ all are the isotopes of hydrogen which stands for hydrogen ion, deuterium ion and tritium ion respectively. Isotopes of any element are that species which are having different atomic masses but the same atomic number as the parent element will have.
Complete Solution :
From the definition of isotopes it is clear that, ${{\text{H}}^{\text{ + }}}{\text{,}}{{\text{D}}^{\text{ + }}}$ and ${{\text{T}}^{\text{ + }}}$ will have same atomic number; but different atomic mass.
-Also we know that, Atomic Number of an atom = no. of electrons or no. of protons.
-And Atomic Mass of an atom = no. of protons + no. of neutrons present in the nucleus.
-Here, ${{\text{H}}^{\text{ + }}}$ is also denoted as ${}_{\text{1}}^{\text{1}}{\text{H}}$, in which upper written quantity is considered as atomic number & lower written quantity is considered as atomic mass.
- So that in hydrogen, no. of electron = 1
No. of proton = 1
& no. of neutron = 0.
-${{\text{D}}^{\text{ + }}}$ is also denoted as ${}_{\text{2}}^{\text{1}}{\text{D}}$,
- So that in Deuterium, no. of electron = 1
No. of proton = 1
& no. of neutrons = 1.
-And, ${{\text{T}}^{\text{ + }}}$ is also denoted as ${}_3^{\text{1}}{\text{D}}$,
- So that in Tritium, no. of electron = 1
No. of proton = 1
& no. of neutron = 2.
From the above calculation it is clear that, ${{\text{H}}^{\text{ + }}}{\text{,}}{{\text{D}}^{\text{ + }}}$ and ${{\text{T}}^{\text{ + }}}$ differ in all except in no. of electrons.
So, the correct answer is “Option A”.
Note: In this question some of you may do wrong calculation by getting confused in the value of no. of neutrons, because in the simple hydrogen atom neutron is not present. So don’t do calculations by keeping in mind that in deuterium & tritium which are the isotopes of hydrogen, neutrons are not present.
Complete Solution :
From the definition of isotopes it is clear that, ${{\text{H}}^{\text{ + }}}{\text{,}}{{\text{D}}^{\text{ + }}}$ and ${{\text{T}}^{\text{ + }}}$ will have same atomic number; but different atomic mass.
-Also we know that, Atomic Number of an atom = no. of electrons or no. of protons.
-And Atomic Mass of an atom = no. of protons + no. of neutrons present in the nucleus.
-Here, ${{\text{H}}^{\text{ + }}}$ is also denoted as ${}_{\text{1}}^{\text{1}}{\text{H}}$, in which upper written quantity is considered as atomic number & lower written quantity is considered as atomic mass.
- So that in hydrogen, no. of electron = 1
No. of proton = 1
& no. of neutron = 0.
-${{\text{D}}^{\text{ + }}}$ is also denoted as ${}_{\text{2}}^{\text{1}}{\text{D}}$,
- So that in Deuterium, no. of electron = 1
No. of proton = 1
& no. of neutrons = 1.
-And, ${{\text{T}}^{\text{ + }}}$ is also denoted as ${}_3^{\text{1}}{\text{D}}$,
- So that in Tritium, no. of electron = 1
No. of proton = 1
& no. of neutron = 2.
From the above calculation it is clear that, ${{\text{H}}^{\text{ + }}}{\text{,}}{{\text{D}}^{\text{ + }}}$ and ${{\text{T}}^{\text{ + }}}$ differ in all except in no. of electrons.
So, the correct answer is “Option A”.
Note: In this question some of you may do wrong calculation by getting confused in the value of no. of neutrons, because in the simple hydrogen atom neutron is not present. So don’t do calculations by keeping in mind that in deuterium & tritium which are the isotopes of hydrogen, neutrons are not present.
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