Q41P - Chemistry: Molecular Nature Of Matter And Change

Question

Do both members of the following pairs have the same number of protons? Neutrons? Electrons?

$\left(a\right)\mathbf{ }{}_{\mathbf{8}}{}^{\mathbf{16}}\mathit{O}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{}_{\mathbf{8}}{}^{\mathbf{17}}\mathit{O}\mathbf{ }\left(b\right)\mathbf{ }{}_{\mathbf{18}}{}^{\mathbf{40}}\mathit{A}\mathit{r}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{}_{\mathbf{18}}{}^{\mathbf{41}}\mathit{K}\mathbf{ }\left(c\right)\mathbf{ }{}_{\mathbf{27}}{}^{\mathbf{60}}\mathit{C}\mathit{o}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{}_{\mathbf{28}}{}^{\mathbf{60}}\mathit{N}\mathit{i}$  

Which pair(s) consist(s) of atoms with the same Z value? N value? A value?

Step-by-Step Solution

Answer

All the elements of the isotopic pair doesn’t acquire an equal number of protons, electrons, and neutrons.

None of the pair(s) consist(s) of atoms with the same atomic number (Z) value, neutrons (N) value, or atomic mass (A) value.

Step 1: Calculation of Z, N, and A in part (a)

${}_8{}^{16}O and {}_8{}^{17}O$   

 ${}_8{}^{16}O$ Atomic number = 8, Protons = 8, Electrons = 8,

Mass number = 16 

Neutrons = Mass number – Protons = 16 – 8 = 8

${}_8{}^{17}O$ Atomic number = 8, Proton = 8, Electron = 8,

Mass number = 17 

Neutrons = Mass number – Protons = 17 – 8 = 9

${}_8{}^{16}O and {}_8{}^{17}O$ both have the same atomic number, same protons, same electrons but different mass numbers, and neutrons.

Step 2: Calculation of Z, N, and A in part (b)

 ${}_{18}{}^{40}Ar and {}_{18}{}^{41}K$

 ${}_{18}{}^{40}Ar$ Atomic number = 18, Protons = 18, Electrons = 18,

Mass number = 40

Neutrons = Mass number – Protons = 40 – 18 = 22  

     

${}_{18}{}^{41}K$Atomic number = 19, Protons = 19, Electrons = 19,

Mass number = 41                                      

Neutrons = Mass number – Protons = 41 – 19 = 22

 ${}_{18}{}^{40}Ar and {}_{18}{}^{41}K$ both have different atomic numbers, protons, and electrons but neutrons are the same.

Step 3: Calculation of Z, N, and A in part (c)

${}_{27}{}^{60}Co and {}_{28}{}^{60}Ni$

${}_{27}{}^{60}Co$ Atomic number = 27, Protons = 27, Electrons = 27,

Mass number = 60 

Neutrons = Mass number – protons = 60 – 27 = 33 

${}_{28}{}^{60}Ni$ Atomic number = 28, Protons = 28, Electrons = 28

Mass number = 60 

Neutrons = Mass number – Protons = 60 – 28 = 32 

${}_{27}{}^{60}Co and {}_{28}{}^{60}Ni$both have different atomic numbers, protons, electrons, and neutrons but the same mass number.