
What would be the electronic configuration of a positively charged sodium ion , $Na$? What would be its atomic number and mass number?
Answer
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Hint: We know that sodium is an alkali metal. It is placed in the $S$ Block of the modern periodic table. It is a very reactive alkali metal. It has a low, low melting point. The mass number of an atom or ion is given by the sum of the number of protons and neutrons of the ion or atom.
Complete step-by-step answer:
As we know that the atomic number of sodium is eleven. The number of electrons and protons of an atom is given by its atomic number. The electronic configuration of $Na$ is $(1{s^2},2{s^2},2{p^6},3{s^1})$. The positively charged sodium ion will have an electron less than its atomic number, we will remove an electron from the last valence shell of sodium that is $3{s^1}$. So the sodium ion will have ten electrons. The electronic configuration of $N{a^ + }$ will be $(1{s^2},2{s^2},2{p^6})$. Removing an electron will not alter the atomic number and mass number of the sodium atom. The sodium atom has eleven protons and twelve neutrons. So the atomic number of the positively charged sodium ion $N{a^ + }$ will be $11$ and the mass number of $N{a^ + }$ will be the sum of the number of protons and neutrons that is $11 + 12 = 23$.
So the final answer can be written as : electronic configuration of a positively charged sodium ion is $(1{s^2},2{s^2},2{p^6})$ . The atomic number of $N{a^ + }$ is $11$ and the mass number is $23$.
Additional information: Sodium is a very important alkali metal. It is a very reactive metal and it reacts with water, snow and ice to form sodium hydroxide and hydrogen. Sodium hydroxide is a very strong base.
Note: Always remember that the atomic number of sodium is $11$ and mass number of sodium atom is $23$. Removing an electron from the sodium atom will not alter the atomic number and mass number. It will just decrease the number of electrons of the atom. The electronic configuration of
$Na$ is $(1{s^2},2{s^2},2{p^6},3{s^1})$ and $N{a^ + }$ is $(1{s^2},2{s^2},2{p^6})$.
Complete step-by-step answer:
As we know that the atomic number of sodium is eleven. The number of electrons and protons of an atom is given by its atomic number. The electronic configuration of $Na$ is $(1{s^2},2{s^2},2{p^6},3{s^1})$. The positively charged sodium ion will have an electron less than its atomic number, we will remove an electron from the last valence shell of sodium that is $3{s^1}$. So the sodium ion will have ten electrons. The electronic configuration of $N{a^ + }$ will be $(1{s^2},2{s^2},2{p^6})$. Removing an electron will not alter the atomic number and mass number of the sodium atom. The sodium atom has eleven protons and twelve neutrons. So the atomic number of the positively charged sodium ion $N{a^ + }$ will be $11$ and the mass number of $N{a^ + }$ will be the sum of the number of protons and neutrons that is $11 + 12 = 23$.
So the final answer can be written as : electronic configuration of a positively charged sodium ion is $(1{s^2},2{s^2},2{p^6})$ . The atomic number of $N{a^ + }$ is $11$ and the mass number is $23$.
Additional information: Sodium is a very important alkali metal. It is a very reactive metal and it reacts with water, snow and ice to form sodium hydroxide and hydrogen. Sodium hydroxide is a very strong base.
Note: Always remember that the atomic number of sodium is $11$ and mass number of sodium atom is $23$. Removing an electron from the sodium atom will not alter the atomic number and mass number. It will just decrease the number of electrons of the atom. The electronic configuration of
$Na$ is $(1{s^2},2{s^2},2{p^6},3{s^1})$ and $N{a^ + }$ is $(1{s^2},2{s^2},2{p^6})$.
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