How do you figure out how many electrons a sodium ion has? Can it be determined with a flame test?
Answer
585.3k+ views
Hint: The atomic number of Na is 11 and the sodium ion is a monovalent ion.
- The calculation of the number of electrons is a quantitative approach.
Complete step by step answer:
- So in the question it is given that how can we figure out the number of electrons in sodium ions and the other part of the question is to comment on whether we could find the number of electrons in a sodium ion through the flame test.
- So first of all let’s compare the two parts of the question, the first part is about the calculation of number of electrons and the second one is about the flame test, both are entirely different approaches, since we have done the flame test during our practical works we are very familiar with the concept that the principle of this flame is the excitation and de-excitation of the electrons by absorbing energy as a source of heat and the flame test is completely a qualitative approach i.e. from the data or result gained from the flame test we could only tell which ion is present in the sample provide for us whereas calculation of electrons is quantitative approach.
- So we can conclude that we cannot find the number of electrons in the sodium ion with the help of flame test.
- Now the first part of the question, we know that the atomic number of Na is 11, therefore there will be a total of 11 electrons in them.
But here we are concerned about sodium ion which is $\text{N}{{\text{a}}^{\text{+}}}$, and $\text{N}{{\text{a}}^{\text{+}}}$ is a sodium ion i.e. it possess a valency of +1 by losing one electrons.
$\text{Na}\to \text{N}{{\text{a}}^{\text{+}}}\text{+}{{\text{e}}^{\text{-}}}$
Therefore number of electrons in $\text{N}{{\text{a}}^{\text{+}}}\text{=10}{{\text{e}}^{\text{-}}}\text{s}$
The $\text{N}{{\text{a}}^{\text{+}}}$ loses one electron and attains the stable noble gas configuration of Ne.
Note: In the flame test, the sample is mixed with con.HCl and the paste is introduced in a non-luminous flame and according to the ions present in the sample it shows various colours. If sodium ion is present in the sample then it gives yellow colour and this colour is due to the excitation and de-excitation of the electrons from $3{{p}^{1}}$ to $3{{s}^{1}}$.
- The calculation of the number of electrons is a quantitative approach.
Complete step by step answer:
- So in the question it is given that how can we figure out the number of electrons in sodium ions and the other part of the question is to comment on whether we could find the number of electrons in a sodium ion through the flame test.
- So first of all let’s compare the two parts of the question, the first part is about the calculation of number of electrons and the second one is about the flame test, both are entirely different approaches, since we have done the flame test during our practical works we are very familiar with the concept that the principle of this flame is the excitation and de-excitation of the electrons by absorbing energy as a source of heat and the flame test is completely a qualitative approach i.e. from the data or result gained from the flame test we could only tell which ion is present in the sample provide for us whereas calculation of electrons is quantitative approach.
- So we can conclude that we cannot find the number of electrons in the sodium ion with the help of flame test.
- Now the first part of the question, we know that the atomic number of Na is 11, therefore there will be a total of 11 electrons in them.
But here we are concerned about sodium ion which is $\text{N}{{\text{a}}^{\text{+}}}$, and $\text{N}{{\text{a}}^{\text{+}}}$ is a sodium ion i.e. it possess a valency of +1 by losing one electrons.
$\text{Na}\to \text{N}{{\text{a}}^{\text{+}}}\text{+}{{\text{e}}^{\text{-}}}$
Therefore number of electrons in $\text{N}{{\text{a}}^{\text{+}}}\text{=10}{{\text{e}}^{\text{-}}}\text{s}$
The $\text{N}{{\text{a}}^{\text{+}}}$ loses one electron and attains the stable noble gas configuration of Ne.
Note: In the flame test, the sample is mixed with con.HCl and the paste is introduced in a non-luminous flame and according to the ions present in the sample it shows various colours. If sodium ion is present in the sample then it gives yellow colour and this colour is due to the excitation and de-excitation of the electrons from $3{{p}^{1}}$ to $3{{s}^{1}}$.
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