What does the superscript mean in electronic configuration notation?
A) No. of neutrons present in a particular shell
B) No. of electrons present in a particular shell
C) No. of protons present in a particular shell
D) None of the above
Answer
569.1k+ views
Hint: We need to know that protons are a positively charged species and the electron is a negatively charged species. In an atom, protons and neutrons lie inside the nucleus of an atom while electrons revolve around the nucleus in circular orbits. Electronic configuration is based on the arrangement of energy levels of \[s,p,d\] and $f$ block elements.
Complete answer:
Before solving this question we will look at the notation of electronic configuration.
Electronic configuration is based on the arrangement of energy levels of $s$, $p$, $d$ and $f$ block elements. To calculate the electronic configuration we will need to find the order of energy levels $s$, $p$, $d$ and $f$ . As \[3p\] orbital has higher energy than \[3s\] orbital and \[3d\] has more energy than \[3p\] .
Filling up of electrons in different orbitals will fill in this line configuration which is as follows:
\[1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s \ldots \ldots \]
Rules for writing electronic configuration:
The maximum number of electrons in main shell= \[2{n^2}\]
The maximum number of electrons in subshell $s$, $p$, $d$ and $f$ =\[2\left( {2l + 1} \right)\]
Where l represents the different orbital as follows:
For s orbital, \[l = 0\]
p orbital, \[l = 1\],
d orbital, \[l = 2\] ,
f orbital, \[l = 3\]
Now we can look at some examples that will give us more clarity.
Consider \[Be\] element which has an atomic number $4$, so we need to fill $4$ electrons in order to get it’s electronic configuration.
\[Be = 1{s^2}2s\]
Consider the \[Fe\]element which has an atomic number 26, so we need to fill 26 electrons in order to get its electronic configuration.
\[Fe = \left[ {Ar} \right]3{d^6}4{s^2}\]
Option A) this is an incorrect option as the superscript in electronic configuration does not represent no. of neutrons as we have seen above.
Option B) this is a correct option as we have seen above that it’s the no. of electrons which represents the superscript in electronic configuration.
Option C) this is an incorrect option as the superscript in electronic configuration does not represent no. of protons as we have seen above.
Option D) this is an incorrect option as we have got a correct option B.
Option B) is the correct answer
Note:
We have to remember that filling of electrons in different orbitals will fill up in this manner
\[1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s \ldots \ldots \]
Since neutrons are neutral species hence they do not carry any charge on them. Protons and electrons carry charges on them.
Complete answer:
Before solving this question we will look at the notation of electronic configuration.
Electronic configuration is based on the arrangement of energy levels of $s$, $p$, $d$ and $f$ block elements. To calculate the electronic configuration we will need to find the order of energy levels $s$, $p$, $d$ and $f$ . As \[3p\] orbital has higher energy than \[3s\] orbital and \[3d\] has more energy than \[3p\] .
Filling up of electrons in different orbitals will fill in this line configuration which is as follows:
\[1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s \ldots \ldots \]
Rules for writing electronic configuration:
The maximum number of electrons in main shell= \[2{n^2}\]
The maximum number of electrons in subshell $s$, $p$, $d$ and $f$ =\[2\left( {2l + 1} \right)\]
Where l represents the different orbital as follows:
For s orbital, \[l = 0\]
p orbital, \[l = 1\],
d orbital, \[l = 2\] ,
f orbital, \[l = 3\]
Now we can look at some examples that will give us more clarity.
Consider \[Be\] element which has an atomic number $4$, so we need to fill $4$ electrons in order to get it’s electronic configuration.
\[Be = 1{s^2}2s\]
Consider the \[Fe\]element which has an atomic number 26, so we need to fill 26 electrons in order to get its electronic configuration.
\[Fe = \left[ {Ar} \right]3{d^6}4{s^2}\]
Option A) this is an incorrect option as the superscript in electronic configuration does not represent no. of neutrons as we have seen above.
Option B) this is a correct option as we have seen above that it’s the no. of electrons which represents the superscript in electronic configuration.
Option C) this is an incorrect option as the superscript in electronic configuration does not represent no. of protons as we have seen above.
Option D) this is an incorrect option as we have got a correct option B.
Option B) is the correct answer
Note:
We have to remember that filling of electrons in different orbitals will fill up in this manner
\[1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s \ldots \ldots \]
Since neutrons are neutral species hence they do not carry any charge on them. Protons and electrons carry charges on them.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

Draw a labelled diagram of the neuron and describe class 11 biology CBSE

