
Write the general electronic configuration of d- block elements.
Answer
557.1k+ views
Hint: In the given question it is asked about the electronic configuration of the given d-block elements. Now to find it we just have to mention the range and the number we obtain regarding the subshells at the end of the configuration like in this case the number and the range of the electrons in each kind of subshell would give the answer as $(n - 1){d^{1 - 10}}n{s^{1 - 2}}$.
Complete answer:
In the given question we have to find out the step by step analysis of how to find the electronic configuration of the d-block elements.
Now by going step by step we need to specify the process of each acquiring in each step:
Step 1: First of all we need to have a variable value as $n$ which denotes the period to which the element belongs. Now specifically for the case of the d-block, this value of $n$ would vary between $4 - 7$.
Step 2: We have to provide the ultimate range of configuration by giving the ranges of both the d and s subshells in the terms of the $n$ and the number of electrons in each subshell.
For the case of d subshell the number of the shell would be $n - 1$ whereas we know that the number of electrons in the shell can also vary from $1 - 10$, so that would give us a resemblance of $(n - 1){d^{1 - 10}}$.
For the case of s subshell the number of the shell would be $n$ whereas we know that the number of electrons in the shell can also vary from $1 - 2$, so that would give us a resemblance of $n{s^{1 - 2}}$.
Therefore that would make the final answer as the $(n - 1){d^{1 - 10}}n{s^{1 - 2}}$.
Note: In atomic physics and quantum chemistry, the electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals. Electronic configurations describe each electron as moving independently in an orbital, in an average field created by all other orbitals.
Complete answer:
In the given question we have to find out the step by step analysis of how to find the electronic configuration of the d-block elements.
Now by going step by step we need to specify the process of each acquiring in each step:
Step 1: First of all we need to have a variable value as $n$ which denotes the period to which the element belongs. Now specifically for the case of the d-block, this value of $n$ would vary between $4 - 7$.
Step 2: We have to provide the ultimate range of configuration by giving the ranges of both the d and s subshells in the terms of the $n$ and the number of electrons in each subshell.
For the case of d subshell the number of the shell would be $n - 1$ whereas we know that the number of electrons in the shell can also vary from $1 - 10$, so that would give us a resemblance of $(n - 1){d^{1 - 10}}$.
For the case of s subshell the number of the shell would be $n$ whereas we know that the number of electrons in the shell can also vary from $1 - 2$, so that would give us a resemblance of $n{s^{1 - 2}}$.
Therefore that would make the final answer as the $(n - 1){d^{1 - 10}}n{s^{1 - 2}}$.
Note: In atomic physics and quantum chemistry, the electron configuration is the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals. Electronic configurations describe each electron as moving independently in an orbital, in an average field created by all other orbitals.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

