
The difference in the number of angular nodes and the number of radial nodes in the orbital to which the last electron of chromium present is _ _ _ _ _.
Answer
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Hint: The angular node and radial nodes are the factors that determine the total electron density present in the orbital. The angular nodes are dependent on the azimuthal quantum number. One can use the concept to find out the difference between the two nodes.
Complete step by step answer:
1) First of all we will learn about the number of angular nodes and the number of radial nodes present in the orbital. The angular nodes and the radical nodes are the nodes where the electron density in an orbital is zero.
2) The angular nodes number is always equal to the value of $l$ where $l$ is the azimuthal quantum number which denotes the subshell present in the orbital. We can find out the value of $l$ finding the orbital in which the last electron is present.
3) The radial nodes are calculated as $n - l - 1$ in which the $n$ stands for the principal quantum number which denotes the shell number in the orbital and the $l$ stands for an azimuthal quantum number.
4) Now let's see the last electrons orbital by calculating the electronic configuration of chromium-
$Cr \to 24 \to {\left[ {Ar} \right]^{18}}4{s^2}3{d^4}$
Which shows the last electron is in the $3d$ orbital of chromium.
Now, in $3d$ the number $3$ denotes the $n$ value which is shell number i.e. $n = 3$ and the $d$ subshell gives the value of $l$ which is $2$ i.e. $l = 2$
So, The number of radial nodes $ = n - l - 1 = 3 - 2 - 1 = 0$
The number of angular nodes $ = l = 2$
5) Therefore, by analyzing the above data the difference in the number of angular nodes and the number of radial nodes in the orbital to which the last electron of chromium present is $2$ which is the answer.
Note:
The subshell in an orbital is denoted by $s,p,d,f$ for which the azimuthal quantum number value i.e. $l$ value is $0,1,2,3$ respectively. The angular nodes present in an orbital are always equal to the value of $l$. The principal quantum number is the value of the number of shells present in the atomic orbital.
Complete step by step answer:
1) First of all we will learn about the number of angular nodes and the number of radial nodes present in the orbital. The angular nodes and the radical nodes are the nodes where the electron density in an orbital is zero.
2) The angular nodes number is always equal to the value of $l$ where $l$ is the azimuthal quantum number which denotes the subshell present in the orbital. We can find out the value of $l$ finding the orbital in which the last electron is present.
3) The radial nodes are calculated as $n - l - 1$ in which the $n$ stands for the principal quantum number which denotes the shell number in the orbital and the $l$ stands for an azimuthal quantum number.
4) Now let's see the last electrons orbital by calculating the electronic configuration of chromium-
$Cr \to 24 \to {\left[ {Ar} \right]^{18}}4{s^2}3{d^4}$
Which shows the last electron is in the $3d$ orbital of chromium.
Now, in $3d$ the number $3$ denotes the $n$ value which is shell number i.e. $n = 3$ and the $d$ subshell gives the value of $l$ which is $2$ i.e. $l = 2$
So, The number of radial nodes $ = n - l - 1 = 3 - 2 - 1 = 0$
The number of angular nodes $ = l = 2$
5) Therefore, by analyzing the above data the difference in the number of angular nodes and the number of radial nodes in the orbital to which the last electron of chromium present is $2$ which is the answer.
Note:
The subshell in an orbital is denoted by $s,p,d,f$ for which the azimuthal quantum number value i.e. $l$ value is $0,1,2,3$ respectively. The angular nodes present in an orbital are always equal to the value of $l$. The principal quantum number is the value of the number of shells present in the atomic orbital.
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