
A neutral atom of an element has 2K,8L,9M, and 2N electrons. The total number of d-electron are:
A. 1
B. 2
C. 3
D. 4
Answer
581.7k+ views
Hint: To find the number of d electrons, we have to first write the electronic configuration of the element. And for that we must know the number of electrons and subshells present in the shells.
Step by step solution:
- Let’s see some of rules of writing electronic configuration-
Afbau principle: according to this principle the lower energy orbitals are filled before higher energy orbitals.
Hund’s rule: It says that one electron goes into each subshell until all of them are half full before pairing up.
Pauli Exclusion principle: no two electrons can be identified by the same set of quantum numbers
- 2K,8L,9M, and 2N electrons are given and we know that K, L, M, N are the shells and the numbers that are 2,8,9,2 represents the number of electrons that are present in the particular shell. And s, p, d, f are the subshells and l represents the number of subshells. So,
- For 2K, n=1 , number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 1-1 \right) \\
& l=0\text{ }to\text{ }\left( 0 \right) \\
& l=0 \\
\end{align}\]
So, we get l=0, that is there is only one value, therefore K shell has only one subshell that is s- subshell.
And we know that s subshell can accommodate two electrons , so we can write the configuration as $1{{s}^{2}}$.
- For 8L, n=2 , number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 2-1 \right) \\
& l=0\text{ }to\text{ }\left( 1 \right) \\
& l=0\text{ }to\text{ 1} \\
\end{align}\] \[\]
So, we get l=0 to 1, that is there are two values, therefore L shell has two subshells that is s- subshell and p subshell.
And we know that s subshell can accommodate two electrons , so we can write the configuration as $2{{s}^{2}}$and p subshell can accommodate six electrons , so we can write the configuration as $2{{p}^{6}}$.
- For 9M, n=3, number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 3-1 \right) \\
& l=0\text{ }to\text{ }\left( 2 \right) \\
& l=0\text{ }to\text{ }2 \\
\end{align}\] \[\]
So, we get l=0 to 2, that is there are three values, therefore L shell has three subshells that is s- subshell, p subshell and d-subshell.
And we know that s subshell can accommodate two electrons , so we can write the configuration as $3{{s}^{2}}$and p subshell can accommodate six electrons , so we can write the configuration as$3{{p}^{6}}$.And d- subshell can accommodate 10 electrons but we have given the total electrons as 9 so , as s contains 2 electrons p contains 6 electrons hence d will contain one electron. The electronic configuration is $3{{d}^{1}}$.
- For 2N, n=4, number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 4-1 \right) \\
& l=0\text{ }to\text{ }\left( 3 \right) \\
& l=0\text{ }to\text{ 3} \\
\end{align}\] \[\]
So, we get l=0 to 3, that is there are four values, therefore M shell has four subshells that is s- subshell, p subshell, d-subshell, and f-subshell.
We are given with two electrons , And we know that s subshell can accommodate two electrons , so we can write the configuration as $4{{s}^{2}}$.
- Now combining all the configurations for each subshell we get,
2K=$1{{s}^{2}}$
8L=$2{{s}^{2}}2{{p}^{6}}$
9M=$3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$
2N=$4{{s}^{2}}$ \[\]
- So we can write the electronic configuration as –$1{{s}^{2}}$$2{{s}^{2}}2{{p}^{6}}$$3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$$4{{s}^{2}}$
Here, atomic number is = 21, which is the atomic number of scandium that is a d block element.
- Now we can clearly see from the electronic configuration that the number of d electrons present is 1 electron.
> Hence, we can conclude that the option (d) is correct.
Additional information:
- Let’s see some basic information of Electron Shell (Orbitals)-
- An atom consists of a number of shells around its nucleus. These shells contain the electrons of the atom. The electron shells mostly consist of one or more subshells. The electronic configuration of the atom is decided by the subshells. The number of electrons that can be accommodated in each shell is given by the formula $2{{n}^{2}}$ where n is the number of the electron shell.
- The electron shells are known as K, L, M, N, O, P, Q going from the innermost electron shell to the outermost shell.
Note: The electronic configuration must be written by the rules of aufbau principle, hund's rule and Pauli exclusion principle. There are some elements that are having exceptions to electronic configuration like chromium, copper.
Step by step solution:
- Let’s see some of rules of writing electronic configuration-
Afbau principle: according to this principle the lower energy orbitals are filled before higher energy orbitals.
Hund’s rule: It says that one electron goes into each subshell until all of them are half full before pairing up.
Pauli Exclusion principle: no two electrons can be identified by the same set of quantum numbers
- 2K,8L,9M, and 2N electrons are given and we know that K, L, M, N are the shells and the numbers that are 2,8,9,2 represents the number of electrons that are present in the particular shell. And s, p, d, f are the subshells and l represents the number of subshells. So,
- For 2K, n=1 , number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 1-1 \right) \\
& l=0\text{ }to\text{ }\left( 0 \right) \\
& l=0 \\
\end{align}\]
So, we get l=0, that is there is only one value, therefore K shell has only one subshell that is s- subshell.
And we know that s subshell can accommodate two electrons , so we can write the configuration as $1{{s}^{2}}$.
- For 8L, n=2 , number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 2-1 \right) \\
& l=0\text{ }to\text{ }\left( 1 \right) \\
& l=0\text{ }to\text{ 1} \\
\end{align}\] \[\]
So, we get l=0 to 1, that is there are two values, therefore L shell has two subshells that is s- subshell and p subshell.
And we know that s subshell can accommodate two electrons , so we can write the configuration as $2{{s}^{2}}$and p subshell can accommodate six electrons , so we can write the configuration as $2{{p}^{6}}$.
- For 9M, n=3, number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 3-1 \right) \\
& l=0\text{ }to\text{ }\left( 2 \right) \\
& l=0\text{ }to\text{ }2 \\
\end{align}\] \[\]
So, we get l=0 to 2, that is there are three values, therefore L shell has three subshells that is s- subshell, p subshell and d-subshell.
And we know that s subshell can accommodate two electrons , so we can write the configuration as $3{{s}^{2}}$and p subshell can accommodate six electrons , so we can write the configuration as$3{{p}^{6}}$.And d- subshell can accommodate 10 electrons but we have given the total electrons as 9 so , as s contains 2 electrons p contains 6 electrons hence d will contain one electron. The electronic configuration is $3{{d}^{1}}$.
- For 2N, n=4, number of subshell is $l=0\text{ }to\text{ }\left( n-1 \right)$
\[\begin{align}
& l=0\text{ }to\text{ }\left( 4-1 \right) \\
& l=0\text{ }to\text{ }\left( 3 \right) \\
& l=0\text{ }to\text{ 3} \\
\end{align}\] \[\]
So, we get l=0 to 3, that is there are four values, therefore M shell has four subshells that is s- subshell, p subshell, d-subshell, and f-subshell.
We are given with two electrons , And we know that s subshell can accommodate two electrons , so we can write the configuration as $4{{s}^{2}}$.
- Now combining all the configurations for each subshell we get,
2K=$1{{s}^{2}}$
8L=$2{{s}^{2}}2{{p}^{6}}$
9M=$3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$
2N=$4{{s}^{2}}$ \[\]
- So we can write the electronic configuration as –$1{{s}^{2}}$$2{{s}^{2}}2{{p}^{6}}$$3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$$4{{s}^{2}}$
Here, atomic number is = 21, which is the atomic number of scandium that is a d block element.
- Now we can clearly see from the electronic configuration that the number of d electrons present is 1 electron.
> Hence, we can conclude that the option (d) is correct.
Additional information:
- Let’s see some basic information of Electron Shell (Orbitals)-
- An atom consists of a number of shells around its nucleus. These shells contain the electrons of the atom. The electron shells mostly consist of one or more subshells. The electronic configuration of the atom is decided by the subshells. The number of electrons that can be accommodated in each shell is given by the formula $2{{n}^{2}}$ where n is the number of the electron shell.
- The electron shells are known as K, L, M, N, O, P, Q going from the innermost electron shell to the outermost shell.
Note: The electronic configuration must be written by the rules of aufbau principle, hund's rule and Pauli exclusion principle. There are some elements that are having exceptions to electronic configuration like chromium, copper.
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