The number of electrons present in the valence shell of an atom with atomic number $38$ is:
(a) $2$
(b) $10$
(c) $1$
(d) $8$
Answer
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:For these questions we have to know about the valence shells, how the electrons are distributed in each shell of any atom and how to calculate the number of electrons present in each shell and the number of electrons in the valence shell.
Complete answer:
Starting with the some basics:
There are different shells in any atom, like K, L, M, N and so on. The K shell is the innermost shell of all the atoms and then comes the L shell moving radially outwards and so on proceeding as K, L, M, N and so on. These shells are filled with the electrons present in the atoms (The atomic number is the number of protons as well as the electrons present in the atom).
Now, how to find the number if electrons present in the each shell:
The highest electrons present in any shell of the atom is calculated by the number of the valence shell from the innermost side and the formula for calculating that is $2{n^2}$ where, $n$ is the number of valence shells. Like any atoms has three shells then the second innermost shell has $8(2{n^2},n = 2)$ electrons in the shell. But let, if there is an atomic number of $20$ , so the innermost shell has $2$ electrons, then $8$ for the second shell but for the third shell it hasn't $18$ electrons. So, the third shell, i.e, M shell will have $8$ electrons and the remaining $2$ electrons will be in the fourth shell, i.e, on N shell and the valence electrons of the atom will be $2$ because it has $2$ electrons in the outermost shell.
Now, what will happen for the atomic number $38$ ,
Its electronic configuration will be $2,8,18,8,2$
The outermost shell has $2$ electrons.
Hence, the number of electrons present in the valence shell of an atom with atomic number $38$ is $2$
Hence, the correct option is (a) $2$ .
Note:
While writing electronic configuration of any atom, we have to keep in mind that the number of electrons present in any shell of the atom will be in the form of $2{n^2}$ except in the valence shell.
Complete answer:
Starting with the some basics:
There are different shells in any atom, like K, L, M, N and so on. The K shell is the innermost shell of all the atoms and then comes the L shell moving radially outwards and so on proceeding as K, L, M, N and so on. These shells are filled with the electrons present in the atoms (The atomic number is the number of protons as well as the electrons present in the atom).
Now, how to find the number if electrons present in the each shell:
The highest electrons present in any shell of the atom is calculated by the number of the valence shell from the innermost side and the formula for calculating that is $2{n^2}$ where, $n$ is the number of valence shells. Like any atoms has three shells then the second innermost shell has $8(2{n^2},n = 2)$ electrons in the shell. But let, if there is an atomic number of $20$ , so the innermost shell has $2$ electrons, then $8$ for the second shell but for the third shell it hasn't $18$ electrons. So, the third shell, i.e, M shell will have $8$ electrons and the remaining $2$ electrons will be in the fourth shell, i.e, on N shell and the valence electrons of the atom will be $2$ because it has $2$ electrons in the outermost shell.
Now, what will happen for the atomic number $38$ ,
Its electronic configuration will be $2,8,18,8,2$
The outermost shell has $2$ electrons.
Hence, the number of electrons present in the valence shell of an atom with atomic number $38$ is $2$
Hence, the correct option is (a) $2$ .
Note:
While writing electronic configuration of any atom, we have to keep in mind that the number of electrons present in any shell of the atom will be in the form of $2{n^2}$ except in the valence shell.
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