
Consider the hypothetical situation where the azimuthal quantum number, l, takes values 0, \[1,{\text{ }}2,{\text{ }}......{\text{ }}n{\text{ }} + {\text{ }}1\], where \[n\] is the principal quantum number. Then, the element with atomic number:
(1) \[13\] has a half-filled valence subshell
(2) \[9\] is the first alkali metal
(3) \[8\] is the first noble gas
(4) \[6\] has a \[2p\] -valence subshell
Answer
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Hint: The element with atomic number \[n\] has certain properties that are determined by the value of \[n\]. These properties include the element's electron configuration, ionisation energy, and atomic radius. In general, as \[n\] increases, so do these properties.
Complete Step by Step Solution:
The electron configuration of an element is determined by the value of\[n\]. The first orbital, or energy level, can hold a maximum of \[2\] electrons. The second level can hold a maximum of \[8\] electrons, the third level can hold a maximum of \[18\] electrons, and so on. The element with atomic number n will have an electron configuration that corresponds to the level that can hold \[n\] electrons. The ionisation energy of an element is the amount of energy required to remove an electron from the atom. The higher the ionisation energy, the more difficult it is to remove an electron. The ionisation energy of an element with atomic number \[n\] is determined by the value of 2, 8, 18. In general, the higher the value of 2, 8, 18, the higher the ionisation energy. The atomic radius of an element is the distance from the centre of the nucleus to the outermost electron. The atomic radius of an element with atomic number \[n\] is determined by the value of \[n\]. In general, the higher the value of\[n\], the larger the atomic radius.
Option 1) \[13{\text{ }}:{\text{ }}1s21p62s21d3\] is not half filled.
Option 2) \[9{\text{ }}:{\text{ }}1s21p62s1\] is the first alkali metal because after losing one electron, it will achieve first noble gas configuration.
Option 3) \[8{\text{ }}:{\text{ }}1s21p6\]is the first noble gas because after \[1p6{\text{ }}e-\] will enter \[2s\] hence new period.
Option 4) \[6{\text{ }}:{\text{ }}1{\text{ }}2{\text{ }}1p4\]has \[1p\] valence subshell.
(1) \[_{13}X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^5}\;-\] Half filled.
(2) \[_9X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^1}\;-\] Not alkali metal
(3) \[_8X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;-\] Second Noble gas
Option 1 is the right answer.
Note: The element with atomic number \[n\] has certain properties that are determined by the value of \[n\]. These properties include the element's electron configuration, ionisation energy, and atomic radius. In general, as \[n\] increases, so do these properties.
Complete Step by Step Solution:
The electron configuration of an element is determined by the value of\[n\]. The first orbital, or energy level, can hold a maximum of \[2\] electrons. The second level can hold a maximum of \[8\] electrons, the third level can hold a maximum of \[18\] electrons, and so on. The element with atomic number n will have an electron configuration that corresponds to the level that can hold \[n\] electrons. The ionisation energy of an element is the amount of energy required to remove an electron from the atom. The higher the ionisation energy, the more difficult it is to remove an electron. The ionisation energy of an element with atomic number \[n\] is determined by the value of 2, 8, 18. In general, the higher the value of 2, 8, 18, the higher the ionisation energy. The atomic radius of an element is the distance from the centre of the nucleus to the outermost electron. The atomic radius of an element with atomic number \[n\] is determined by the value of \[n\]. In general, the higher the value of\[n\], the larger the atomic radius.
Option 1) \[13{\text{ }}:{\text{ }}1s21p62s21d3\] is not half filled.
Option 2) \[9{\text{ }}:{\text{ }}1s21p62s1\] is the first alkali metal because after losing one electron, it will achieve first noble gas configuration.
Option 3) \[8{\text{ }}:{\text{ }}1s21p6\]is the first noble gas because after \[1p6{\text{ }}e-\] will enter \[2s\] hence new period.
Option 4) \[6{\text{ }}:{\text{ }}1{\text{ }}2{\text{ }}1p4\]has \[1p\] valence subshell.
(1) \[_{13}X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^5}\;-\] Half filled.
(2) \[_9X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;1{d^1}\;-\] Not alkali metal
(3) \[_8X{\text{ }} = {\text{ }}1{s^2}\;1{p^6}\;-\] Second Noble gas
Option 1 is the right answer.
Note: The element with atomic number \[n\] has certain properties that are determined by the value of \[n\]. These properties include the element's electron configuration, ionisation energy, and atomic radius. In general, as \[n\] increases, so do these properties.
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