
How many orbitals are filled in the second period?
Answer
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Hint: An orbital is the region in the atom where probability of finding an electron is the maximum. The sub shells that are s, p, d, f consist of atomic orbitals that can accommodate 2 electrons each. The number of orbitals in s, p, d, f subshells is 1, 3, 5, 7 respectively. A period consists of elements having the same number of shells, so the number of orbitals can be calculated accordingly.
Complete answer:
The modern periodic table consists of 7 periods or horizontal rows. The elements in these periods are present in the number in accordance with the filling of the shells K, L, M, and N. The periods consist of elements having the same number of shells.
The first period of the periodic table has only 2 elements according to the K shell that can have 2 electrons. Now, the second period consists of elements from Li to Ne. These are total 8 in number according to the L shell. So, the orbitals that will be filled in the second period will be inferred by the electronic configuration of the last element.
The electrons in an atom are present in atomic orbitals. The number of orbitals in the subshells s, p, d, and f are 1, 3, 5, and 7 respectively. Now according to the configuration of Ne that is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$, there are s and p subshells occupied by the elements of 2nd period. So, the orbitals will be 1 of s and 3 of p that is 4 orbitals.
So, 4 orbitals are filled in the second period.
Note:
A trend may be noted that the 2nd period has their valence electrons entering into the 2nd shell, while in the 3rd period the last shell is 3 and so on. According to the shells K, L, M and N, there are 2 elements in 1st period, 8 elements in 2nd and 3rd period, 18 elements in the 4th and 5th period, and 32 elements in the 6th and 7th period.
Complete answer:
The modern periodic table consists of 7 periods or horizontal rows. The elements in these periods are present in the number in accordance with the filling of the shells K, L, M, and N. The periods consist of elements having the same number of shells.
The first period of the periodic table has only 2 elements according to the K shell that can have 2 electrons. Now, the second period consists of elements from Li to Ne. These are total 8 in number according to the L shell. So, the orbitals that will be filled in the second period will be inferred by the electronic configuration of the last element.
The electrons in an atom are present in atomic orbitals. The number of orbitals in the subshells s, p, d, and f are 1, 3, 5, and 7 respectively. Now according to the configuration of Ne that is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$, there are s and p subshells occupied by the elements of 2nd period. So, the orbitals will be 1 of s and 3 of p that is 4 orbitals.
So, 4 orbitals are filled in the second period.
Note:
A trend may be noted that the 2nd period has their valence electrons entering into the 2nd shell, while in the 3rd period the last shell is 3 and so on. According to the shells K, L, M and N, there are 2 elements in 1st period, 8 elements in 2nd and 3rd period, 18 elements in the 4th and 5th period, and 32 elements in the 6th and 7th period.
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