
If each orbital can hold a maximum of 3 electrons, the number of elements in ${2^{nd}}$ period of the periodic table (long form) will be:
(A) 27
(B) 9
(C) 18
(D) 12
Answer
476.7k+ views
Hint: The formula to find the number of types of orbitals present in the atom is $\dfrac{{(n + 2)}}{2}$ . This formula is applicable only when the value of n is even. ‘n’ is the principal quantum number here.
Complete step by step solution:
Here, we are given to expect that each orbital has a capacity to fill three electrons. In that way, we need to calculate the number of elements in the second period of the periodic table.
- Let’s find the number of orbitals present in an element in the case where n is given. Here, ‘n’ is the principal quantum number.
So, in the second period of the periodic table, the principal quantum number of all the elements will be 2. So, number of orbitals present in all the elements will be$\dfrac{{(n + 2)}}{2} = \dfrac{{(2 + 2)}}{{}} = \dfrac{4}{2} = 2$
- So, we know that the two types of orbitals will be ‘s’ and ‘p’ orbitals.
- Now, we know that s has one orbital and p has three orbitals. We are given that each orbital has the ability to include three electrons.
- So, S orbital can have $3 \times 1 = 3$ electrons and p orbitals will have$3 \times 3 = 9$ electrons. So, total number of electrons in both the types of orbitals will be 12.
- So, we can say that the second shell will have the capacity to have 12 electrons. So, as the capacity of 12 electrons is there, there will be 12 different electronic configurations available and hence the number of elements in the period will be 12.
- Thus, we can say that if an orbital could hold a maximum of 3 electrons, then there would be a total 12 elements in the second period of the periodic table.
So, the correct answer is (D).
Note: Note that as we are told to give the number of elements in the second period only, we need not to take any other shell into consideration except the second shell. If we add the electrons of other shells also, then there will be error in calculation.
Complete step by step solution:
Here, we are given to expect that each orbital has a capacity to fill three electrons. In that way, we need to calculate the number of elements in the second period of the periodic table.
- Let’s find the number of orbitals present in an element in the case where n is given. Here, ‘n’ is the principal quantum number.
So, in the second period of the periodic table, the principal quantum number of all the elements will be 2. So, number of orbitals present in all the elements will be$\dfrac{{(n + 2)}}{2} = \dfrac{{(2 + 2)}}{{}} = \dfrac{4}{2} = 2$
- So, we know that the two types of orbitals will be ‘s’ and ‘p’ orbitals.
- Now, we know that s has one orbital and p has three orbitals. We are given that each orbital has the ability to include three electrons.
- So, S orbital can have $3 \times 1 = 3$ electrons and p orbitals will have$3 \times 3 = 9$ electrons. So, total number of electrons in both the types of orbitals will be 12.
- So, we can say that the second shell will have the capacity to have 12 electrons. So, as the capacity of 12 electrons is there, there will be 12 different electronic configurations available and hence the number of elements in the period will be 12.
- Thus, we can say that if an orbital could hold a maximum of 3 electrons, then there would be a total 12 elements in the second period of the periodic table.
So, the correct answer is (D).
Note: Note that as we are told to give the number of elements in the second period only, we need not to take any other shell into consideration except the second shell. If we add the electrons of other shells also, then there will be error in calculation.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

State the laws of reflection of light

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
