Answer
Verified
109.2k+ views
Hints: The Number of orbitals can be easily calculated according to the orbit concerned. First, we need to calculate the number of subshells, then the number of orbitals, and the number of elements.
Formula used:
Number of orbitals = ${n^2}$, where $n$ is the number of orbit(s).
Complete step by step solution:
Firstly, Rutherford stated that the electrons revolve around the nucleus. Later, Bohr came to the conclusion that the electrons revolve around the nucleus but in a stationary orbit and never jump into other orbits until they are forced to do so, like supplying energy to it or absorbing energy from it. After that Bohr stated that there are orbitals associated with respective orbits namely s (sharp), p (principle), d (diffused), f (fundamental), and so on. These orbitals have their respective limits of holding electrons such as s=2, p=6, d=10, f=14, and so on.
Each orbital will contain half shells to their limited electron numbers. Each shell of orbitals contains 2 electrons of opposite spin. The first orbit contains only s orbital; 2nd orbit contains s, p orbitals; 3rd orbit contains s, p, d orbitals; 4th orbit contains s, p, d, f orbitals and it continues further. Now the number of orbitals concerned for a particular can be determined by the square of the number of that orbit.
Since, for the first orbit, $n = 1$
Number of subshell $ = {n^2} = {1^2} = 1$
Hence, the 9 orbitals are associated with the 3rd orbit i.e. one s, three p, and five d orbitals.
For 2nd orbit, $n = 2$
Number of subshell $ = {2^2} = 4$
For 3rd orbit, $n = 3$
Number of subshell $ = {3^2} = 9$
Note: There is only one orbital in the 2s subshell. But as there are three directions in which a p-orbital can point, there are three orbitals in the 2p subshell. One component of these orbitals is oriented along the X-axis, another along the Y-axis, and the last one is along the Z-axis of a coordinate system and these orbitals are known as the 2px, 2py, and 2pz orbitals.
Formula used:
Number of orbitals = ${n^2}$, where $n$ is the number of orbit(s).
Complete step by step solution:
Firstly, Rutherford stated that the electrons revolve around the nucleus. Later, Bohr came to the conclusion that the electrons revolve around the nucleus but in a stationary orbit and never jump into other orbits until they are forced to do so, like supplying energy to it or absorbing energy from it. After that Bohr stated that there are orbitals associated with respective orbits namely s (sharp), p (principle), d (diffused), f (fundamental), and so on. These orbitals have their respective limits of holding electrons such as s=2, p=6, d=10, f=14, and so on.
Each orbital will contain half shells to their limited electron numbers. Each shell of orbitals contains 2 electrons of opposite spin. The first orbit contains only s orbital; 2nd orbit contains s, p orbitals; 3rd orbit contains s, p, d orbitals; 4th orbit contains s, p, d, f orbitals and it continues further. Now the number of orbitals concerned for a particular can be determined by the square of the number of that orbit.
Since, for the first orbit, $n = 1$
Number of subshell $ = {n^2} = {1^2} = 1$
Hence, the 9 orbitals are associated with the 3rd orbit i.e. one s, three p, and five d orbitals.
For 2nd orbit, $n = 2$
Number of subshell $ = {2^2} = 4$
For 3rd orbit, $n = 3$
Number of subshell $ = {3^2} = 9$
Note: There is only one orbital in the 2s subshell. But as there are three directions in which a p-orbital can point, there are three orbitals in the 2p subshell. One component of these orbitals is oriented along the X-axis, another along the Y-axis, and the last one is along the Z-axis of a coordinate system and these orbitals are known as the 2px, 2py, and 2pz orbitals.
Recently Updated Pages
If x2 hx 21 0x2 3hx + 35 0h 0 has a common root then class 10 maths JEE_Main
The radius of a sector is 12 cm and the angle is 120circ class 10 maths JEE_Main
For what value of x function fleft x right x4 4x3 + class 10 maths JEE_Main
What is the area under the curve yx+x1 betweenx0 and class 10 maths JEE_Main
The volume of a sphere is dfrac43pi r3 cubic units class 10 maths JEE_Main
Which of the following is a good conductor of electricity class 10 chemistry JEE_Main