Question

In a multielectron atom which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields?
(I)$n = 1,l = 0,m = 0$
(ii) $n = 2,l = 0,m = 0$
(iii) $n = 2,l = 2,m = 0$
(iv) $n = 3,l = 2,m = 1$
(v) $n = 3,l = 2,m = 0$
a) (i) and (ii)
b) (ii) and (iii)
c) (iii) and (iv)
d) (IV) and (v)

Answer 1

Having knowledge about quantum numbers is necessary. The relation between the four quantum numbers should be clear. Knowing about the $n + l$rule is also important to get the correct answer.

Complete answer:
Quantum numbers are those numbers which tells us about the complete information of an electron in an atom.
There are four quantum numbers that we study-principal quantum number, azimuthal quantum number, magnetic quantum number and spin quantum number.
Principal quantum number ($n$ ) specifies the shell number. It is a positive integer with value $n = 1,2,3.....$ . The maximum value of a shell in an electronic configuration will be equal to the value of$n$ which is known as the valence shell of an electron.
Azimuthal quantum number ($l$) specifies about the subshell. Its value lies between $0$ to$n - 1$. These are the values for the respective orbitals, $l = 0;s:l = 1;p:l = 2;d:l = 3;f$ .
Magnetic quantum number (${m_l}$) tells about the orientation in space of an orbital with $n$ and$l$. Its value lies from$- l,0, + l$.
For $n = 1,l = 0,m = 0$, the orbital will be “$1s$”;
For $n = 2,l = 0,m = 0$, the orbital will be “$2s$”;
For $n = 2,l = 2,m = 0$, the orbital will be “$2d$”;
For $n = 3,l = 2,m = 1$, the orbital will be “$3d$”;
For $n = 3,l = 2,m = 0$, the orbital will be “$3d$”.
‘Aufbau principle’ states that the ground state of atom orbitals are filled in the order of increasing energy. It is represented by-
$1s,2s,2p,3s,3p,4s,3d,4p,5s.......$
According to the $n + l$ rule, orbitals having the same $n + l$ value will have the same energy in the absence of magnetic and electric fields.
Thus $3d$ have $n + l$ value equals $5$ and so $n = 3,l = 2,m = 1$ and $n = 3,l = 2,m = 0$ have the same energy in the absence of magnetic and electric fields.

Hence, the correct option is (d).
Note:
Spin quantum number gives us the orientation of electrons in space which can be either anticlockwise denoted by $s = \dfrac{{ - 1}}{2}or$ clockwise denoted by$s = \dfrac{1}{2}$ .