Test: Quantum Numbers (April 24) - JEE MCQ

Given a set of quantum numbers, n=2,l=0 for a multi-electron atom refers to 2s orbital. 

The next higher allowed set of 'n' and 'l' quantum numbers for this atom in the ground state is n=2,l=1. This corresponds to 2p orbital.

Note: The orbital with a higher value of the sum (n+l) has higher energy.

For 2s orbital (n+l)=(2+0)=2

For 2p orbital (n+l)=(2+1)=3

4l -orbital 
Thus, n = 4, l = 3
m, (anyone) = - 3, - 2, - 1, 0, +1, + 2, + 3

 

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electrons in it.

Since 3p is closer to nucleus than 3d so it will experience greater Z

Hence C

(a) Cr (24) = [Ar] 4s¹3d⁵ Correct
(b) K = 1s²2s²2 p⁶3s²3p⁶4s¹

(c) p-suborbit has three orbitals, each orbital can have maximum of two electrons thus, correct.
(d) s-suborbit has one orbital and this orbital can have maximum of two electrons thus, correct.

When the wave function for an atom is solved using the Schrodinger Wave Equation, the solutions obtained are called the Quantum Number which are basically n, l and m.

In option (b), for n=1, the value of l must be 0, but here l=1, which is not possible. Therefore, this set of quantum numbers is not allowed.

(9) n = 3 can have suborbits

Total nine electrons.