Question

Match List-I with List-II. List-I contains quantum numbers and List-II contains orbitals. 

List-I	n	l
A	2	1
B	4	0
C	5	3
D	3	2

 

List-II	Orbital
I	3d
II	2p
III	4s
IV	5f

• A. A-II, B-III, C-I, D-IV
• B. A-II, B-III, C-IV, D-I
• C. A-IV, B-II, C-III, D-I
• D. A-I, B-II, C-III, D-IV

Hint:

Remember orbital symbols from \(l\)-values: \(l=0\Rightarrow s\), \(l=1\Rightarrow p\), \(l=2\Rightarrow d\), \(l=3\Rightarrow f\).

Answer 1

The Correct Option is B

Step 1: Recall the relation between \(l\) value and orbital type.
The azimuthal quantum number \(l\) decides the type of orbital. \[ l=0 \Rightarrow s\text{-orbital} \] \[ l=1 \Rightarrow p\text{-orbital} \] \[ l=2 \Rightarrow d\text{-orbital} \] \[ l=3 \Rightarrow f\text{-orbital} \]

Step 2: Match A.
For A: \[ n=2,\quad l=1 \] Since: \[ l=1 \Rightarrow p \] Therefore: \[ A=2p \] So: \[ A-II \]

Step 3: Match B.
For B: \[ n=4,\quad l=0 \] Since: \[ l=0 \Rightarrow s \] Therefore: \[ B=4s \] So: \[ B-III \]

Step 4: Match C.
For C: \[ n=5,\quad l=3 \] Since: \[ l=3 \Rightarrow f \] Therefore: \[ C=5f \] So: \[ C-IV \]

Step 5: Match D.
For D: \[ n=3,\quad l=2 \] Since: \[ l=2 \Rightarrow d \] Therefore: \[ D=3d \] So: \[ D-I \] Thus, the correct matching is: \[ A-II,\quad B-III,\quad C-IV,\quad D-I \]