Column -I Column-II

A. $2 s$ (P) $n = 4, l = 2, m = 0$

B. $2 p_{z}$ (Q) $n = 4, l = 2, m = - 2 o r + 2$

C. $4 d_{x^{2} - y^{2}}$ (R) $n = 2, l = 1, m = 0$

D. $4 d_{z}^{2}$ (S) $n = 2, l = 0, m = 0$

Match the column -I with column -II

A - S, B - R, C - Q, D - P

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A - Q, B - P, C - S, D - R

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A - R, B - S, C - P, D - Q

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A - S, B - R, C - P, D - Q

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Solution

The correct option is A $A - S, B - R, C - Q, D - P$
We know that,
$n$ = Principal quantum number which represents the energy level of the orbital.
$l$ = Azimuthal quantum number which have the value $n - 1$
m= magnetic quantum number which have the value $- l t o + l$
For, $2 s \Rightarrow n = 2, l = 0, m = 0 2 p_{z} \Rightarrow n = 2, l = 1, m = 0 4 d_{x^{2} - y^{2}} \Rightarrow n = 4, l = 2, m = - 2 o r + 2 4 d_{z}^{2} \Rightarrow n = 4, l = 2, m = 0$

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Question 18

Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals.

(a) (i) $n = 3, l = 2, m_{l} = - 2, m_{s} = - \frac{1}{2}$ (ii) $n = 3, l = 2, m_{l} = - 1, m_{s} = - \frac{1}{2}$

(b) (i) $n = 3, l = 1, m_{l} = 1, m_{s} = + \frac{1}{2}$ (ii) $n = 3, l = 2, m_{l} = - 1, m_{s} = + \frac{1}{2}$

(d) (i) $n = 3, l = 2, m_{l} = + 2, m_{s} = - \frac{1}{2}$ (ii) $n = 3, l = 2, m_{l} = + 2, m_{s} = + \frac{1}{2}$

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