Question

In an atom the order of increasing energy of electrons with quantum numbers (i) n=4, l=1 (ii) n=4, l=0 (iii) n=3, l=2 and (iv) n=3, l=1 is :

• A. $\left(iii\right)<\left(i\right)<\left(iv\right)<\left(ii\right)$
• B. $\left(ii\right)<\left(iv\right)<\left(i\right)<\left(iii\right)$
• C. $\left(i\right)<\left(iii\right)<\left(ii\right)<\left(iv\right)$
• D. $\left(iv\right)<\left(ii\right)<\left(iii\right)<\left(i\right)$

Answer 1

The correct option is D $\left(iv\right)<\left(ii\right)<\left(iii\right)<\left(i\right)$

Energy of electrons with quantum numbers depends upon the value of n + l.

Hence , the value of n + l is more there will be more energy.

(i) n = 4, l = 1

n + l = 4 + 1 = 5

(ii) n = 4 , l = 0

n + l = 4 + 0 = 4

(iii) n = 3 , l = 2

n + l = 3 + 2 = 5

(iv) n = 3 , l = 1 is

n + l = 3 + 1 = 4

If the value of n + l is same, then the atom whose value of n is more have greater value of energy .

Thus according to above points, we get atoms in the increasing order :-

(iv) < (ii) < (iii) < (i)

Hence , option D is correct .