Engineering
Chemistry
Quantum Numbers
Question

The electrons identified by quantum numbers n and ℓ :

(a) n = 4, ℓ = 1               (b) n = 4, ℓ = 0 

(c) n = 3, ℓ = 2               (d) n = 3 , ℓ = 1

can be placed in order of increasing energy as :

(c) < (d) < (b) < (a)

(d) < (b) < (c) < (a)

(b) < (d) < (a) < (c)

(a) < (c) < (b) < (d)

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Solution

                                               Value of (n + )

            (a) n = 4, = 1 4p          4 + 1 = 5

            (b) n = 4, = 0 4s          4 + 0 = 4

            (c) n = 3, = 2 3d          3 + 2 = 5

            (d) n = 3, = 1 3p          3 + 1 = 4

            orbital having more (n + ) value has more energy if value of (n + ) is same then orbital having lower value of n has less energy.

            3p < 4s < 3d < 4p

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