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Quantum Numbers

The quantum n...

Question

The quantum number of four electrons are given below:
I. $n = 4, l = 2, m_{l} = - 2, m_{s} = - 1 / 2$
II. $n = 3, l = 2, m_{l} = 1, m_{s} = + 1 / 2$
III. $n = 4, l = 1, m_{l} = 0, m_{s} = + 1 / 2$
IV. $n = 3, l = 1, m_{l} = 1, m_{s} = - 1 / 2$
The correct order of their increasing energies will be:

I < I I I < I I < I V

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I < I I < I I I < I V

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I V < I I < I I I < I

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I V < I I I < I I < I

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Solution

The correct option is C $I V < I I < I I I < I$
For energy calculations, more is the $n + l$ value greater is the energy. If two electrons have the same value of $n + l$ , the orbital with higher $n$ value will have a higher energy. So the correct order will be $I V < I I < I I I < I$

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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) $n = 3, l = 2, m_{l} = + 2, m_{s} = + \frac{1}{2}$

n = 4, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}

n = 3, l = 1, m_{l} = 1, m_{s} = - \frac{1}{2}

n = 3, l = 2, m_{l} = 0, m_{s} = + \frac{1}{2}

n = 3, l = 0, m_{l} = 0, m_{s} = - \frac{1}{2}

n = 4, l = 2, m_{1} = - 2, m_{s} = - 1 / 2

n = 3, l = 2, m_{1} = 1, m_{s} = + 1 / 2

n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2

n = 3, l = 2, m_{1} = - 2, m_{s} = - 1 / 2

n = 3, l = 1, m_{1} = - 1, m_{s} = + 1 / 2

n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2

n = 0

l = 0

m_{1} = 0

m_{s} = + 1 / 2

n = 1

l = 0

m_{1} = 0

m_{s} = - 1 / 2

n = 1

l = 1

m_{1} = 0

m_{s} = + 1 / 2

n = 2

l = 1

m_{1} = 0

m_{s} = - 1 / 2

n = 2

l = 3

m_{1} = - 3

m_{s} = + 1 / 2

n = 3

l = 1

m_{1} = 0

m_{s} = + 1 / 2

n

l

n = 4, l = 1

n = 4, l = 0

n = 3, l = 2

n = 3, l = 1

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