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Standard X

Chemistry

Quantum Numbers

The quantum n...

Question

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.
1. $n = 4, l = 2, m_{1} = - 2, m_{s} = - 1 / 2$
2. $n = 3, l = 2, m_{1} = 1, m_{s} = + 1 / 2$
3. $n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2$
4. $n = 3, l = 2, m_{1} = - 2, m_{s} = - 1 / 2$
5. $n = 3, l = 1, m_{1} = - 1, m_{s} = + 1 / 2$
6. $n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2$

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Solution

1. The quantum numbers $n = 4, l = 2, m_{1} = - 2, m_{s} = - 1 / 2$ represents 4d orbital.

2. The quantum numbers $n = 3, l = 2, m_{1} = 1, m_{s} = + 1 / 2$ represents 3d orbital.

3. The quantum numbers $n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2$ represents 4p orbital.

4. The quantum numbers $n = 3, l = 2, m_{1} = - 2, m_{s} = - 1 / 2$ represents 3d orbital.

5. The quantum numbers $n = 3, l = 1, m_{1} = - 1, m_{s} = + 1 / 2$ represents 3p orbital.

6. The quantum numbers $n = 4, l = 1, m_{1} = 0, m_{s} = + 1 / 2$ represents 4p orbital.

The order of increasing energies is $(v) < (i i) = (i v) < (v i) = (i i i) < (i)$ .

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Similar questions

Q. Explain giving reasons which of the following sets of quantum numbers are not possible.
(a )

n = 0

l = 0

m_{1} = 0

m_{s} = + 1 / 2

(b)

n = 1

l = 0

m_{1} = 0

m_{s} = - 1 / 2

(c)

n = 1

l = 1

m_{1} = 0

m_{s} = + 1 / 2

(d)

n = 2

l = 1

m_{1} = 0

m_{s} = - 1 / 2

(e)

n = 2

l = 3

m_{1} = - 3

m_{s} = + 1 / 2

(f)

n = 3

l = 1

m_{1} = 0

m_{s} = + 1 / 2

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

Q. 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:

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = –2 , ms = –1/2
2. n = 3, l = 2, ml= 1 , ms = +1/2
3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = –2 , ms = –1/2
5. n = 3, l = 1, ml = –1 , ms= +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2

Q. Arrange the electrons represented by the following sets of quantum numbers in decreasing order of energy.
(i)

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

(ii)

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

(iii)

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

(iv)

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

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The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.1. n=4,l=2,m1=−2,ms=−1/22. n=3,l=2,m1=1,ms=+1/23. n=4,l=1,m1=0,ms=+1/24. n=3,l=2,m1=−2,ms=−1/25. n=3,l=1,m1=−1,ms=+1/26. n=4,l=1,m1=0,ms=+1/2