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Byju's Answer

Standard XII

Chemistry

Electronic Configuration

Which of the ...

Question

Which of the following is not a permissible set of quantum numbers for an electron?

n = 4,

l

= 3,

m_{l}

= 0

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n = 4,

l

= 2,

m_{l}

= 1

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n = 4,

l

= 4,

m_{l}

= 1

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n = 4,

l

= 0,

m_{l}

= 0

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Solution

The correct option is C n = 4, $l$ = 4, $m_{l}$ = 1
Each electron of an atom can be described by four quantum numbers-
$∙$ Principle quantum number (n)
$∙$ Azimuthal quantum number ( $l$ )
$∙$ Magnetic quantum number ( $m_{l}$ )
$∙$ Spin quantum number ( $m_{s}$ )

The possible values of azimuthal quantum number for a given value of n is given by ( $l$ ) = 0,1,2,3, ….., (n – 1)

Also, the permissible values of magnetic quantum number ranges from - $l$ to + $l$ .

When n = 4,
$l$ = 0, 1, 2, 3
and $m_{l}$ = -3, -2,-1, 0, 1 ,2, 3

Therefore, option: n = 4, l = 4, $m_{l}$ = 1 is not possible because l=4 is not permissible for n=4.

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Similar questions

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}

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
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3. n = 4, l = 1, ml = 0 , ms = +1/2
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Q. The quantum number of four electrons are given below:
I.

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

II.

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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:

Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0 l = 0 ml = 0 $m, = + \frac{1}{2}$
b) n = 1 l = 0 ml = 0 $m, = - \frac{1}{2}$
c) n = 1 l = 1 ml = 0 $m, = + \frac{1}{2}$
d) n = 2 l = 1 ml = 0 $m, = - \frac{1}{2}$
e) n = 3 l = 3 ml = – 3 $m, = + \frac{1}{2}$
f) n = 3 l = 1 ml = 0 $m, = + \frac{1}{2}$

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