Ncert Exercises & Solutions

Given below are the quantum numbers for two cases:
(a) $( n = 4,~ m_s = -\frac{1}{2} ) $
(b) $( n = 3, l = 0 ) $
How many electrons in an atom can have these quantum numbers, respectively?

1. 16, 3
2. 18, 2
3. 16, 2
4. 2, 16

Hint: Total number of electrons in an atom for a value of n =

2 n^{2}

(a) Total number of electrons in an atom for a value of n = $2 n^{2}$
$∴$ For n = 4,
Total number of electrons = 2 (4)2 = 32
The given element has a fully filled orbital as $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$ .
Hence, all the electrons are paired.
$∴$ Number of electrons (having n = 4 and m, $= - \frac{1}{2}$ ) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

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