What are the four quantum numbers of 19th electron in $S c (Z = 21)$ ?

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $n = 4, l = 0, m = 0, m_{s} = + \frac{1}{2}$
$S c (21) = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{1}$

19th electron goes into $4 s$ .

Hence, $n = 4, l = 0,$

$m = 0, m_{s} = + \frac{1}{2}$

Therefore, option A is correct.

Suggest Corrections

Similar questions

Q. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.

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

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