Question

How would you write a full set of quantum numbers for the first, third, and fourteenth electron removed from the ground state of a Cu atom?

Answer 1

Quantum number are the numbers that represent the position and energy of an electron in any orbit of the atom. There are four categories of quantum numbers: Principal, Azimuthal, Magnetic and Spin quantum numbers. Quantum system’s conserved quantities are given by these quantum numbers.

Complete answer:
The exact position of an electron in an atom is given by these four quantum numbers. Principal Quantum number is given as ‘n’. Azimuthal quantum number is given by ‘l’ , magnetic quantum number as ${m_l}$ and spin quantum number as ${m_s}$ .
The shell or energy level of an atom is given by Principal quantum number. The shape of the atomic orbital and its angular momentum is given by the Azimuthal quantum number. The spatial orientation is given by the Magnetic quantum number and the position of an electron’s intrinsic angular momentum is given by spin quantum number

Quantum Number	Symbol	Possible Values
Principal Quantum Number	n	1,2,3,4,5…..n
Azimuthal Quantum Number	$l$	0,1,2,3,….. $(n - 1)$
Magnetic quantum number	${m_l}$	$- l.... - 1,0, + 1.... + l$
Spin quantum number	${m_s}$	$+ 1/2, - 1/2$

The electronic configuration of Cu atom is given as: $Cu:1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^1}$
The first electron that will be removed will be from the 4s orbital. It is the fourth energy level. The quantum number sets for the electron to be removed from the 4s orbital will be:
$n = 4 \to$ electron in the fourth energy level
$l = 0 \to$ located in s-orbital
${m_l} = 0 \to$ electron located in s-orbital
${m_s} = + 1/2 \to$ electron has a up spin
Removing the electron from the 4s orbital will lead to the formation of $C{u^ + }$ . The electronic configuration now becomes: $C{u^ + }:1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}$
The removal of the third electron from the neutral atom will be the same as removing the second electron from the $C{u^ + }$ cation. The electron will be removed from the 3d orbital only. It is in the third energy level. The quantum numbers will be:
$n = 3 \to$ electron in the third energy level
$l = 2 \to$ located in d-orbital
${m_l} = - 1 \to$ electron located in $3{d_{xy}}$ -orbital
${m_s} = - 1/2 \to$ electron has a down spin, assuming that the first electron in the empty orbital had a spin of ${m_s} = + 1/2$
The quantum number of the fourteenth electron has to be found out. The 3d orbital has 10 electrons and the 4s has 1 electron. Removing 11 electrons will leave us with: $C{u^{ + 11}}:1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}$
Removing 2 more electrons will leave us with $C{u^{ + 13}}:1{s^2}2{s^2}2{p^6}3{s^2}3{p^4}$
Therefore, we can say that the fourteenth electron will be removed from the 3p orbital only. The quantum number for the same will be:
$n = 3 \to$ electron in the third energy level
$l = 1 \to$ located in p-orbital
${m_l} = + 1 \to$ electron located in $2{p_z}$ -orbital
${m_s} = - 1/2 \to$ electron has a down spin since it is the first electron to be removed from the $2{p_z}$ orbital.

Note:
Quantum numbers are necessary to figure out the exact location of an electron and also to know the atom’s electron structure. Properties like ionisation energy, atomic radius have the significance of quantum numbers. Note that no atom has all four quantum numbers the same.