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19) Be's 4th electron will have four quantum numbers:

\begin{tabular}{|l|l|l|l|l|}
\hline
Option & [tex]$n$[/tex] & [tex]$l$[/tex] & [tex]$m_l$[/tex] & [tex]$s$[/tex] \\
\hline
A & 1 & 0 & 0 & [tex]$1/2$[/tex] \\
\hline
B & 1 & 1 & 1 & [tex]$1/2$[/tex] \\
\hline
C & 2 & 0 & 0 & [tex]$-1/2$[/tex] \\
\hline
D & 2 & 1 & 0 & [tex]$+1/2$[/tex] \\
\hline
\end{tabular}



Answer :

To determine the correct set of quantum numbers for the fourth electron of Beryllium (Be), we need to follow several steps by understanding the electron configuration and quantum numbers. Let's break this down:

1. Electron Configuration of Beryllium:
- Beryllium has an atomic number of 4, meaning it has 4 electrons.
- The electron configuration for Beryllium is: 1s² 2s².

2. Identifying the 4th Electron:
- Using the electron configuration, we can identify the distribution of electrons:
- The 1s orbital is fully occupied by 2 electrons.
- The 2s orbital holds the remaining 2 electrons.
- Thus, the 4th electron is located in the 2s orbital.

3. Quantum Numbers: Four quantum numbers uniquely describe an electron's state within an atom:
- Principal Quantum Number (n): Indicates the main energy level or shell. For the 2s orbital, [tex]\( n = 2 \)[/tex].
- Azimuthal Quantum Number (l): Defines the subshell or orbital type (s, p, d, f), where:
- For s orbitals, [tex]\( l = 0 \)[/tex].
- Magnetic Quantum Number (m_l): Specifies the orientation of the orbital in space. For:
- s orbitals (l = 0), [tex]\( m_l \)[/tex] can only be 0.
- Spin Quantum Number (m_s): Describes the spin of the electron. It can be [tex]\( +1/2 \)[/tex] for an electron with an upward spin or [tex]\( -1/2 \)[/tex] for an electron with a downward spin:
- In the 2s orbital, the first electron will have a spin [tex]\( +1/2 \)[/tex] and the second (4th overall) will have a spin [tex]\( -1/2 \)[/tex].

Combining all these quantum numbers for Beryllium's 4th electron, we get:
- [tex]\( n = 2 \)[/tex]
- [tex]\( l = 0 \)[/tex]
- [tex]\( m_l = 0 \)[/tex]
- [tex]\( m_s = -1/2 \)[/tex]

Therefore, the correct set of quantum numbers for Beryllium's 4th electron is given by option:
- C [tex]\((n = 2, l = 0, m_l = 0, m_s = -1/2)\)[/tex].

The correct answer is: C