Questions: Complete the table by inputting in the n value, l value, and number of orbitals for each given sublevel. sublevel n value l value number of orbitals ------------ 6 d 5 s 2 s

Complete the table by inputting in the n value, l value, and number of orbitals for each given sublevel.

sublevel  n value  l value  number of orbitals
------------
6 d      
5 s      
2 s
Transcript text: Complete the table by inputting in the $n$ value, $l$ value, and number of orbitals for each given sublevel. \begin{tabular}{|c|c|c|c|} \hline sublevel & $n$ value & $l$ value & number of orbitals \\ \hline $6 d$ & $\square$ & $\square$ & $\square$ \\ \hline $5 s$ & $\square$ & $\square$ & $\square$ \\ \hline $2 s$ & $\square$ & $\square$ & $\square$ \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Determine the $n$ value for each sublevel

The principal quantum number \( n \) is the first number in the sublevel notation.

  • For \( 6d \), \( n = 6 \)
  • For \( 5s \), \( n = 5 \)
  • For \( 2s \), \( n = 2 \)
Step 2: Determine the $l$ value for each sublevel

The azimuthal quantum number \( l \) is determined by the letter in the sublevel notation:

  • \( s \) corresponds to \( l = 0 \)

  • \( p \) corresponds to \( l = 1 \)

  • \( d \) corresponds to \( l = 2 \)

  • \( f \) corresponds to \( l = 3 \)

  • For \( 6d \), \( l = 2 \)

  • For \( 5s \), \( l = 0 \)

  • For \( 2s \), \( l = 0 \)

Step 3: Determine the number of orbitals for each sublevel

The number of orbitals in a sublevel is given by \( 2l + 1 \).

  • For \( 6d \), \( l = 2 \), so the number of orbitals is \( 2(2) + 1 = 5 \)
  • For \( 5s \), \( l = 0 \), so the number of orbitals is \( 2(0) + 1 = 1 \)
  • For \( 2s \), \( l = 0 \), so the number of orbitals is \( 2(0) + 1 = 1 \)

Final Answer

\[ \begin{array}{|c|c|c|c|} \hline \text{sublevel} & n \text{ value} & l \text{ value} & \text{number of orbitals} \\ \hline 6d & \boxed{6} & \boxed{2} & \boxed{5} \\ \hline 5s & \boxed{5} & \boxed{0} & \boxed{1} \\ \hline 2s & \boxed{2} & \boxed{0} & \boxed{1} \\ \hline \end{array} \]

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