To find the maximum value of \( n + l + m \) for the last electron present in an element that belongs to the fifth period and group number 15, we can follow these steps:
### Step 1: Identify the Element
The fifth period of the periodic table includes elements with atomic numbers from 37 (Rubidium) to 54 (Xenon). Group number 15 elements in this period include:
- Arsenic (As, atomic number 33)
- Selenium (Se, atomic number 34)
- Bromine (Br, atomic number 35)
- Krypton (Kr, atomic number 36)
- Rubidium (Rb, atomic number 37)
- Strontium (Sr, atomic number 38)
- Yttrium (Y, atomic number 39)
- Zirconium (Zr, atomic number 40)
- Niobium (Nb, atomic number 41)
- Molybdenum (Mo, atomic number 42)
- Technetium (Tc, atomic number 43)
- Ruthenium (Ru, atomic number 44)
- Rhodium (Rh, atomic number 45)
- Palladium (Pd, atomic number 46)
- Silver (Ag, atomic number 47)
- Cadmium (Cd, atomic number 48)
- Indium (In, atomic number 49)
- Tin (Sn, atomic number 50)
- Antimony (Sb, atomic number 51)
- Tellurium (Te, atomic number 52)
- Iodine (I, atomic number 53)
- Xenon (Xe, atomic number 54)
The element in group 15 of the fifth period is **Antimony (Sb)** with atomic number 51.
### Step 2: Determine the Electron Configuration
The electron configuration of Antimony (Sb) is:
\[
\text{[Kr]} 5s^2 4d^{10} 5p^3
\]
This shows that the last electron fills the 5p orbital.
### Step 3: Identify Quantum Numbers
For the last electron in the 5p orbital:
- The principal quantum number \( n \) is **5** (since it is in the fifth period).
- The azimuthal quantum number \( l \) for a p orbital is **1**.
- The magnetic quantum number \( m \) for the three p orbitals can be -1, 0, or +1. The maximum value of \( m \) for p orbitals is **1**.
### Step 4: Calculate \( n + l + m \)
Now we can calculate:
\[
n + l + m = 5 + 1 + 1 = 7
\]
### Conclusion
The maximum value of \( n + l + m \) for the last electron present in an element which belongs to the fifth period and group number 15 is **7**.
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