The number of orbitals having (n + l) < 5 is — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Understanding the Quantum Number Constraints In atomic structure, the state of an electron in an atom is defined by a set of quantum numbers: * $n$: The principal quantum number, which represents the main energy shell ($n = 1, 2, 3, \dots$). * $l$: The azimuthal quantum number, which represents the subshell and defines its shape. For any given shell $n$, $l$ can take integer values from $0$ to $n - 1$. The subshells are conventionally designated as: * $s$-subshell: $l = 0$ * $p$-subshell: $l = 1$ * $d$-subshell: $l = 2$ * $f$-subshell: $l = 3$ The mathematical condition given in the problem is: $$n + l < 5$$ --- ### Step 2 - Determining Orbitals per Subshell Each subshell characterized by $l$ contains a specific number of degenerate orbitals given by the formula: $$\text{Number of orbitals} = 2l + 1$$ Therefore: * Any $s$-subshell ($l = 0$) contains: $2(0) + 1 = 1$ orbital. * Any $p$-subshell ($l = 1$) contains: $2(1) + 1 = 3$ orbitals. * Any $d$-subshell ($l = 2$) contains: $2(2) + 1 = 5$ orbitals. --- ### Step 3 - Systematic Listing and Evaluation of Subshells Let us systematically evaluate all possible combinations of $(n, l)$ starting from the lowest energy level ($n = 1$) to find which combinations satisfy the condition $n + l < 5$: 1. **For $n = 1$:** * $l = 0$ ($1s$ subshell): $$n + l = 1 + 0 = 1 < 5 \quad \text{(Valid)}$$ Number of orbitals = $1$ 2. **For $n = 2$:** * $l = 0$ ($2s$ subshell): $$n + l = 2 + 0 = 2 < 5 \quad \text{(Valid)}$$ Number of orbitals = $1$ * $l = 1$ ($2p$ subshell): $$n + l = 2 + 1 = 3 < 5 \quad \text{(Valid)}$$ Number of orbitals = $3$ 3. **For $n = 3$:** * $l = 0$ ($3s$ subshell): $$n + l = 3 + 0 = 3 < 5 \quad \text{(Valid)}$$ Number of orbitals = $1$ * $l = 1$ ($3p$ subshell): $$n + l = 3 + 1 = 4 < 5 \quad \text{(Valid)}$$ Number of orbitals = $3$ * $l = 2$ ($3d$ subshell): $$n + l = 3 + 2 = 5 \not< 5 \quad \text{(Excluded)}$$ 4. **For $n = 4$:** * $l = 0$ ($4s$ subshell): $$n + l = 4 + 0 = 4 < 5 \quad \text{(Valid)}$$ Number of orbitals = $1$ * $l = 1$ ($4p$ subshell): $$n + l = 4 + 1 = 5 \not< 5 \quad \text{(Excluded)}$$ For any shell $n \ge 5$, since $l \ge 0$, the value of $n + l$ will always be at least $5$. Thus, no subshells from $n \ge 5$ can satisfy the inequality. --- ### Step 4 - Summing the Total Number of Orbitals The total number of orbitals satisfying the given inequality is the sum of the orbitals in all the valid subshells identified in Step 3: $$\text{Total Orbitals} = \text{Orbitals in } (1s + 2s + 2p + 3s + 3p + 4s)$$ $$\text{Total Orbitals} = 1 + 1 + 3 + 1 + 3 + 1$$ $$\text{Total Orbitals} = \boxed{10}$$ --- ### Step 5 - Evaluation of Options * **Option (A) is incorrect:** $9$ orbitals would be obtained if the $4s$ subshell ($1$ orbital) was mistakenly omitted from the sum. * **Option (B) is incorrect:** $8$ orbitals does not account for all of the degenerate orbitals present in the $p$-subshells. * **Option (C) is incorrect:** $4$ represents only the number of valid $s$-subshells ($1s, 2s, 3s, 4s$), completely ignoring the $p$-subshells. * **Option (D) is correct:** There are exactly $10$ orbitals that satisfy the condition $(n + l) < 5$.