Question

Match the atoms/ions given in Column-I with the number of unpaired electrons possessed by them as given in Column-II. \[ \begin{array}{llll} (a) & Cr & (p) & 1 \\[6pt] (b) & Mn^{2+} & (q) & 3 \\[6pt] (c) & N & (r) & 5 \\[6pt] (d) & Sc^{2+} & (s) & 6 \end{array} \]

• A. \((a)-(s),\ (b)-(r),\ (c)-(q),\ (d)-(p)\)
• B. \((a)-(r),\ (b)-(s),\ (c)-(q),\ (d)-(p)\)
• C. \((a)-(s),\ (b)-(r),\ (c)-(p),\ (d)-(q)\)
• D. \((a)-(s),\ (b)-(q),\ (c)-(r),\ (d)-(p)\)

Hint:

\[ \begin{aligned} Cr &[Ar]\,3d^5\,4s^1 &&\Rightarrow 6\ \text{unpaired electrons} \\ Mn^{2+} &[Ar]\,3d^5 &&\Rightarrow 5\ \text{unpaired electrons} \\ N &1s^2\,2s^2\,2p^3 &&\Rightarrow 3\ \text{unpaired electrons} \\ Sc^{2+} &[Ar]\,3d^1 &&\Rightarrow 1\ \text{unpaired electron} \end{aligned} \]

Answer 1

The Correct Option is A

Concept: The number of unpaired electrons is determined using electronic configuration and Hund's rule. Electrons occupy degenerate orbitals singly before pairing occurs.

Step 1: Determine the number of unpaired electrons in each species. For chromium, \[ Cr=[Ar]\,3d^5\,4s^1 \] All five \(d\)-electrons and one \(s\)-electron are unpaired. \[ (a)\rightarrow(s) \] \[ \text{Unpaired electrons}=6 \] For manganese ion, \[ Mn^{2+}=[Ar]\,3d^5 \] \[ (b)\rightarrow(r) \] \[ \text{Unpaired electrons}=5 \]

Step 2: Determine the remaining matches. For nitrogen, \[ N=1s^2\,2s^2\,2p^3 \] According to Hund's rule, all three \(p\)-electrons remain unpaired. \[ (c)\rightarrow(q) \] \[ \text{Unpaired electrons}=3 \] For scandium ion, \[ Sc^{2+}=[Ar]\,3d^1 \] \[ (d)\rightarrow(p) \] \[ \text{Unpaired electrons}=1 \] \[ \begin{aligned} Cr &: 6 \text{ unpaired electrons} \\ Mn^{2+} &: 5 \text{ unpaired electrons} \\ N &: 3 \text{ unpaired electrons} \\ Sc^{2+} &: 1 \text{ unpaired electron} \end{aligned} \] Therefore, \[ (a-s),\quad (b-r),\quad (c-q),\quad (d-p) \] \[ \boxed{(a-s),\quad (b-r),\quad (c-q),\quad (d-p)} \] Hence, option \(\mathbf{(A)}\) is correct.