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

Step 1: Plan the work.
Write the electron arrangement of each species, then count electrons sitting alone in orbitals.

Step 2: Chromium (a).
$Cr$ is $[Ar]3d^5 4s^1$. All five $d$ electrons are unpaired and the $4s$ one is unpaired too, giving $6$ unpaired electrons. So $(a)$ goes with $(s)$.

Step 3: Manganese ion (b).
$Mn^{2+}$ is $[Ar]3d^5$. The five $d$ electrons each sit alone, giving $5$ unpaired electrons. So $(b)$ goes with $(r)$.

Step 4: Nitrogen (c).
$N$ is $1s^2 2s^2 2p^3$. The three $p$ electrons are unpaired, giving $3$. So $(c)$ goes with $(q)$.

Step 5: Scandium ion (d).
$Sc^{2+}$ is $[Ar]3d^1$. Only one electron sits alone, giving $1$. So $(d)$ goes with $(p)$.

Step 6: Collect the pairs.
Putting it together: $(a)-(s),\ (b)-(r),\ (c)-(q),\ (d)-(p)$, the first option.
\[ \boxed{(a)-(s),\ (b)-(r),\ (c)-(q),\ (d)-(p)} \]