Q79 / 200 · NEET 2002 · Chemistry
Atomic number of $\mathrm{Cr}$ and $\mathrm{Fe}$ are respectively 24 and 26 , which of the following is paramagnetic?
- A$\left[\mathrm{Cr}(\mathrm{CO})_6\right]$
- B$\left[\mathrm{Fe}(\mathrm{CO})_5\right]$
- C$\left[\mathrm{Fe}\left(\mathrm{CN}_6\right]^{4-}\right.$
- Correct answerD$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
Solution
Odd electrons, ions and molecules are paramagnetic.
In \(\mathrm{Cr}(\mathrm{CO})_6\) molecule 12 electrons are contributed by CO group and it contains no odd electron.
\(\mathrm{Cr} \rightarrow 3 d^5 4 s^1\)
\(\mathrm{Fe}(\mathrm{CO})_5\) molecule also does not contain odd electron.
\(\begin{aligned}
& \mathrm{Fe} \rightarrow 3 d^6 4 s^2 \\
& \mathrm{In}\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4 \text { ion } \mathrm{Fe}(+\mathrm{II}) \rightarrow 3 d^6 4 s^0
\end{aligned}\)
\(\therefore\) No odd electrons.
\(\text {In }\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} \text { ion } \mathrm{Cr}(+\mathrm{III}) \rightarrow 3 d^3 4 s^0\)
This ion contains odd electron so it is paramagnetic.
Hence, the correct answer is option: 'D'.