Search any question & find its solution
Question:
Answered & Verified by Expert
In which of the following octahedral complexes of $\operatorname{Co}($ At. no. 27$)$, will the magnitude of $\Delta_{0}$ be the highest?
Options:
Solution:
1416 Upvotes
Verified Answer
The correct answer is:
$\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}$
Spectrochemical series. According to the strength of splitting is as:
$$
\begin{array}{l}
\mathrm{CO}>\mathrm{CN}^{-}>\mathrm{NO}_{2}^{-}>\text {en }>\mathrm{NH}_{3}>\text { py }>\mathrm{NCS}^{-} \\
>\mathrm{H}_{2} \mathrm{O}>\mathrm{O}^{2-}>\mathrm{Ox}^{2-}>\mathrm{OH}^{-}>\mathrm{F}>\mathrm{Cl}^{-} \\
>\mathrm{SCN}^{-}>\mathrm{S}^{2-}>\mathrm{Br}^{-}>\mathrm{I}^{-}
\end{array}
$$
Crystal field stabilisation energy (CFSE) for octahedral complex, $\Delta^{\circ}$ dependson the strength of negative ligand. [Higher the strength more will be theCFSE]
$$
\begin{array}{l}
\mathrm{CO}>\mathrm{CN}^{-}>\mathrm{NO}_{2}^{-}>\text {en }>\mathrm{NH}_{3}>\text { py }>\mathrm{NCS}^{-} \\
>\mathrm{H}_{2} \mathrm{O}>\mathrm{O}^{2-}>\mathrm{Ox}^{2-}>\mathrm{OH}^{-}>\mathrm{F}>\mathrm{Cl}^{-} \\
>\mathrm{SCN}^{-}>\mathrm{S}^{2-}>\mathrm{Br}^{-}>\mathrm{I}^{-}
\end{array}
$$
Crystal field stabilisation energy (CFSE) for octahedral complex, $\Delta^{\circ}$ dependson the strength of negative ligand. [Higher the strength more will be theCFSE]
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.