NEET UG 2008 — Chemistry Inorganic Chemistry

2008mcqmedium

In which of the following coordination entities the magnitude of $\Delta_o$ (CFSE in octahedral field) will be maximum? (Atomic number $\mathrm{Co}=27$ )

Official previous-year question

Held on 30 Apr 2008 · Verified 9 Jul 2026.

Options

A
$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$
B
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
C
$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$
D
$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

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Solution

Key Idea: The magnitude of $\Delta_{\propto \mathrm{c} t}$ (the orbital splitting energy) is decided by the nature of ligand Strong field ligand has highest $\Delta_{\text {oct }}$ . The increasing field strength is as $\mathrm{I}^{-} < \mathrm{Br}^{-} < \mathrm{Cl}^{-} < \mathrm{F}^{-} < \mathrm{OH}^{-} < \mathrm{H}_2 \mathrm{O} < \mathrm{C}_2 \mathrm{O}_4^{2-} < \mathrm{NH}_3 < \mathrm{en} < \mathrm{NO}_2^{-} < \mathrm{CN}^{-}$ The $\mathrm{CN}^{-}$ is the strongest ligand among these, hence the magnitude of $\Delta_{\text {oct }}$ will be maximum in $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ . Note : The magnitude of $\Delta_{\text {oct }}$ is also decided by oxidation state of the metal ion. Thus, greater the ionic charge on the central metal ion, the greater the value of $\Delta_{\text {oct }}$ . But here the oxidatinon state of Co metal ion is same in all complexes.

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Options

$\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}$

Solution

NEET UG 2008 — Chemistry Inorganic Chemistry

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NEET UG 2008 — Chemistry Inorganic Chemistry

Options

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Solution

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NEET UG Inorganic Chemistry in other years

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Other Chemistry topics for NEET UG

Physical Chemistry

Organic Chemistry

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