JEE Main 2020 — Chemistry Inorganic Chemistry

2020mcqmedium

Among the statements $(a)-(d)$ , the incorrect ones are:

$(a)$ Octahedral $Co(III)$ complexes with strong field ligands have very high magnetic moments

$(b)$ When ${\Delta }_{0}<P$ , the d-electron configuration of $Co(III)$ in an octahedral complex is ${t}_{eg}^{4}{e}_{g}^{2}$

$(c)$ Wavelength of light absorbed by ${[Co{(en)}_{3}]}^{3+}$ is lower than that of ${[{CoF}_{6}]}^{3-}$

$(d)$ If the ${\Delta }_{0}$ for an octahedral complex of $Co(III)$ is $18,000c{m}^{-1}$ , the ${\Delta }_{t}$ for its tetrahedral complex with the same ligand will be $16,000c{m}^{-1}$ .

Official previous-year question

Held on 7 Jan 2020 · Verified 6 Jul 2026.

Options

A
$(a)$ and $(d)$ only
B
$(c)$ and $(d)$ only
C
$(a)$ and $(b)$ only
D
$(b)$ and $(c)$ only

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Solution

JEE Main 2020 Chemistry, Coordination Compounds — solution figure

So if there are strong field ligands in complex, pairing of electrons will take place, due to which number of unpaired electrons decrease due to which magnetic moment decrease.

$(c)$ Splitting power of ethylenediamine (en) is greater than fluoride $( {\text{F}}^{-} )$ ligand therefore more energy absorbed by ${ [ \text{Co}( \text{en})]}^{ 3+}$ as compared to ${[{\text{CoF}}_{\text{6}} ]}^{ 3-}$

So wave length of light absorbed by ${ [ \text{Co}( \text{en})]}^{ 3+}$ is lower than that of ${ [ { \text{CoF}}_{\text{6}} ]}^{ 3-}$

$\text{(d) }{\Delta }_{t}=\frac{4}{9}{\Delta }_{0}$

$\text{so if }{\Delta }_{0}=18,000{\text{cm}}^{-1}$

${\Delta }_{\text{t}}=\frac{4}{9}\times 18000=8000{\text{cm}}^{-1}$

Statement $(a)$ and $(d)$ are incorrect.

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Solution

JEE Main 2020 Chemistry, Coordination Compounds — solution figure

So if there are strong field ligands in complex, pairing of electrons will take place, due to which number of unpaired electrons decrease due to which magnetic moment decrease.

So wave length of light absorbed by ${ [ \text{Co}( \text{en})]}^{ 3+}$ is lower than that of ${ [ { \text{CoF}}_{\text{6}} ]}^{ 3-}$

$\text{(d) }{\Delta }_{t}=\frac{4}{9}{\Delta }_{0}$

$\text{so if }{\Delta }_{0}=18,000{\text{cm}}^{-1}$

${\Delta }_{\text{t}}=\frac{4}{9}\times 18000=8000{\text{cm}}^{-1}$

Statement $(a)$ and $(d)$ are incorrect.

JEE Main 2020 — Chemistry Inorganic Chemistry

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Solution

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JEE Main 2020 — Chemistry Inorganic Chemistry

Options

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Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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