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Q

Which of these statements about [Co(CN)6]3- is true?

• Option 1)

[Co(CN)6]3 - has four unpaired electrons and will be in a high-spin configuration.

• Option 2)

[Co(CN)6]3 - has no unpaired electrons and will be in a high-spin configuration.

• Option 3)

[Co(CN)6]3 - has no unpaired electrons and will be in a low-spin configuration.

• Option 4)

[Co(CN)6]3 - has four unpaired electrons and will be in a low-spin configuration.

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As we discussed in the concept

Inner sphere complex -

Complexes in which (n-1)d orbitals are used in hyberdisation.

- wherein

For example: $d^{2} sp^{3} hybrid \left [ Fe\left ( CN \right )_{6} \right ]$

Strong Field Ligand -

These are the ligands which pair the electron in inner d-orbital and makes required d-orbitals empty/available.

- wherein

For example: CO, CN etc

$CN^-$  is a strong field ligand which pairs the all unpaired electrons of $Co^{3+}\: in\:[Co(CN)_6]^{3-}$ which results no unpaired electron in the $Co^{3+}$ and it form low spin complex.

Option 1)

[Co(CN)6]3 - has four unpaired electrons and will be in a high-spin configuration.

Incorrect

Option 2)

[Co(CN)6]3 - has no unpaired electrons and will be in a high-spin configuration.

Incorrect

Option 3)

[Co(CN)6]3 - has no unpaired electrons and will be in a low-spin configuration.

Correct

Option 4)

[Co(CN)6]3 - has four unpaired electrons and will be in a low-spin configuration.

Incorrect

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