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Statement-1: [NiF(6)]^(4-)" and "[NiF(6)...

Statement-1: `[NiF_(6)]^(4-)" and "[NiF_(6)]^(2-)` both are high spin complexes.
Statement-2:`F^(-)` is weak field ligand.

A

Statement-1 is True, Statement-2 is True , Statement-2 is a correct explanation for Statement-1.

B

Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

Text Solution

Verified by Experts

The correct Answer is:
D
`[NiF_(6) ]^(2-)` is spin paired complex because of `Ni^(4+)`
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Statement-1: [FeCl_(2)(SCN)_(2)]^(2-) , (high spin complex) can exist as diastereoisomers. Statement-2: In tetrahedral geometry, all positions are identical to each other.

Statement-1: [Co(NO_(2))_(6)]^(4-)" and "[Co(NO_(2))_(6)]^(3-) both complex involve d^(2)sp^(3) hybridization but former is paramagnetic and later one diamagnetic. Statement-2: In [Co(NO_(2))_(6)]^(4-) one of the 3d electrons jumps to 4d orbital making the two d-orbitals empty for d^(2)sp^(3) hybridization. NO_(2) is a stronger ligand which favours the pairing of electrons and therefore, d^(2)sp^(3) hybridization.

Statement-1: [Ni(CN)_(4)]^(2-) is a paramagnetic complex. Statement-2: The complex is inner orbital complex.

Statement-1: Among the complexes [Fe(CN)_(6)]^(4-)" and "[Fe(NH_(3))_(6)]^(2+) , former complex absorbs the light of shorter wavelength. Statement-2: The crystal field splitting depends upon the strength of the ligands and charge on the metal ion.

Explain the following giving reasons (a) [NiCI_(4)]^(2-) is tetrahedral and paramagnetic whereas [Ni(CN)_(4)]^(2-) is square plannar and dimagnetic (b) [Fe(H_(2)O)_(6)]^(3+) ion is more paramagnetic than [Fe(CN)_(6)]^(3-) ion (c ) Ni(CO)_(4) is tetrahedral while [Ni(CN)_(4)]^(2-) ion is square planar (d) [Co(F_(6))]^(3-) is a high spin complex whereas [Co(CN)_(6)]^(3-) ion is a low spin complex .

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