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Coordination Compounds

Short Answer Type

1. Write the formula for the following coordination compound:
Tetraaminediaquacobalt(lll)chloride

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2. Write the formula for the following coordination compound:
Potassium tetracyanidonickelate(II)

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3. Write the formula for the following coordination compound:
Tris(ethane -1, 2-diamine) chromium(III) chloride

2064 Views

4. Write the formula for the following coordination compound:
Amminebromidochloridonitrito-N-platinate (II)

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5. Write the formula for the following coordination compound:
Dichloridobis(ethane-1, 2-diamine)platinum(lV) nitrate.

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6. Write the formula for the following coordination compound:
Iron(III) hexacyanidoferrate(II)

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7. Write the IUPAC names of the following coordination compounds:
(i) [Co(NH₃)₆]Cl₃(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)_3]
(v) K₂[PdCl4]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl.

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Indicate the type of isomerism exhibited by the following complexes and draw structures for these isomers:
(i) K[Cr(H₂O)₂(C₂O₄)₂], (ii) [Co(en)₃Cl₃,
(iii) [Co(NH₃)₅(NO₂)]|NO₃]₂, (iv) [Pt(NH₃)(H₂O)Cl₂]

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Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers.

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Long Answer Type

10. Explain on the basis of valence bond theory that [Ni(CN)₄]^2– ion with square planar is diamagnetic and the [NiCl₄]^2– ion with tetrahedral geometry is paramagnetic.

(i) [Ni(CN)₄]2–

Ni is in the +2 oxidation state i.e., in d⁸ configuration.
There are 4 CN− ions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CN− ion is a strong field ligand, it causes the pairing of unpaired 3d electrons.

CN^– will cause pairing of electrons. It is diamagnetic in nature due to the unpaired electron.

(ii) [Ni(Cl₄)]2–

In case of [NiCl4] ²⁻, Cl⁻ ion is a weak field ligand. Therefore, it does not lead to the pairing of unpaired 3d electrons. Therefore, it undergoes sp3 hybridization. Since there are 2 unpaired electrons in this case, it is paramagnetic in nature.