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In coordination compounds… (Class 12 Chemistry 2023, Q32)

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ChemistryCBSE Class 12 · Board exam 2023, set 56/1/1 · 4 marks · Case-based

CBSE Class 12 Chemistry 2023: question 32

Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section D, question 32, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 4 marks · Case-basedCompetency-basedCoordination Compounds

In coordination compounds, metals show two types of linkages, primary and secondary. Primary valencies are ionisable and are satisfied by negatively charged ions. Secondary valencies are non-ionisable and are satisfied by neutral or negative ions having lone pair of electrons. Primary valencies are non-directional while secondary valencies decide the shape of the complexes.

  1. (i)

    If does not react with , what will be its formula ?

    1 mark
  2. (ii)

    What is the secondary valency of ?

    1 mark
  3. (iii)

    (1) Write the formula of Iron(III)hexacyanidoferrate(II).
    (2) Write the IUPAC name of .

    2 marks

  4. OR

    (iii)

    Write the hybridization and magnetic behaviour of . [Atomic number : Ni = 28]

    2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

AgNO₃ precipitates only the chloride ions outside the coordination sphere.

Hint 2

Count donor atoms, not ligands, for the coordination number.

Hint 3

For [Ni(CN)₄]²⁻, find the d-electron count of Ni²⁺ and remember CN⁻ is a strong-field ligand.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) No precipitate with AgNO₃ means no ionisable Cl⁻, so both Cl are inside the coordination sphere: ****.

    1 mark

  2. (ii) Ethylenediamine is bidentate, so three en ligands give 3 × 2 donor atoms. The secondary valency (coordination number) is 6.

    1 mark

  3. (iii)(a)(1) Iron(III) hexacyanidoferrate(II): **** (Prussian blue).

    1 mark

  4. (iii)(a)(2) : pentaamminechloridocobalt(III) chloride.

    1 mark

  5. (iii)(b) In , Ni is +2, so Ni²⁺ is [Ar]3d⁸. CN⁻ is a strong-field ligand, so the eight d electrons pair up in four 3d orbitals. One 3d, one 4s and two 4p orbitals then mix to form dsp² hybrid orbitals (square planar). With no unpaired electrons the complex is diamagnetic.
    (Alternative (iii)(b) carries the same 2 marks: hybridisation 1, magnetic behaviour 1.)

Answer: (i) [Pt(NH₃)₂Cl₂]; (ii) 6; (iii)(a) Fe₄[Fe(CN)₆]₃; pentaamminechloridocobalt(III) chloride; (iii)(b) dsp², diamagnetic

Where the marks are

  • (i) [Pt(NH₃)₂Cl₂] (1 mark)
  • (ii) Secondary valency 6 (1 mark)
  • (iii)(a) Fe₄[Fe(CN)₆]₃ (1 mark); pentaamminechloridocobalt(III) chloride (1 mark)
  • OR (iii)(b) dsp² (1 mark); diamagnetic (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Coordination Compounds, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 56/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.