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ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 2 marks · Very short answer

CBSE Class 12 Chemistry 2025: question 19(a)

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section B, question 19(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q19(a) · 2 marks · Very short answerCoordination Compounds

Write the IUPAC name of the complex . Draw the structure of geometrical isomer of this complex which is optically inactive.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

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

Hint 1

Find the oxidation state of Pt from the overall charge of +2.

Hint 2

Ligands are named alphabetically; use 'bis' for en.

Hint 3

The isomer with a plane of symmetry has the two Cl atoms opposite each other.

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. IUPAC name: Dichloridobis(ethane-1,2-diamine)platinum(IV) ion. The oxidation state of Pt is , which gives .

    1 mark
  2. Optically inactive isomer: the trans isomer. Pt sits at the centre of an octahedron. The two ligands occupy opposite (axial) positions at 180° to each other. The two bidentate en ligands span the four equatorial positions, each en joining two adjacent corners of the square plane. This isomer has a plane of symmetry, so it is optically inactive. (The cis isomer is optically active.)

    1 mark

Answer: Dichloridobis(ethane-1,2-diamine)platinum(IV) ion; the trans isomer (Cl–Pt–Cl at 180°, both en in the equatorial plane) is optically inactive.

Where the marks are

  • Correct IUPAC name with Pt(IV) (1 mark)
  • Correct structure of the trans isomer (1 mark)

Revise the chapter

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.