The number of statements correct from the following for Copper (at. no. 29) is/are ____________.
(A) Cu(II) complexes are always paramagnetic.
(B) Cu(I) complexes are generally colourless
(C) Cu(I) is easily oxidized
(D) In Fehling solution, the active reagent has Cu(I)

Solution

Chemistry ma’am · 3 steps

Write the d-electron count for each copper ion and test every statement against it.

1. Electron counts. Copper is [Ar] 3d¹⁰4s¹. So Cu⁺ is 3d¹⁰ and Cu²⁺ is 3d⁹.

   Statement	Check	Verdict
   (A) Cu(II) always paramagnetic	3d⁹ is an odd count, so one electron is always unpaired, whatever the ligand	Correct
   (B) Cu(I) generally colourless	3d¹⁰ is a full set, so no d–d transition is possible	Correct
   (C) Cu(I) easily oxidised	In water 2Cu⁺ → Cu²⁺ + Cu, because Cu²⁺ has a much larger hydration enthalpy	Correct
   (D) Fehling's active reagent has Cu(I)	The reagent is a Cu(II) tartrate complex. Cu(I) appears only in the red product Cu₂O	Incorrect

2. Fehling's test. The aldehyde reduces the blue Cu(II) complex to a red precipitate of Cu₂O. Cu(II) is the reagent and Cu(I) is the product, so (D) has them the wrong way round.
3. So. Statements (A), (B) and (C) are correct, which gives 3.

The common slip: 4. Counting (D) as correct gives 4. Fehling's reagent contains Cu(II), and Cu(I) shows up only in the red Cu₂O precipitate that forms.

The idea to keep. For copper, think d⁹ versus d¹⁰. d⁹ always has an unpaired electron and is coloured. d¹⁰ is colourless and diamagnetic. In water, Cu(I) does not last long and goes to Cu(II).

Answer from the NTA final answer key. Solution by PrepWiser.