Among CrO, Cr₂ O₃ and CrO₃, the sum of spin-only magnetic moment values of basic and amphoteric oxides is _________ 10⁻² BM (nearest integer).
(Given atomic number of Cr is 24 )

Solution

Chemistry ma’am · 3 steps

Sort the oxides by oxidation state first; the low-valent ones are basic and amphoteric, and only they count.

1. Oxide character. For one metal, the oxide gets more acidic as its oxidation state rises. So CrO (+2) is basic, Cr₂O₃ (+3) is amphoteric and CrO₃ (+6) is acidic, which leaves CrO₃ out of the sum.
2. Unpaired electrons. Cr is [Ar] 3d⁵ 4s¹. Remove the 4s electron first, then 3d electrons. That gives 3d⁴ for Cr²⁺ and 3d³ for Cr³⁺, with every d electron unpaired.

   Oxide	Cr ion	Unpaired n	Nature	μ = √n(n+2) (BM)
   CrO	Cr²⁺, 3d⁴	4	basic	√24 = 4.899
   Cr₂O₃	Cr³⁺, 3d³	3	amphoteric	√15 = 3.873
   CrO₃	Cr⁶⁺, 3d⁰	0	acidic	0 (not counted)

3. So. The sum is 4.899 + 3.873 = 8.772 BM = 877.2 × 10⁻² BM, which rounds to 877.

The common slip: 387. Taking the amphoteric and acidic oxides (Cr₂O₃ and CrO₃) instead of the basic and amphoteric ones gives 3.873 + 0, which is 387.

The idea to keep. For oxides of one metal, low oxidation state means basic, middle means amphoteric and high means acidic. Then μ = √n(n+2), using the unpaired d electrons of the ion.

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