Question 61 of 62intermediate🔧 ApplyShort Answer2 marks

Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

(a)
Answer

P (Z=15): Configuration is [Ne]3s23p3[Ne] 3s^2 3p^3. Since p3p^3 is half-filled, it has 3 unpaired electrons.

(b)
Answer

Si (Z=14): Configuration is [Ne]3s23p2[Ne] 3s^2 3p^2. The unpaired electrons are present in the 3p3p orbital, giving 2 unpaired electrons.

(c)
Answer

Cr (Z=24): It adopts the d5d^5 configuration ([Ar]4s13d5[Ar] 4s^1 3d^5) for extra stability. All 6 electrons are unpaired, giving 6 unpaired electrons.

(d)
Answer

Fe (Z=26): Configuration is [Ar]4s23d6[Ar] 4s^2 3d^6. The five 3d3d orbitals are progressively occupied, leaving 4 unpaired electrons.

(e)
Answer

Kr (Z=36): Configuration is [Ar]4s23d104p6[Ar] 4s^2 3d^{10} 4p^6. All orbitals are fully filled, so it has 0 unpaired electrons.

Explanation

The number of unpaired electrons is determined by writing the electronic configuration and applying Hund's rule. For Cr, the exception of half-filled d5d^5 stability is applied as mentioned in the text. For Si, the text explicitly states unpaired electrons are in the 3p3p orbital. P has a half-filled p3p^3 configuration. Fe has a 3d63d^6 configuration with 4 unpaired electrons. Kr is a noble gas with fully filled orbitals.

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