Paramagnetic means unpaired electrons.
So, Na+, F-, and Ar are diamagnetic, because all of their electrons are paired.
Ar+ and F would be expected to be paramagnetic. Both only exist in the gas state. Elemental fluorine is F2, which is diamagnetic.
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Answer:
Ar⁺, S and Co are paramagnetic in nature
Explanation:
Paramagnetism is a property of a given chemical substance and it is due to the presence of unpaired electrons.
Zinc (Zn)-
Electron configuration in ground state- Zn: [Ar] 3d¹⁰4s²
Since the transition metal zinc has completely filled (n-1)d and ns orbital.
Therefore, the number of unpaired electrons = 0
⇒ Zn is not paramagnetic
Sulphide (S²⁻)- anion of sulfur
Reduction reaction: S + 2e⁻ → S²⁻
Electron configuration in ground state- S: [Ne] 3s² 3p⁴
Electron configuration in excited state- S²⁻ : [Ne] 3s² 3p⁶
Since sulfide ion, S²⁻ has completely filled ns and np orbitals.
Therefore, the number of unpaired electrons = 0
⇒ S²⁻ is not paramagnetic
Ar⁺- cation of argon
Oxidation reaction: Ar → Ar⁺ + e⁻
Electron configuration in ground state- Ar: [Ne] 3s² 3p⁶
Electron configuration in excited state- Ar⁺ : [Ne] 3s² 3p⁵
Since Ar⁺ has completely filled ns orbital and 5 electrons in np orbital.
Therefore, the number of unpaired electrons = 1
⇒ Ar⁺ is paramagnetic
Sulfur (S)-
Electron configuration in ground state- S: [Ne] 3s² 3p⁴
Since S has completely filled ns orbital and 4 electrons in the np orbital.
Therefore, the number of unpaired electrons = 2
⇒ S is paramagnetic
Cobalt (Co)-
Electron configuration in ground state- Co: [Ar] 3d⁷ 4s²
As, the transition metal cobalt has completely filled ns orbital and 7 electrons in the (n-1)d orbital.
Therefore, the number of unpaired electrons = 3
⇒ Co is paramagnetic
