The paramagnetic property of the oxygen molecule is due to the presence of unpaired electrons present in

Question

The paramagnetic property of the oxygen molecule is due to the presence of unpaired electrons present in (A) (σ 2px)1 and (σ 2px)1 (B) (σ 2px)1 and (π2py)1 (C) (pi*2px)1 and (π*2px)1 (D) (pi*2py)1 and (π2py)1

Tardigrade Admin · Accepted Answer

The electronic configuration of oxygen molecule is as :

O_{2} : KK σ 2 s^{2}, σ^{*} 2 s^{2}, σ^{*} 2 s_{z}^{2}, π 2 p_{x}^{2} \approx π 2 p_{y}^{2}, π^{*} 2 p_{x}^{1}, π^{*} 2 p_{y}^{1}

The molecule has paramagnetic character due to the presence of two unpaired electrons

(π^{*} 2 p_{x})^{1} an d (π^{*} 2 p_{y})^{1}

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Q. The paramagnetic property of the oxygen molecule is due to the presence of unpaired electrons present in

$(σ 2 p_{x})^{1} an d (σ 2 p_{x})^{1}$

$(σ 2 p_{x})^{1} an d (π 2 p_{y})^{1}$

$(p i^{} 2 p_{x})^{1} an d (π^{} 2 p_{x})^{1}$

$(p i^{*} 2 p_{y})^{1} an d (π 2 p_{y})^{1}$

$(p i^{*} 2 p_{z})^{1} an d (π 2 p_{z})^{1}$

Q. The paramagnetic property of the oxygen molecule is due to the presence of unpaired electrons present in

(σ2px​)1and(σ2px​)1

(σ2px​)1and(π2py​)1

(pi∗2px​)1and(π∗2px​)1

(pi∗2py​)1and(π2py​)1

(pi∗2pz​)1and(π2pz​)1

The electronic configuration of oxygen molecule is as : O2​:KKσ2s2,σ∗2s2,σ∗2sz2​,π2px2​≈π2py2​,π∗2px1​,π∗2py1​ The molecule has paramagnetic character due to the presence of two unpaired electrons (π∗2px​)1and(π∗2py​)1.

$(σ 2 p_{x})^{1} an d (σ 2 p_{x})^{1}$

$(σ 2 p_{x})^{1} an d (π 2 p_{y})^{1}$

$(p i^{} 2 p_{x})^{1} an d (π^{} 2 p_{x})^{1}$

$(p i^{*} 2 p_{y})^{1} an d (π 2 p_{y})^{1}$

$(p i^{*} 2 p_{z})^{1} an d (π 2 p_{z})^{1}$