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

Question

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

Tardigrade Admin · Accepted Answer

The electronic configuration of O 2 molecule on the basic of MOT is: (σ 1 s)2(σ* 1 s)2(σ 2 s)2(σ* 2 s)2(σ 2 pz)2(π 2 px)2(π 2 py)2(π* 2 px)1(π* 2 py)1 Thus, due to the presence of an unpaired electron each in (π* 2 py)1 and (π* 2 px)1, O2 molecule is paramagnetic.

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

$(σ 2 p_{x})^{1}$ and $(σ^{*} 2 p_{x})^{1}$

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

$(π^{} 2 p_{y})^{1}$ and $(π^{} 2 p_{x})^{1}$

$(π^{*} 2 p_{y})^{1}$ and $(π 2 p_{y})^{1}$

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

(σ2px​)1 and (σ∗2px​)1

(σ2px​)1 and (π2py​)1

(π∗2py​)1 and (π∗2px​)1

(π∗2py​)1 and (π2py​)1

The electronic configuration of O2​ molecule on the basic of MOT is : (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz​)2(π2px​)2(π2py​)2(π∗2px​)1(π∗2py​)1 Thus, due to the presence of an unpaired electron each in (π∗2py​)1 and (π∗2px​)1,O2​ molecule is paramagnetic.

$(σ 2 p_{x})^{1}$ and $(σ^{*} 2 p_{x})^{1}$

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

$(π^{} 2 p_{y})^{1}$ and $(π^{} 2 p_{x})^{1}$

$(π^{*} 2 p_{y})^{1}$ and $(π 2 p_{y})^{1}$