Paramagnetism of oxygen is explained on the basis of its electronic configuration of:

Question

Paramagnetism of oxygen is explained on the basis of its electronic configuration of: (A) (2 ππ*)(2 πy)1 (B) (π* 2 py)1(π* 2 pz1) (C) (2 σs*)1(2 πy)1 (D) (2 σs*)1(2 πy)1

Tardigrade Admin · Accepted Answer

Write configuration of oxygen according to molecular orbital theory

O_{2} = 16 = σ 1 s^{2}, σ 1^{*} s^{2}, σ 2 s^{2}, σ^{*} 2 s^{2}, σ 2 p_{x}^{2}

π 2 p_{y}^{2}, π 2 p_{z}^{2}, π^{*} 2 p_{y}^{1}, π^{*} 2 p_{z}^{1}

∴

It has two unpaired electrons.

∴ O_{2}

is paramagnetic.

Q. Paramagnetism of oxygen is explained on the basis of its electronic configuration of:

$(2 π_{π}^{*}) (2 π_{y})^{1}$

$(π^{} 2 p_{y})^{1} (π^{} 2 p_{z}^{1})$

$(2 σ_{s}^{*})^{1} (2 π_{y})^{1}$

$(2 σ_{s}^{*})^{1} (2 π_{y})^{1}$

Write configuration of oxygen according to molecular orbital theory
$O_{2} = 16 = σ 1 s^{2}, σ 1^{} s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{x}^{2}$
$π 2 p_{y}^{2}, π 2 p_{z}^{2}, π^{} 2 p_{y}^{1}, π^{} 2 p_{z}^{1}$
$∴$ It has two unpaired electrons.
$∴ O_{2}$ is paramagnetic.

Q. Paramagnetism of oxygen is explained on the basis of its electronic configuration of:

(2ππ∗​)(2πy​)1

(π∗2py​)1(π∗2pz1​)

(2σs∗​)1(2πy​)1

(2σs∗​)1(2πy​)1

Write configuration of oxygen according to molecular orbital theory O2​=16=σ1s2,σ1∗s2,σ2s2,σ∗2s2,σ2px2​ π2py​2,π2pz​2,π∗2py​1,π∗2pz​1 ∴ It has two unpaired electrons. ∴O2​ is paramagnetic.

$(2 π_{π}^{*}) (2 π_{y})^{1}$

$(π^{} 2 p_{y})^{1} (π^{} 2 p_{z}^{1})$

$(2 σ_{s}^{*})^{1} (2 π_{y})^{1}$

$(2 σ_{s}^{*})^{1} (2 π_{y})^{1}$

Write configuration of oxygen according to molecular orbital theory
$O_{2} = 16 = σ 1 s^{2}, σ 1^{} s^{2}, σ 2 s^{2}, σ^{} 2 s^{2}, σ 2 p_{x}^{2}$
$π 2 p_{y}^{2}, π 2 p_{z}^{2}, π^{} 2 p_{y}^{1}, π^{} 2 p_{z}^{1}$
$∴$ It has two unpaired electrons.
$∴ O_{2}$ is paramagnetic.