KVPY 2019 SB SX Exam Previous Papers

Question : 47 of 120

Marks: +1, -0

The diamagnetic species among the following is

Solution:

According to MOT
(a)

{O}_{2}^{+}(15 {e}^{-})=σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 {p}_{2}}^{2}(π_{2 {p}_{ {x}}}^{2}=π_{2 {p}_{ {y}}}^{2})(π_{2 {p}_{ {z}}}^{*_{1}}=π_{2 {p}_{y}}^{* 0})

Number of unpaired

{e}^{-s}=1 ⇒

Paramagnetic
(b)

{O}_{2}^{-}(17 {e}^{-})=σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{*_{2}} σ_{2 {p}_{2}}^{2}(π_{2 p_{ {x}}}^{2}=π_{2 p_{y}}^{2})(π_{2 p_{ {x}}}^{*_{2}}=π_{2 {p}_{y}}^{*_{1}})

Number of unpaired

{e}^{-s}=1

(c)

{O}_{2}(16 {e}^{-})=σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{2}}^{2}(π_{2 p_{x}}^{2}=π_{2 p_{y}}^{2})(π_{2 p_{ {x}}}^{*_{1}}=π_{2 p_{y}}^{*_{1}})

Number of unpaired

{e}^{-s}=2 ⇒

paramagnetic
(d)

{O}_{2}^{2-}(18 {e}^{-})=σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 {p}_{2}}^{2}(π_{2 p_{ {x}}}^{2}=π_{2 p_{y}}^{2})(π_{2 p_{x}}^{*_{2}}=π_{2 p_{y}}^{*_{2}})

Number of unpaired

{e}^{- {s}}=0=>

diamagnetic

Go to Question:

Prev Question

Next Question