JEE Main 2019 — Chemistry Inorganic Chemistry

According to molecular orbital theory

Bond order $=\frac{({e}^{-} in bonding molecular orbital) -({e}^{-} in Anti bonding Molecular orbital)}{2}$=2(e−inbondingmolecularorbital)−(e−inAntibondingMolecularorbital)​

Molecular orbital configuration.

$(A)$(A) ${O}_{2}\Longrightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\overset{ }{\sigma }2{p}_{x}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}<\overset{*}{\sigma }2{p}_{y}^{1}=\overset{*}{\pi }2{p}_{y}^{1}$O2​⟹σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2​<π2py2​=π2pz2​<σ∗2py1​=π∗2py1​

$$B.O.=\frac{6-2}{2}=2$$B.O.=26−2​=2

$${O}_{2}^{+}\Longrightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\overset{ }{\sigma }2{p}_{x}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}<\overset{*}{\pi }2{p}_{y}^{1}$$O2+​⟹σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2​<π2py2​=π2pz2​<π∗2py1​

$$B.O.=\frac{6-1}{2}=2.5$$B.O.=26−1​=2.5

Both are paramagnetic because having unpaired ${e}^{-}$e−

bond order increase ${O}_{2}\rightarrow {O}_{2}^{+}$O2​→O2+​

$$(B) NO= \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\overset{ }{\sigma }2{p}_{x}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}<\overset{*}{\pi }2{p}_{y}^{1}=\overset{*}{\pi }2{p}_{z}^{0}$$(B)NO=σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2​<π2py2​=π2pz2​<π∗2py1​=π∗2pz0​

$$B.O.=\frac{6-1}{2}=2.5$$B.O.=26−1​=2.5

$1$1 Unpaired ${e}^{-}$e− so paramagnetic

$$N{O}^{+}\Longrightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\sigma 2p{x}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}$$NO+⟹σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2<π2py2​=π2pz2​

$$B.O.=\frac{6-0}{2}=3$$B.O.=26−0​=3

$N{O}^{+}$NO+ doesn't have unpaired ${e}^{-}$e− so diamagnetic

$NO\rightarrow N{O}^{+}$NO→NO+ {Bond order increases and change magnetic character paramagnetic to diamagnetic character change}

$(C)$(C) ${O}_{2}\rightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\overset{ }{\sigma }2{p}_{x}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}<\overset{*}{\pi }2{p}_{y}^{1}=\overset{*}{\pi }2{p}_{y}^{1}$O2​→σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2px2​<π2py2​=π2pz2​<π∗2py1​=π∗2py1​

$$B.O.=\frac{6-2}{2}=2$$B.O.=26−2​=2

Having $2-$2− unpaired ${e}^{-}$e− so paramagnetic

$${O}_{2}^{-}\rightarrow \sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\overset{ }{\sigma }2{s}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}<\overset{*}{\pi }2{p}_{y}^{2}=\overset{*}{\pi }2{p}_{y}^{1}$$O2−​→σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2s2<π2py2​=π2pz2​<π∗2py2​=π∗2py1​

$$B.O.=\frac{6-3}{2}=1.5$$B.O.=26−3​=1.5

Having 1- unpaired ${e}^{-}$e− so paramagnetic bond order decrease

$(D)$(D) ${N}_{2}=\sigma 1{s}^{2}<\overset{*}{\sigma }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2p{y}^{2}=\pi 2{p}_{y}^{2}<\sigma 2{p}_{x}^{2}$N2​=σ1s2<σ∗1s2<σ2s2<σ∗2s2<π2py2=π2py2​<σ2px2​

$$B.O.=\frac{6-0}{2}=3$$B.O.=26−0​=3

No unpaired ${e}^{-}$e− so diamagnetic

$${N}_{2}^{+}\rightarrow \sigma 1{s}^{2}<\overset{*}{\pi }1{s}^{2}<\sigma 2{s}^{2}<\overset{*}{\sigma }2{s}^{2}<\pi 2{p}_{y}^{2}=\pi 2{p}_{z}^{2}<\sigma 2{p}_{x}^{1}$$N2+​→σ1s2<π∗1s2<σ2s2<σ∗2s2<π2py2​=π2pz2​<σ2px1​

$$B.O.=\frac{5-0}{2}=2.5$$B.O.=25−0​=2.5

Having one unpaired ${e}^{-}$e− so paramagnetic

Bond order increases ${N}_{2}$N2​ to ${N}_{2}^{+}$N2+​