Question

Write the molecular orbital configurations of
NO+, NO and NO-. Calculate their bond
orders and predict which of them will be
paramagnetic.

Answer 1

Answer : The molecule NO is paramagnetic.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 7 electrons present in nitrogen and 6 electrons in oxygen.

(a) The number of electrons present in $NO^+$ molecule = 7 + 6 - 1 = 12

The molecular orbital configuration of $NO^+$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^0,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The formula of bond order = $\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})$

The bonding order of $NO^+$ = $\frac{1}{2}\times (8-4)=2$

The number of unpaired electron in $NO^+$ molecule is, 0. So, this is diamagnetic. That means, more the number of unpaired electrons, more paramagnetic.

(b) The number of electrons present in $NO$ molecule = 7 + 6 = 13

The molecular orbital configuration of $NO$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^1,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The bonding order of $NF^+$ = $\frac{1}{2}\times (9-4)=2.5$

The number of unpaired electron in $NO$ molecule is, 1. So, this is paramagnetic.

(c) The number of electrons present in $NO^-$ molecule = 7 + 6 + 1 = 14

The molecular orbital configuration of $NO^-$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^2,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0$

The bonding order of $NO^-$ = $\frac{1}{2}\times (10-4)=3$

The number of unpaired electron in $NO^-$ molecule is, 0. So, this is diamagnetic.

Hence, the molecule NO is paramagnetic.

Answer 2

Explanation:

The molecule NO is paramagnetic.

According to the molecular orbital theory, the general molecular orbital configuration will be,

(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)(σ

1s

),(σ

1s

∗

),(σ

2s

),(σ

2s

∗

),[(π

2p

x

)=(π

2p

y

)],(σ

2p

z

),[(π

2p

x

∗

)=(π

2p

y

∗

)],(σ

2p

z

∗

)

As there are 7 electrons present in nitrogen and 6 electrons in oxygen.

(a) The number of electrons present in NO^+NO

+

molecule = 7 + 6 - 1 = 12

The molecular orbital configuration of NO^+NO

+

molecule will be,

(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^0,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0(σ

1s

)

2

,(σ

1s

∗

)

2

,(σ

2s

)

2

,(σ

2s

∗

)

2

,[(π

2p

x

)

2

=(π

2p

y

)

2

],(σ

2p

z

)

0

,[(π

2p

x

∗

)

0

=(π

2p

y

∗

)

0

],(σ

2p

z

∗

)

0

The formula of bond order = \frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})

2

1

×(Number of bonding electrons−Number of anti-bonding electrons)

The bonding order of NO^+NO

+

= \frac{1}{2}\times (8-4)=2

2

1

×(8−4)=2

The number of unpaired electron in NO^+NO

+

molecule is, 0. So, this is diamagnetic. That means, more the number of unpaired electrons, more paramagnetic.

(b) The number of electrons present in NONO molecule = 7 + 6 = 13

The molecular orbital configuration of NONO molecule will be,

(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^1,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0(σ

1s

)

2

,(σ

1s

∗

)

2

,(σ

2s

)

2

,(σ

2s

∗

)

2

,[(π

2p

x

)

2

=(π

2p

y

)

2

],(σ

2p

z

)

1

,[(π

2p

x

∗

)

0

=(π

2p

y

∗

)

0

],(σ

2p

z

∗

)

0

The bonding order of NF^+NF

+

= \frac{1}{2}\times (9-4)=2.5

2

1

×(9−4)=2.5

The number of unpaired electron in NONO molecule is, 1. So, this is paramagnetic.

(c) The number of electrons present in NO^-NO

−

molecule = 7 + 6 + 1 = 14

The molecular orbital configuration of NO^-NO

−

molecule will be,

(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],(\sigma_{2p_z})^2,[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0(σ

1s

)

2

,(σ

1s

∗

)

2

,(σ

2s

)

2

,(σ

2s

∗

)

2

,[(π

2p

x

)

2

=(π

2p

y

)

2

],(σ

2p

z

)

2

,[(π

2p

x

∗

)

0

=(π

2p

y

∗

)

0

],(σ

2p

z

∗

)

0

The bonding order of NO^-NO

−

= \frac{1}{2}\times (10-4)=3

2

1

×(10−4)=3

The number of unpaired electron in NO^-NO

−

molecule is, 0. So, this is diamagnetic.

Hence, the molecule NO is paramagnetic.