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A gaseous species 'X' has bond order 2.5. It is paramagnetic and isoelectronic with...

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user8235304 | Student, Grade 11

Posted August 2, 2013 at 1:45 AM via web

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A gaseous species 'X' has bond order 2.5. It is paramagnetic and isoelectronic with nitric oxide.This is cationic in nature.It is colourless.This species is found in many super oxide crystal.

Q.How many number of anti-bonding electrons are there in the species?

a)1

b)2

c)3

d)4

1 Answer | Add Yours

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted August 2, 2013 at 7:09 AM (Answer #1)

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To solve this, we should first construct the molecular orbital diagram for the unknown species ‘X’. It has been given that the unknown is isoelectronic to NO which means that it has a valence of 11 electrons (N= 5; O=6).

MO diagram for unknown species ‘X’

`sigma_2 *`    ___

`pi *`      ___ ___

`sigma_1 *`    ___

`pi`        ___ ___

`sigma_2`       ___

`sigma_1`       ___

Fill the orbitals with electron pairs starting from the lowest orbital sigma_1.

`sigma_2 *`    ___

`pi *`         `uarr`  ___

`sigma_1 *`    `uarr darr`

`pi `     `uarr darr`    `uarr darr`

`sigma_2`       `uarr darr`

`sigma_1`       `uarr darr`

Confirm the bond order by solving it:

Bond order = (number of bonding electrons – number of anti-bonding electrons)/2

antibonding electrons = those electrons located in the orbitals with 'star' `*`  (`sigma *` and `pi *` ) 

Bond order = (8-3)/2 = 2.5

Finally, you can now answer the question. There are three anti-bonding electrons present in the molecule. Answer c) 3

Sources:

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