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  • Answer please! According to MO theory which of the following lists ranks the nitrogen species in terms of increasingbond order?

According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?

  • Option 1)

    N_{2}^{2-}< N_{2}^{-}< N_{2}

  • Option 2)

    N_{2}< N_{2}^{2-}< N_{2}^{-}

  • Option 3)

    N_{2}^{-}< N_{2}^{2-}< N_{2}

  • Option 4)

    N_{2}^{-}< N_{2}< N_{2}^{2-}

 

Answers (2)

best_answer

As we learnt in 

Bond Order -

Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals.

- wherein

Bond \:Order=\frac{N_{b}-{N_{a}}}{2}

 

Electronic config  of  N_{2}^{2-}:\left [ -3 \sigma 2s^{2}\:\sigma^{*} 2s^{2}\:\pi 2p_x^{2}\:\pi 2p_y^{2}\:\sigma 2p_z^{2}\:{\pi^{*}_{2px}}^{1}}\:{\pi^{*}_{2py}}^{1} \right ]

\therefore B.O =\frac{1}{2}\left ( 8-4 \right )=2

Electronic config of N_{2}^{-}=\left [ -3\sigma^{2}_{2s}\:{\sigma^{*}_{2s}}^{2}\: \pi ^{2}_{2px}\: \pi ^{2}_{2py}\:\sigma^{2}_{2pz}\:\pi ^{*1}_{2px} \right ]

\therefore B.O  = \frac{1}{2}\left ( 8-3 \right )=2.5

Electronic config of N_{2} = \left [ -3\sigma^{2}_{2s} \sigma^{*2}_{2s}\:\pi^{2}_{2px}\:\pi ^{2}_{2py}\:\sigma^2p^{2}_{2pz} \right ]

\therefore B.O  = \frac{1}{2}\left ( 8-2 \right )=3

\therefore  iuoreasing leand order : N_{2}^{2-}< N_{2}^{-}< N_{2}


Option 1)

N_{2}^{2-}< N_{2}^{-}< N_{2}

Correct

Option 2)

N_{2}< N_{2}^{2-}< N_{2}^{-}

Incorrect

Option 3)

N_{2}^{-}< N_{2}^{2-}< N_{2}

Incorrect

Option 4)

N_{2}^{-}< N_{2}< N_{2}^{2-}

Incorrect

Posted by

perimeter

View full answer

According to MOT, the molecular orbital
electronic configuration of
N2 : (cr lsi (cr* lsi (cr2s)2 (cr*2s)2 (1T2p/ (n2p/ (cr2pi
:. B.0 = 10 - 4 =3
2
N,- : (cr ls)2 (cr* lsi (cr2s)2 (cr*2si (n2pi (np/
- (cr2p/ (n*2p)'
:. B.O= l0-5 = 2.5
2
N2
2
- : (crl s/ (cr* 1 s)2 (cr2s)2 (cr*2s)2 (n2p,i (n2p/
( cr2p / (n*2pi (n*2p/
:. B.O. = lO - 6 = 2.
2
Hence the order : N2
2- < N2
- < N2

Posted by

ashish singh

View full answer

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