Read the passage given below and answer the question that follow :
The distribution of electrons among various molecular orbitals is called the electronic configuration of the molecule which provides us the following very important information about the molecule.
(I)Stability of molecule : The molecule is stable if number of bonding molecular orbital electrons (Nb ) is greater than the number of anti bonding molecular orbital electrons(Na).
(II)Bond order = 12 ( Nb – Na )A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.
(III)Nature of the bond : Bond order 1, 2 and 3 corresponds to single, double and triple bonds respectively.
(IV)Bond length : Bond length decreases as bond order increases.
(V)Magnetic nature : If molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.
Which of the following statement is incorrect ?
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Solution
(i)Bond order ? 1/bond length Bond order \(O_{2}^{+}\) = 2.5, O2= 2, \(O_{2}^{-}\) = 1.5 So correct order of bond length is \(O_{2}^{-}\) > O2 > \(O_{2}^{+}\)
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Which of the following is the wrong statement ?
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Which one of the following molecules is expected to exhibit diamagnetic behaviour ?
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In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are) :
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Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl– , CN– and H2O, respectively, are
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The species having pyramidal shape is :
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The shapes of SO3, BrF3 and \(SiO_{3}^{2-}\) are triangular planar respectively.
Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is:
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The species having bond order different from that in CO is
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