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 (N_b ) is greater than the number of anti bonding molecular orbital electrons(N_a).
(II)Bond order = 12 ( N_b – N_a )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.

N₂ has greater dissociation energy than \(N_{2}^{+}\), whereas O₂ has a lower dissociation energy than \(O_{2}^{+}\) it is because –g

bond order is reduced when O₂ is ionized to \(O_{2}^{+}\) and bond order is increased when N₂ is ionized to \(N_{2}^{+}\)
bond order is increased when O₂ is ionized to \(O_{2}^{+}\) and bond order is decreased when N₂ is ionized to \(N_{2}^{+}\)
bond order is decreased when O₂ and N₂ are ionized to \(O_{2}^{+}\) and \(N_{2}^{+}\) respectively
None of these

Solution

In which of the following molecules / ions BF₃, \(NO_{2}^{-}\),\(NH_{2}^{-}\)and H₂O , the central atom is sp² hybridized ?

\(NH_{2}^{-}\) and H₂O
\(NO_{2}^{-}\) and H₂O
BF₃ and \(NO_{2}^{-}\)
\(NO_{2}^{-}\) and \(NH_{2}^{-}\)

Solution

On determining hybridisation from H =1/2 (V + M – C + A).The hybridisation of BF₃, \(NO_{2}^{-}\), \(NH_{2}^{-}\) and H₂O are sp2, sp2, sp3 and sp3 respectively.

The correct order of increasing bond angles in the following triatomic species is

\(NO_{2}^{-}\) < \(NO_{2}^{+}\) < No₂
\(NO_{2}^{-}\) < No₂ < \(NO_{2}^{+}\)
\(NO_{2}^{+}\) < No₂ < \(NO_{2}^{-}\)
\(NO_{2}^{+}\) < \(NO_{2}^{-}\) < No₂

Solution

Which of the following represents the correct order of Cl–O bond lengths in ClO^–,\(ClO_{2}^{-}\),\(ClO_{3}^{-}\),\(ClO_{4}^{-}\)?

\(ClO_{4}^{-}\) = \(ClO_{3}^{-}\) = \(ClO_{2}^{-}\) = ClO^-
ClO^- < \(ClO_{2}^{-}\) < \(ClO_{3}^{-}\) < \(ClO_{4}^{-}\)
\(ClO_{4}^{-}\) < \(ClO_{3}^{-}\) < \(ClO_{2}^{-}\) < ClO^-
\(ClO_{3}^{-}\) < \(ClO_{4}^{-}\) < \(ClO_{2}^{-}\) < ClO^-

Solution

The b.o in Cl – O^- is 1
The b.o in O =Cl – O^- is 1.5

Select the most stable cation

\(NH_{4}^{+}\)
\(CH_{5}^{+}\)
\(OH_{3}^{+}\)
\(FH_{2}^{+}\)

Solution

Its conjugate base (i.e. NH₃) is most stable.

According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding \(O_{2}^{+}\)

Paramagnetic and Bond order < O₂
Paramagnetic and Bond order > O₂
Diamagnetic and Bond order < O₂
Diamagnetic and Bond order > O₂

Solution

Pb⁴⁺ is less stable than Sn⁴⁺. It is because of

Higher value of IE₁ + IE₂+ IE₃+ IE₄ for Pb than Sn
Lesser polarising power of Pb⁴⁺ than Sn⁴⁺
Inert pair effect in Pb
None of these

Solution

ns² electrons in the havier elements, like lead, remain paired and do not take part in bonding. Thus, Pb²⁺ is more stable than Pb⁴⁺.

Which is correct about the directional orientation of the orbital?

p > sp
p < sp
p = sp
none of these

Solution

The p-orbital has equal sized lobes whereas an sp hybrid orbital has more probability density on one side. Hence, sp orbital is more directional in character.

The bond order in NO is 2.5 while that in NO⁺ is 3.Which of the following statements is true for these two species ?

Bond length in NO⁺ is equal to that in NO
Bond length in NO is greater than in NO⁺
Bond length in NO⁺ is greater than in NO
Bond length is unpredictable

Solution

Now since bond order of NO⁺(3) is higher than that of NO(2.5). Thus bond length of NO⁺ will be shorter.

H₂O is dipolar, whereas BeF₂ is not. It is because

the electronegativity of F is greater than that of O
H₂O involves hydrogen bonding whereas BeF2 is a discrete molecule
H₂O is linear and BeF₂ is angular
H₂O is angular and BeF₂ is linear

Solution

In a linear symmetrical molecule like BeF₂, the bond angle between three atoms is 180°, hence the polarity due to one bond is cancelled by the equal polarity due to other bond, while it is not so in angular molecules, like H₂O

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