JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 2 - JEE MCQ

O₂ = Oxygen (Z = 8) has following molecular orbital configuration of O₂.

 i.e., 2 unpaired and 14 paired electrons.

Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus

In NF₃ :
Thus  N in NF₃ is sp³ hybridized as 4 orbitals are involved in bonding.
In NO₃^– :
Thus N in NO₃^– is sp² hybridized as 3 orbitals are involved in bonding

In BF₃ :
Thus B in BF₃ is sp² hybridized and 3 orbitals are involved in bonding

In H₃O⁺ :
Thus O in H₃O⁺ is sp³ hybridized as 4 orbitals are involved in bonding.
Thus, isostructural pairs are [NF₃, H₃O⁺] and [ NO₃^-, BF₃]

Calcium carbide is an ionic compound (Ca²⁺ C^2–) which produces acetylene on reacting with water. Thus the structure of C^2– is [C ≡ C]^2–. It has one s and two p bonds. [∵ A triple bond consists of one s and two p-bonds]

(a)     It has ionic and non-polar

(b) H – C ≡ N - It has ionic and polar covalent bonds.

(c) 

It has polar and non polar both type of covalent bonds.

(d)  

It has non polar covalent bonds only.

Critical temperature of water is higher than O₂ because H₂O molecule has dipole moment which is due to its V-shape.

TIPS/Formulae :  H = (V + M – C + A)
(i) CO₂; H = (4 + 0 – 0 + 0) = 2
∴ sp hybridisation.
(ii) SO₂; H =  (6 + 0 – 0 + 0) = 3
∴ sp2 hybridisation.
(iii) CO; H = (4 + 0 – 0 + 0) = 2
∴ sp hybridisation.

H =(3 + 3 + 0 - 0) = 3 
∴ Boron, in BF₃ , is sp² hybridised leading to trigonal planar shape.

KEY CONCEPT

(i) Bond length 

(ii) Bond order is calcula ted by eith er the h elp of molecular orbital theory or by resonance.

(i) Bond order of CO as calculated by molecular orbital

theory  = 

(ii) Bond order of CO₂ (by resonance method)

(iii) Bond order in CO₃^2– (by resonance method)

∴ Order of bond length of C – O is CO < CO₂ < CO₃^2–

Hybridisation of S in H₂S =(6 + 2 + 0 – 0) = 4
∴ S has sp³ hybridisation and 2 lone pair of electrons in H₂S
∴ It has angular geometry and so it has non-zero value of dipole moment.

The structure of species can be predicted on the basis of hybridisation which in turn can be known by knowing the number of hybrid orbitals (H) in that species

For SF₄ : S is sp³d hybridised in SF₄.
Thus SF₄ has 5 hybrid orbitals of which only four are used by F, leaving one lone pair of electrons on sulphur.
For CF₄ :
  ∴ sp³ hybridisaion
Since all the four orbitals of carbon are involved in bond formation, no lone pair is present on C having four valence electrons
For XeF₄ :
 ∴ sp³d² hybridization  of the six hybrid orbitals, four form bond with F, leaving behind two lone pairs of electrons on Xe.

For  

∴ sp hybridisation

For  

∴ sp² hybridisation

For 

∴ sp³ hybridisation

Number of electrons in each species are CN^- = 6 + 7 +1= 14 , CO = 6+ 8= 14
NO⁺ = 7 +8 - 1=14
Each of the species has 14 electrons which are distributed in MOs as below

Bond order = 

TIPS/Formulae : H = [V + M – C + A]
Hybridisation of N in NH₃
=  [5 + 3 – 0 + 0] = 4     ∴ sp³
Hybridisation of Pt in [PtCl₄]₂^–
=  [2 + 4 – 0 + 2] = 4      ∴ dsp²
Hybridisation of P in PCl₅
=  [5 + 5 – 0 + 0] = 5      ∴ sp³d
Hybridisation of B in BCl₃
=  [3 + 3 – 0 + 0] = 3      ∴ sp²

H₃N → BF₃ where both N, B are attaining tetrahedral geomerty.

NOTE THIS STEP : Write configuration of all species.
Half filled and full filled orbitals are more stable as compared to nearly half filled and nearly full filled orbitals.

Li^– = 1s², 2s² ; Be^– = 1s², 2s², 2p1
B^– = 1s², 2s², 2p²  ;  C– = 1s², 2s², 2p³
∴ Be^– will be least stable. It has lowest I.E.  

N₂(7+7=14); σ1s², σ1^*s², σ2s², σ^*2s², 

F₂(9 + 9 = 18); σ1s², σ^*1s², σ2s², σ^*2s², σ2p_x², 

O₂^– (8 + 8 +1 = 17); σ1s², σ^*1s^2, σ2s², σ^*2s², σ2p_x²,

O₂^2- (8 + 8 + 2 = 18); σ1s², σ^*1s², σ2s², σ^*2s², σ²p_x²,

∴ O₂^– is the only species having unpaired electron.

NOTE : Isoelectronic species have same number of electrons and isostructural species have same type of hybridisation at central atom.
NO₃^– ; No. of e^– = 7 + 8 × 3 + 1 = 32, hybridisation of N in NO₃^– is sp³
CO₃^2– ; No. of e^– = 6 + 8 × 3 + 2 = 32, hybridisation of C in
CO₃^2– is sp³ ClO₃^– ; No. of e^– = 17 + 8 × 3 + 1 = 42, hybridisation of Cl in
ClO₃^– is sp³ SO₃; No. of e^– = 16 + 8 × 3 = 40, hybridisation of S in
SO₃ is sp²
∴ NO₃– and CO₃^2– are isostructural and isoelectronic.

O₂ : σ1s² , σ^*1s² , σ2s², σ^*2s²,σ2p_x² ,

Bond order = 

(two unpaired electrons in antibonding molecular orbital)

Bond order = 

(One unpaired electron in antibonding molecular orbital) Hence O₂ as well as O⁺₂ is paramagnetic, and bond order of O⁺₂ is greater than that of O₂.

(i) In Na₂O₂, we have O₂^2- ion. Number of valence elctrons of the two oxygen in O₂^2- ion = 8 × 2 + 2 =18 which are present as follows

∴ Number of unpaired electrons = 0, hence, O₂^2-is diamagnetic.

(ii) No. of valence electrons of all atoms in O₃ = 6 × 3 = 18.
Thus, it also, does not have any unpaired electron, hence it is diamagnetic. (iii)

No. of valence electrons of all atom in N₂O = 2 × 5 + 6  = 16.
Hence, here also all electrons are paired. So it is diamagnetic.

(iv) In KO₂, we have O₂^- No. of valence electrons of all atoms in O₂^- = 2 × 6 + 1 = 13,
Thus it has one unpaired electron, hence it is paramagnetic.

Molecular electronic configuration of

Therefore, bond order = 

Bond order  = 

Bond order = 

Bond order  = 

Bond order  = 

∴ NO^– has different bond order from that in CO.

Molecular orbital configuration of B₂(10) as per the condition will be

Bond order of B₂ =,B₂ will be diamagnetic.

OSF₂ : .  It has 1 lone pair.

(Shape is trigonal pyramidal)

The shapes of SO₃, BrF₃ and SiO₃^2- are triangular planar respectively.

[NiCl₄]^2–. = 3d⁸ configuration with nickel in +2 oxidation state, Cl^– being weak field ligand does not compel for pairing of electrons.
So,
 [NiCl₄]^2–

Hence, complex has tetrahedral geometry

[Ni(CN)₄]^2– = 3d⁸ configuration with nickel in + 2 oxidation state, CN^– being strong field ligand compels for pairing of electrons.
So,

Hence, complex has square planar geometry.

[Ni(H₂O)₆] = 3d⁸ configuration with nickel in +2 oxidation state. As with 3d⁸ configuration two dorbitals are not available for d²sp³ hybridisation. So, hybridisation of Ni (II) is sp³d² and Ni (II) with six co-ordination will have octahedral geometry.

Note : With water as ligand, Ni (II) forms octahedral complexes.

Be₂ = σ1s² σ*1s² σ2s² σ*2s²
B₂ = σ1s² σ*1s2 σ2s² σ*2s² σ²pz²
C₂ = σ1s² σ*1s² σ2s² σ*2s² σ²p_z² π²p_x¹ π²p_y¹
N₂ = σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π²p_x² π²p_y²
Thus only C₂ will be paramagnetic

Ni²⁺ with NH₃ shows CN = 6 forming [Ni(NH₃)₆]²⁺ (Octahedral)

Pt²⁺ with NH₃ shows CN = 4 forming [Pt(NH₃)₄]²⁺ (5d series CMA, square planner)

Zn²⁺ with NH₃ shows CN = 4 forming [Zn(NH₃)₄]²⁺ (3d¹⁰ configuration, tetrahedral)