JEE Advanced (Single Correct MCQs): Chemical Bonding & Molecular Structure- 1 - JEE MCQ

In KCN, ionic bond is present between K⁺ and CN^– and covalent bonds are present between carbon and nitrogen C ≡ N

∵ after forming the bonds, C has only 6 e^– in its valence shell

X⁺ Y ^– ∵ Electropositive elements forms cation and electronegative elements forms anion.
Except this all compounds are ionic.

H₂             H – H

N ≡ N               

∵ It forms hydrogen bonds with water

NOTE : Dipole moment is vector quantity In trigonal planar geometry (for sp² hybridisation), the vector sum of two bond moments is equal and opposite to the dipole moment of third bond.

NOTE : Isoelectronic species have same number of electrons.
Electrons in CO = 6 + 8 = 14
Electrons in CN^–  = 6 + 7 + 1 = 14
Electrons in O₂^– = 8 + 8 + 1 = 17
Electrons in O₂⁺ = 8 + 8 – 1 = 15
∴ CO and CN^– are isoelectronic.

TIPS/Formulae : Hybridisation = (V + M – C + A)
where; V = no. of electron in valence shell of central atom M = no. of monovalent atoms, C = charge on cation, A = charge on anion
(i) CO₂, H = (4 + 0 – 0 + 0) = 2
∴ sp hybridisation
(ii) SO₂, H = (6 + 0 – 0 + 0) = 3
∴ sp² hybridisation
(iii) NO₂ has V shaped structure.
(iv) ClO₂ has V shaped structure.
∴ CO₂ having sp hybridation has linear shape.

TIPS/Formulae : (i) Dipole moment is vector quantity. When vector sum of all dipoles in molecule will be zero, then molecule will not have net dipole moment.
(ii) NOTE : For net dipole moment to be equal to zero, all the atoms attached to central atom must be identical and geometry must be regular.

∴ Carbon tetrachloride having regular geometry and identical atoms attached to bonds has zero dipole moment.

In regular tetrahedral structure, dipole moment of one bond is cancelled by opposite dipole moment of the other bonds.

TIPS/Formulae : Hydrogen bonding is formed in those compounds in which F or O or N atoms are attached to hydrogen atom.
∵ HCl does not have F or N or O
∴ It does not form hydrogen bond.

Ionic bond or electrovalent between Cu²⁺ and SO₄^2- , covalent and coordinate in   ion.

TIPS/Formulae : sp type of hybridization involves the intermixing of one s and one p (say p_x) orbitals to give two equivalent hybrid orbitals, known as sp hybrid orbitals.
The two sp hybrid orbitals are directed diagonally, i.e., in a straight line with an angle of 180º (collinear orbitals). The other two p orbitals (say p_y and p_z) remain pure.

NOTE THIS STEP : Write the electronic configuration of each species according to molecular orbital theory.

1 unpaired electron.

no unpaired electron

In covalent bonds between two identical non-metal atoms share the pair of electrons equally between them, e.g. : F₂, O₂, N₂.

NOTE : Gr eater the difference between electronegativities of two covalently bonded atoms more will be strength of hydrogen bond.
∴ F – H ..........F bond is strongest due to largest difference in electronegativity of atoms and smallest size of F atom.

TIPS/Formulae : H = (V + M – C + A)
For SO₂ , H = (6 + 0 + 0 – 0) = 3
∴ sp² hybridisation.

NOTE : Compounds having F or O or N attached to H form hydrogen bond.

∴ Ethanol having H attached to O atom will form hydrogen bond. Rest of the compounds do not hydrogen bonds.

From amongst given species PH₃, NH₃ and SbH₃ are all sp³ hybridised. Their central atom has both bond pair as well as lone pair of electrons. The lone pair occupy the fourth orbital. CH₃⁺ has only three pairs of electrons so it is sp² hybridised.

TIPS/Formulae : Compound having sp hybridisation will have linear shape. ∴CO₂ or (O=C=O) which has C in sp hybrid state has linear shape.

TIPS/Formulae : Dipole moment of compound having regular geometry and same type of atoms is zero. It is vector quantity.
The zero dipole moment of BF₃ is due to its symmetrical (triangular planar) structure. The three fluorine atoms lie at the corners of an equilateral triangle with boron at the centre.
NOTE : The vectorial addition of the dipole moments of the three bonds gives a net sum of zero because the resultant of any two dipole moments is equal and opposite to the third. The dipole moment of NH₃ is 1.46 D indicating its unsymmetrical structure. The dipole moment of CH₂Cl₂ (the molecule uses sp³ hybridisation but is not symmetric) is 1.57D.

TIPS/Formulae : Molecule having sp³ hybridisation and one lone pair of electron will have pyramidal structure. (i) CO₃^2– and NO₃^– have tetrahedron structure. (ii) In PCl₃, P is sp³ hybridised and has one lone pair of electrons, hence it is pyramidal in shape.

TIPS/Formulae : 4s bonds – sp³ hybridisation 2s and 2p bonds – sp² hybridisation

1σ and 3π bonds – sp hybridisation [For hybridization only s-bonds are considered]

(a)3σ, 1π   (b) 3σ, 1π   (c) 4σ   (d)  3σ, 1π

∴ (CH₃)₃COH has 4σ bonds and thus it has sp³ hybridisation

Thus, O^-₂ has one unpaired electron; hence it is paramagnetic. Other species have no unpaired electron.
All of them have 14 electrons.

H = (V + M – C + A)
where H = No. of orbitals involved in hybridisation (viz. 2, 3, 4, 5, 6) and hence nature of hybridisation (viz. sp², sp³, sp³d, sp³d²) can be ascertained.
V = No. of electrons in valence shell of the central atom, M = No. of monovalent atoms, C = Charge on cation, A = Charge on anion,
For ClO₂^- , we have, H =  (7 + 0 - 0 + 1)  
⇒ H = (7 + 1) = 4  or  sp³ hybridisation as 4 orbitalsare involved

H₂O molecule can for m four hydr ogen bonds per molecule, two via lone pairs and two via hydrogen atoms.

In N₂, similar atoms are linked to each other and thus there is no polarity.

No. of e^– in CH₃⁺ = 6 + 3 – 1 = 8
No. of e^– in H₃O⁺ = 3 + 8 – 1 = 10
No. of e^– in NH₃ = 7 + 3 = 10
No. of e^– in CH₃^– = 6 + 3 + 1 = 10
∴ H₃O⁺, NH₃ and CH₃^– are isoelectronic.

TIPS/Formulae : (i) Non metallic oxides are more covalent (or less ionic) as compared to metallic oxides.
(ii) Higher the polarising power of cation (higher for higher oxidation state of similar size cations) more will be covalent character.

(i)P₂O₅ will be more covalent than other metallic oxides.
(ii) Oxidation state of Mn is + 7 in Mn₂O₇, oxidation state of Cr in CrO₃ is + 6 and oxidation state of Mn is + 2 in MnO.
∴ MnO is most ionic.
NOTE : P₂O₅, being a non-metallic oxide will definitely be more covalent than the other metallic oxides. Further, we know that higher the polarising power of the cation (higher for higher oxidation state of the similar size cations) more will be the covalent character. Here Mn is in +7 O.S in Mn₂O₇, Cr in +6 in CrO₃ and Mn in +2 in MnO. So MnO is the most ionic and Mn₂O₇ is the most covalent.