Chemical Bonding and Molecular Structure Quiz-1

Hint

\(\mathrm{CHCl}_3\) is the only polar compound given among the two.
\(\mathrm{CH}_4=\mathrm{CCl}_4<\mathrm{CHCl}_3\). In other words we can also say that methane and carbon tetrachloride have zero dipole moment.

Hint

A covalent bond is formed between two nonmetals and an Ionic bond is formed between one metal and one nonmetal.
So, in \(\mathrm{KCl}, \mathrm{K}\) is metal and \(\mathrm{Cl}\) is nonmetal. Hence, It is an ionic bond. \(\mathrm{B}_2 \mathrm{H}_6, \mathrm{PH}_3\) and \(\mathrm{H}_2 \mathrm{SO}_4\) contains covalent bonds.

Hint

Dipole moment \(=(\) Distance between opposite charges \() \times(\) charge,\(q)\) \(\mu=q \times d\)
So, the greater the distance between the opposite charges higher the dipole. Due to the resonance, the greater charge separation occurs between charges due to linearity.

Hint

(c)

\(
\begin{aligned}
& \mathrm{NO}^{+}=7+8-1=14 \mathrm{e}^{-} . \\
& \mathrm{O}_2=16 \mathrm{e}^{-} \text {i.e., not isoelectronic. }
\end{aligned}
\)
Boron forms only covalent compounds. This is due to its extremely high ionisation energy. Compounds of \(\mathrm{Tl}^{+}\)are much more stable than those of \(\mathrm{Tl}^{3+}\).
\(\mathrm{LiAlH}_4\) is a versatile reducing agent in organic synthesis.

Hint

(c) The bond order of \(\mathrm{N}_2, \mathrm{O}_2\), and \(\mathrm{O}_2^{-}\)are 3,2 and 1.5 respectively. Since higher bond order implies higher bond dissociation energy, hence the correct order will be
\(
\mathrm{N}_2>\mathrm{O}_2>\mathrm{O}_2^{-}
\)

Hint

(b) The geometry of \(\mathrm{IF}_5\) is square pyramide with an unsymmetric charge distribution, therefore this molecule is polar.

Hint

(a) Hybridisation of \(\mathrm{S}\) in \(\mathrm{H}_2 \mathrm{~S}=\frac{1}{2}(6+2+0-0)=4\)
\(\therefore \mathrm{S}\) has \(s p^3\) hybridisation and 2 lone pair of electrons in \(\mathrm{H}_2 \mathrm{~S}\).
\(\therefore\) It has angular geometry and so it has non-zero value of dipole moment.

Hint

The critical temperature is directly proportional to the intermolecular force of attraction. \(\mathrm{H}_2 \mathrm{O}\) is a polar molecule, has greater intermolecular force of attraction than \(\mathrm{O}_2\), hence higher critical temperature.
The critical temperature of the water is higher than \(\mathrm{O}_2\) because \(\mathrm{H}_2 \mathrm{O}\) molecule has dipole moment which is due to its V-shape.

Hint

Hint

(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.
(a) \(\mathrm{P}_2 \mathrm{O}_5\) will be more covalent than other metallic oxides.
(b) Oxidation state of \(\mathrm{Mn}\) is +7 in \(\mathrm{Mn}_2 \mathrm{O}_7\), oxidation state of \(\mathrm{Cr}\) in \(\mathrm{CrO}_3\) is +6 and oxidation state of \(\mathrm{Mn}\) is +2 in \(\mathrm{MnO}\).
\(\therefore \mathrm{MnO}\) is most ionic.

NOTE : \(\mathrm{P}_2 \mathrm{O}_5\), 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 \(\mathrm{Mn}\) is in +7 O.S in \(\mathrm{Mn}_2 \mathrm{O}_7\), Cr in +6 in \(\mathrm{CrO}_3\) and \(\mathrm{Mn}\) in +2 in \(\mathrm{MnO}\). So, \(\mathrm{MnO}\) is the most ionic and \(\mathrm{Mn}_2 \mathrm{O}_7\) is the most covalent.

Hint

(d) No. of \(\mathrm{e}^{-}\)in \(\mathrm{CH}_3^{+}=6+3-1=8\)

No. of \(\mathrm{e}^{-}\)in \(\mathrm{H}_3 \mathrm{O}^{+}=3+8-1=10\)
No. of \(\mathrm{e}^{-}\)in \(\mathrm{NH}_3=7+3=10\)
No. of \(\mathrm{e}^{-}\)in \(\mathrm{CH}_3^{-}=6+3+1=10\)
\(\therefore \mathrm{H}_3 \mathrm{O}^{+}, \mathrm{NH}_3\) and \(\mathrm{CH}_3^{-}\)are isoelectronic.

Hint

(b) In \(\mathrm{N}_2\), similar atoms are linked to each other and thus there is no polarity.

The cyanide ions, \(\mathrm{CN}^{-}\)and \(\mathrm{N}_2\) are isoelectronic, but in contrast to \(\mathrm{CN}^{-}, \mathrm{N}_2\) is chemically inert because of the absence of polarity of bond. Cyanide ion is negatively charged but \(\mathrm{N}_2\) is a neutral molecule.

Hint

(b) The dipole moment of compound having regular geometry and same type of atoms is zero. It is vector quantity. The zero dipole moment of \(\mathrm{BF}_3\) 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 \(\mathrm{NH}_3\) is \(1.46 \mathrm{D}\) indicating its unsymmetrical structure. The dipole moment of \(\mathrm{CH}_2 \mathrm{Cl}_2\) (the molecule uses \(s p^3\) hybridisation but is not symmetric) is \(1.57 \mathrm{D}\).

Hint

(d) In covalent bonds, between two identical non-metal, atoms share the pair of electrons equally between them, e.g. : \(\mathrm{F}_2, \mathrm{O}_2, \mathrm{~N}_2\).

Hint

(c) Ionic bond or electrovalent between \(\mathrm{Cu}^{2+}\) and \(\mathrm{SO}_4^{2-}\), covalent and coordinate in \(\mathrm{SO}_4^{2-}\);

Hint

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

Hint

(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.

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

Hint

(a) NOTE : Isoelectronic species have same number of electrons. Electrons in \(\mathrm{CO}=6+8=14\)
Electrons in \(\mathrm{CN}^{-}=6+7+1=14\)
Electrons in \(\mathrm{O}_2^{-}=8+8+1=17\)
Electrons in \(\mathrm{O}_2^{+}=8+8-1=15\)
\(\therefore \mathrm{CO}\) and \(\mathrm{CN}^{-}\)are isoelectronic.

Hint

(c) NOTE: Dipole moment is vector quantity.

In trigonal planar geometry (for \(s p^2\) hybridisation), the vector sum of two bond moments is equal and opposite to the dipole moment of the third bond.

Hint

(b) It forms hydrogen bonds with water

Explanation for the correct option:
b. \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\)
– Water is a polar solvent with hydrogen bonding, ethyl alcohol molecules should have polarity or \(\mathrm{H}\)-bond in order to dissolve in water.
– Ethyl alcohol possesses hydrogen bonding as well as being polar due to the strong electronegativity of oxygen; as a result, it is water-soluble.

Explanation for incorrect options:
a. \(\mathrm{CS}_2\)
– \(\mathrm{CS}_2\) is non-polar.
– It won’t dissolve in water.
c. \(\mathrm{CCl}_4\)
– \(\mathrm{CCl}_4\) is non-polar in the third scenario. Any bond dipoles \((\mathrm{C}-\mathrm{Cl})\) will cancel out because of the molecule’s tetrahedral symmetry.
– It will be water-insoluble.
d. \(\mathrm{CHCl}_3\)
– Chloroform is a moderately polar chemical.
– However, it is unable to form any strong bonds (such as hydrogen bonds) with water.
– Its solubility in water is quite low due to the lack of a strong contact with water.

Hint

(c)

\(\mathrm{N}_2\) has triple bond and each covalent bond is associated with one pair of electrons, therefore, six electrons are involved in forming bonds in \(\mathrm{N}_2\)

Hint

Hydrogen \(\left(\mathrm{H}_2\right)\) is a covalent molecule as it is formed by sharing a pair of electron between two atoms of same element. \(\mathrm{CaO}, \mathrm{KCl}\) and \(\mathrm{Na}_2 \mathrm{~S}\) are ionic as they are formed between atoms of elctropositive and electronegative elements.

Hint

(a) \(X^{+} Y^{-}\)
\(\because\) Electropositive elements forms cation and electronegative elements forms anion. Both are held together by electrostatic forces of attraction.

Hint

Hint

(c) In \(\mathrm{KCN}\), ionic bond is present between \(\mathrm{K}^{+}\)and \(\mathrm{CN}^{-}\)and covalent bonds are present between carbon and nitrogen \(\mathrm{C} \equiv \mathrm{N}\).

Hint

Three centred two-electron bonds or banana bond;
NOTE : The formation of three centred two-electron bond is due to one empty \(s p^3\) orbital of one of the B atom, \(1 s\) orbital of the bridge hydrogen atom and one of the \(s p^3\) (filled) orbital of the other B-atom. This forms a delocalized orbital covering the three nuclei giving the shape of a banana. Thus, also known as banana bonds.

Hint

\(\mathrm{N} \equiv \mathrm{N}\left(\mathrm{N}_2\right)\) has \(1 \sigma\) and \(2 \pi\) bonds. (A triple bond consists of \(1 \sigma\) and \(2 \pi\) bonds)

Hint

\(\mathrm{CH}_4\) has the largest \(\mathrm{H}-\mathrm{M}-\mathrm{H}\) bond angle which is \(109.5^{\circ}\).
The bond angle of other species are given below:
\(\mathrm{H}_2 \mathrm{O}-104.5^{\circ}\left(s p^3\right.\) with 2 lone pair at \(\left.\mathrm{O}\right)\)
\(\mathrm{NH}_3-107^{\circ}\left(s p^3\right.\) with 1 lone pair at \(\left.\mathrm{N}\right)\)
\(
\mathrm{CH}_4-109.5^{\circ}\left(s p^3\right)
\)
\(\mathrm{H}_2 \mathrm{~S}-92^{\circ}\left(s p^3\right.\) with 2 lone pair at \(\left.\mathrm{O}\right)\)
Lone pair-bond pair repulsion in \(\mathrm{H}_2 \mathrm{~S}\) will increase because ‘ \(\mathrm{S}\) ‘ has lower electronegativity than ‘O’. So there will be lesser electron density on ‘S’ and thus \(\mathrm{H}-\mathrm{S}-\mathrm{H}\) bond angle will be smaller than \(\mathrm{H}_2 \mathrm{O}\).

Hint

(a) \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}=d s p^2\)
(b) \(\mathrm{BrF}_5=s p^3 d^2\)
(c) \(\mathrm{XeF}_4=s p^3 d^2\)
(d) \(\left[\mathrm{CrF}_6\right]^{3-}=d^2 s p^2\)
The molecule in which hybrid MOs involve only one d-orbital of the central atom is \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\) as in this complex \(\mathrm{Ni}^{2+}\) undergoes \(dsp^2\) hybridization

Hint

For \(\mathrm{AB}_4\) compound possible geometry are
\(
\begin{array}{ccc}
\text { No. of Bond pair } & \text { No. of lone pair } & \text { Hybridisation } \\
4 & 0 & s p^3 \\
4 & 1 & s p^3 d \\
4 & 2 & s p^3 d^2
\end{array}
\)
Structure with \(s p^3 d^2\) hybridisation is polar due to lone pair moment while in other possibilities molecules is non-polar.
Square pyramidal can be polar due to lone pair moment as the bond pair moments will get cancelled out.

Hint

(i) \(\mathrm{XeF}_5^{-} \quad\) St. No. = Bond pair + Lone Pair
\(
=(5+2)=7
\)

So, hybridisation is \(=s p^3 d^\beta\)
and the structure is pentagonal planar.
\(\begin{array}{ll}\text { (ii) } \mathrm{XeO}_3 \mathrm{~F}_2 & \text { St. No. }=5\end{array}\)
So, hybridisation is \(=s p^3 d\) and structure is trigonal bipyramidal.

Hint

(b) \(\mathrm{SF}_4\)

Bond pair \(=4\)
Lone pair \(=1\)
Steric number \(=5\),
So, hybridisation is \(s p^3 d\).

Geometry is trigonal bipyramidal but the shape is “See Saw”.

Hint

(d) \(\mathrm{ICl}_5\) is \(s p^3 d^2\) hybridised ( \(\left.5 b p, 1 l p\right)\)

\(\mathrm{ICl}_4^{-}\)is \(s p^3 d^2\) hybridised \((4 b p, 2 l p)\)

Hint

\(
\begin{array}{ll}
\text { (a) Species } & \text { Hybridisation } \\
\mathrm{ICl}_2^{-} & s p^3 d \\
\mathrm{ICl}_4^{-} & s p^3 d^2 \\
\mathrm{BrF}_2^{-} & s p^3 d \\
\mathrm{IF}_6^{-} & s p^3 d^3
\end{array}
\)

Hint

Total number of lone pair of electrons is 9.

Hint

(a) \(\mathrm{NF}_3\) has trigonal pyramidal geometry. \(\mathrm{N}\) atom has one lone pair and three bond pairs of electrons. The electron pair geometry is tetrahedral and molecular geometry is trigonal pyramidal. The bond angles are lower than tetrahedral bond angles due to lone pair – lone pair and lone pair – bond pair repulsions. \(\mathrm{N}\) atom is \(s p^3\) hybridised.

Hint

\(
\mathrm{XeOF}_2
\)

\(
\mathrm{XeOF}_4
\)

Hint

Hint

(c) Hybridisation \((\mathrm{H})=[\) No. of valence electrons of central atom + No. of monovalent atoms attached to it + (-ve charge if any) \(-(+\mathrm{ve}\) charge if any)]
\(
\mathrm{NO}_2^{+}=\text {i.e. } s p \text { hybridisation }
\)
\(
\mathrm{NO}_2^{-}=\text {i.e. } s p^2 \text { hybridisation }
\)
\(
\mathrm{NO}_3^{-}=\text {i.e. } s p^2 \text { hybridisation }
\)
The Lewis structure of \(\mathrm{NO}_2\) shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridization will be \(s p^2\).

Hint

(d) \(\mathrm{ClF}_3 \longrightarrow\) Hybridisation \(=3+\frac{1}{2}[7-3]=5\left(s p^3 d\right)\)
\(
\begin{aligned}
& \mathrm{XeOF}_2 \longrightarrow \text { Hybridisation }=3+\frac{1}{2}[8-4]=5\left(s p^3 d\right) \\
& \mathrm{XeF}_3^{+} \longrightarrow \text { Hybridisation }=3+\frac{1}{2}[8-3-1]=5\left(s p^3 d\right)
\end{aligned}
\)

All molecules have \(s p^3 d\) hybridisation and 2 lone pairs. Hence all have identical (T-shape).

Hint

\(
\mathrm{C}_2^{2-} \text { ion } \rightarrow[\mathrm{C} \equiv \mathrm{C}]^{2-}
\)
Dicarbide ion \(\mathrm{C}_2^{2-}\) has 1 sigma and 2 \(\pi\) bonds. There are two \(\pi\) bonds and one \(\sigma\) bond in a triple bond.

Hint

The shape of \(\mathrm{IF}_6^{-}\)is trigonally distorted octahedron. The central iodine atom has 7 valence electrons. It gains one electron from negative charge. Out of 8 valence electrons, 6 are involved in formation of \(6 \mathrm{I}-\mathrm{F}\) bonds and one lone pair of electrons is present. \(6 \mathrm{~F}\) atoms are present at 6 corners of distorted octahedron. The lone pair of electrons distorts the octahedral shape due to electron-electron repulsion. This is due to the presence of a “weak” lone pair.

Hint

(d) \(\mathrm{OSF}_2: H=\frac{6+2}{2}=4\). It has 1 lone pair.
\(
\text { The shapes of } \mathrm{SO}_3, \mathrm{BrF}_3 \text { and } \mathrm{SiO}_3^{2-} \text { are triangular planar respectively. }
\)

Hint

\(\mathrm{I}_3^{-}\)has the maximum number of lone pair of electron on the central atom. \(\mathrm{I}_3^{-}, \mathrm{XeF}_4, \mathrm{SF}_4\) and \(\mathrm{ClO}_3^{-}\)have \(3,2,1,1\) lone pair of electron respectively.

Hint

(a) NOTE : Isoelectronic species have same number of electrons and isostructural species have same type of hybridisation at central atom.
\(
\mathrm{NO}_3^{-} \text {; No. of } \mathrm{e}^{-}=7+8 \times 3+1=32 \text {, hybridisation of } \mathrm{N} \text { in } \mathrm{NO}_3^{-} \text {is } s p^3
\)
\(
\mathrm{CO}_3^{2-} ; \text { No. of } \mathrm{e}^{-}=6+8 \times 3+2=32 \text {, hybridisation of } \mathrm{C} \text { in } \mathrm{CO}_3^{2-} \text { is } s p^3
\)
\(
\mathrm{ClO}_3^{-} \text {; No. of } \mathrm{e}^{-}=17+8 \times 3+1=42 \text {, hybridisation of } \mathrm{Cl} \text { in } \mathrm{ClO}_3^{-} \text {is } s p^3
\)
\(\mathrm{SO}_3 ;\) No. of \(\mathrm{e}^{-}=16+8 \times 3=40\), hybridisation of \(\mathrm{S}\) in \(\mathrm{SO}_3\) is \(s p^2\) \(\therefore \mathrm{NO}_3^{-}\)and \(\mathrm{CO}_3^{2-}\) are isostructural and isoelectronic.

Hint

\(
\text { (a) } \mathrm{H}_3 \mathrm{N} \rightarrow \mathrm{BF}_3 \text { where both } \mathrm{N} \text {, } \mathrm{B} \text { are attaining tetrahedral geomerty. }
\)

Hint

\(
\text { Formulae : } H=\frac{1}{2}[V+M-C+A]
\)
\(
\begin{gathered}
\text { Hybridisation of } \mathrm{N} \text { in } \mathrm{NH}_3 \\
=\frac{1}{2}[5+3-0+0]=4 \quad \therefore s p^3
\end{gathered}
\)

Hybridisation of \(\mathrm{Pt}\) in \(\left[\mathrm{PtCl}_4\right]^{2-}\)
\(
=\frac{1}{2}[2+4-0+2]=4 \quad \therefore d s p^2
\)

Hybridisation of \(\mathrm{P}\) in \(\mathrm{PCl}_5\)
\(
=\frac{1}{2}[5+5-0+0]=5 \quad \therefore s p^3 d
\)

Hybridisation of \(\mathrm{B}\) in \(\mathrm{BCl}_3\)
\(
=\frac{1}{2}[3+3-0+0]=3 \quad \therefore s p^2
\)

Hint

(b) For \(\mathrm{NO}_2^{+}: H=\frac{1}{2}(5+0+0-1)=2\);
\(\therefore s p\) hybridisation

For \(\mathrm{NO}_3^{-}: H=\frac{1}{2}[5+0+1-0)=3\);
\(\therefore s p^2\) hybridisation
For \(\mathrm{NH}_4^{+}: H=\frac{1}{2}[5+4+0-1]=4\);
\(\therefore s p^3\) hybridisation

Hint

(d) 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

\(
=\frac{1}{2}(6+4+0-0)=5
\)

For \(\mathbf{S F}_4: \mathrm{S}\) is \(s p^3 d\) hybridised in \(\mathrm{SF}_4\). Thus \(\mathrm{SF}_4\) has 5 hybrid orbitals of which only four are used by \(\mathrm{F}\), leaving one lone pair of electrons on sulphur.

For \(\mathbf{C F}_4: H=\frac{1}{2}[4+4+0-0]=4 \therefore s p^3\) hybridisaion
Since all the four orbitals of carbon are involved in bond formation, no lone pair is present on \(\mathrm{C}\) having four valence electrons
For \(\mathrm{XeF}_4: H=\frac{1}{2}(8+4+0-0)=6, \therefore s p^3 d^2\) hybridization of the six hybrid orbitals, four form bond with \(\mathrm{F}\), leaving behind two lone pairs of electrons on Xe.

Hint

(b) \(\quad H=\frac{1}{2}(3+3+0-0)=3\)
\(\therefore\) Boron, in \(\mathrm{BF}_3\), is \(s p^2\) hybridised leading to trigonal planar shape.