Chemical Bonding And Molecular Structure - Class 11 Engineering Chemistry - Extra Questions

Predict which out of the following molecules will have higher dipole moment and why ?
$$CS_{2}$$ and $$OCS$$

$$S = C = S$$ and $$O - C = S$$ both are linear molecules but bond moments in $$CS_{2}$$ cancel out so that net dipole moment = $$0.$$ But in OCS, bond moment of $$C = O$$ is not equal to that of $$C = S.$$ Hence it has a net dipole moment. Thus dipole moment of OCS is higher .


Arrange the following in the increasing order of $$C-C$$ bond length $$C_2H_6, C_2H_4, C_2H_2$$.

As the bond order of $$C-C$$ increases bond length decreases. So, the incresing order of bond length is:

$$C_2H_2 < C_2H_4 < C_2H_6$$


Fill in the blanks:
For hydrogen bonding to occur, hydrogen atom must be linked to .......... atom.

For hydrogen bonding to occur, hydrogen atom must be linked to a highly electronegative atom.


Explain the hydrogen bonding in molecules of $$HF,$$ $$H_2O$$ and $$NH_3$$.

$${ H }_{ 2 }O>HF>{ NH }_{ 3 }$$
This is because one water molecule can form $$4$$ hydrogen bonds whereas $$HF$$ and $${ NH }_{ 3 }$$ only two.Also $${ P }^{ - }$$ is more electronegative than $${ N }^{ - }$$.


In $$ CuSO_4.5H_2O $$ all Cu-O bond length are not equal. Find the total number of shorter Cu-O bonds.

  • In $$CuSO_4.5H_2O$$ structure is distorted octahedral.
  • There are four bonds between $$Cu$$ and oxygen which is $$1.958\overset {o}{A}$$.
  • Two bonds between oxygen atom of $$SO_4^{-2}$$ and $$Cu$$, which is $$2.385\overset {o}{A}$$.
  • Therefore, total number of shorter $$Cu-O$$ bonds will be four.


What is the formal charge on atom in carbonate and nitric acid?

The charge assigned to an atom in a molecule is a formal charge.
Formal charge $$(FC)$$ is given by the formula,
$$FC=V-N-B/2$$
Where,
$$V=$$ Number of valence electrons
$$N=$$ Number of non-bonding electrons
$$B=$$ Total number of electrons shared in covalent bonds.
The formal charge of $$N$$ in $$NO2^-$$ is
$$FC = 5 - 2 - 6/2  =0$$
For the LHS oxygen the $$FC$$ is-
$$FC = 6 - 6 - 2/2 = -1$$
For the RHS oxygen the FC is-
$$FC = 6 - 4 - 4/2 =0$$
The carbonate ion $$[CO_3{^{ 2-}}]$$ has an overall charge of -2. In the carbonate ion, the carbon atom is bonded with a double bond to an oxygen atom, and with single bonds to two oxygen atoms.
Formal charge on oxygen bonded with the double bond = $$6 - 4 - 1/2 (4) = 0$$
Formal charge on oxygen atoms bonded with single bond = $$6 - 6 - 1/2 (2) = -1$$
In $$HNO_3$$
The nitrogen atom carries a charge of $$+1$$, and one oxygen atom carries a charge of $$-1.$$

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On the basis of ground electronic configuration, arrange the following molecules in the order of increasing $$O-O$$ bond lengths: $$ KO_2 ,\ O_2 ,\ O_2[AsF_6] .$$

Using MOT concept:

                            $$ O^{2-}_2 $$      $$ O^-_2 $$           $$ O_2 $$          $$ O^+_2 $$
Bond order:         1            1.5              2             2.5
Magnetic P.:       Dia          Para      Para          Para

The larger the bond order, the smaller the bond length.

Therefore , $$ O^+_2[ AsF_6]^-  < O_2 < K^+ O^-_2 $$


Although $$ CO_2 $$ has no dipole moment, $$ SO_2 $$ and $$ H_2O $$ have considerable dipole moments-explain.

$$ CO_2 $$ has linear structure while $$SO_2 $$ and $$ H_2O $$ have V_shaped structures. Hence, dipole moment in $$CO_2$$ is cancelled out but $$SO_2 $$ and $$ H_2O $$ posses a dipole moment.
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Name two intermolecular forces that exist between $$HF$$ molecules in a liquid state.

In liquid HF two intermolecular forces that are exist are as follows:-
 $$\textbf{Hydrogen bond and Dipole-Dipole interactions}$$ are present.


Answer the following question:
Dipole moment of phenol is smaller than that of methanol. Why?

In phenol, $$C-O$$ bond is less polar due to the electron-withdrawing effect of benzene ring whereas, in methanol, $$C-O$$ bond is more polar due to electron-releasing effect of  $$-CH_3$$ group.


Write a short note on Hydrogen bond.

Hydrogen bond:

A hydrogen bond is a type of attractive (dipole-dipole) interaction between an electronegative atom and a hydrogen atom bonded to another electronegative atom. This bond always involves a hydrogen atom. Hydrogen bonds can occur between molecules or within parts of a single molecule. A hydrogen bond tends to be stronger than van-der Waal's forces, but weaker than covalent bonds or ionic bonds.
There are two types of hydrogen bonds:
1. Intermolecular hydrogen bond:
This type of bond formed between two or more same or differnt molecules.
Example: water, ammonia etc.
2. Intramolecular hydrogen bond:
This type of bond is formed between two differnt atoms of same molecule.
Example: O-nitrophenol.

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Which of the following has maximum dipole moment; $$Na^+, K^+, Li^+$$ and $$Cs^+$$?

Dipole moment is directly proportional to electronegativity. Electronegativity decreases with increasing atomic size. Thus, the ion having smallest size has maximum electronegativity and thus, maximum dipole moment. Also atomic size increase down the group. Therefore, in the above ions, $$Li^+$$ has smallest size.
Hence, it has maximum dipole moment.


Dipole moment of $$CO_{2}$$ molecule is zero where as $$SO_{2}$$ has some dipole moment. Explain the reason.


The best way to understand geometry is by drawing the Lewis dot structure of the molecules.
The geometry of $$CO_2$$ is linear (electrons geometry is linear too) because the two bonds are the farthest apart and the bond angle is $$180^o$$. This is what makes of $$CO_2$$ a non polar molecules because the dipole moment of both bonds cancel out.
The geometry of $$SO_2$$ is bent because the lone pair on the central sulphur atom pushes the two bonds down for a bond angle of $$120^o$$ due to the electrons repulsion. This is what makes of $$SO_2$$ a polar molecule because the dipole moments of both bonds of not cancel out, the molecule will have a net dipole moment.

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Fill in the blanks:
Dipole moment of a diatomic polar molecule is the product of .............

Dipole moment of a diatomic polar molecule is the product of charge, internuclear distance.


Explain how VB theory differs from Lewis concept.

(i) According to Lewis concept, a covalent bond is formed by mutual sharing of electrons whereas according to VB theory a covalent bond is formed by the overlap of half-filled atomic orbitals containing electrons which opposite spin.
(ii) VB theory can explain the shapes of molecules whereas Lewis concept cannot.
(iii) VB theory can explain the strength of bonds whereas Lewis concept cannot.


What requirement should a molecule fulfil for the formation of a hydrogen bond ?


$$\text { The two requirements are } \\$$
$$\text { a) it should be more electronegative } \\$$
$$\text { than hydrogen. } \\$$
$$\text { b) it should be small sized atom } \\$$
$$\text { The molecules who fulfil these needs are } \\$$
$$\text { oxygen, nitrogen and fluorine. }$$



Which of the following has larger dipole moment? Explain.
1-Butyne or 1-Butene

The direction of dipole moments of individual bonds in $$1-$$ butyne and $$1-$$butene are shown in the figure.
The resultant dipole moment of both $$1-$$butyne and $$1-$$butene is due to the dipole moment of $$CH_3CH_2 -C$$ and $$C-H$$ bonds which oppose each other. 

Since a $$sp$$ carbon is more electronegative than a $$sp^2$$ carbon, therefore the dipole moment of $$CH_3CH_1(sp^3)-C(sp)$$ bond in $$1-$$butyne is more than that of $$CH_3CH_1(sp^3)-C(sp^2)$$ bond in $$1-$$butene. As a result, the dipole moments of $$1-$$butene is more than that of $$1-$$butene.

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Find the number of delta bonding molecular orbital from the following set if z is the inter-nuclear axis :
$$ d_{x^2\,-\,y^2}$$ and $$d_{xz}, \,d_{x^2\,-\,y^2}$$ or $$d_{xz}$$

The number of delta bonding molecular orbital from the following set if z is the inter-nuclear axis is $$1$$.
Delta bonding molecular orbital is formed by the overlap of  $$ d_{x^2\,-\,y^2}$$ and $$d_{x^2\,-\,y^2}$$ orbitals when z is the internuclear axis.


Assuming all the four valencies of carbon atom in propane pointing towards the corners of a regular tetrahedron; the distance (in $$\dot{A}$$) between the terminal carbon atoms in propane will be: (write the value of the nearest integer) 
Given: $$C-C$$ single bond length is $$1.54$$ $$\dot{A}$$.

Since each $$C$$ atom has tetrahedral geometry, the $$C-C$$ bond angle will be $$\displaystyle 109.5^o$$
$$\displaystyle \theta = \dfrac {180-109.5}{2} = 35.25^o$$
The distance between terminal carbon atoms $$\displaystyle = 2 cos \theta \times $$ C-C single bond length
It will be $$\displaystyle 2\:cos\:35.25 \times 1.54 = 2 \times 0.82 \times 1.54 = 2.514\:A^o$$


Write two differences between $$O^{2}$$ and $$O^{2-}$$.

Two atoms of oxygen element are represented as '$$2O$$'. The 2 oxygen  atoms are unreactive and unstable readily reacts and form molecules.

$${ O }_{ 2 }$$ represents one molecule oxygen. When the two oxygen atoms react with each other forms one mole of oxygen. 


Arrange Halides (MeX) in order of decreasing bond length.  

MeF, MeCl, MeBr, MeI.


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Find the sum of nodal plane of $$ \pi^* _{d-d} $$ anti-bonding molecular orbit and $$ \pi^*_{p-d} $$ anti-bonding molecular orbital.


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Why the repulsions between non-bonded orbitals are greater than between the bonded orbitals?

The non-bonded orbitals relatively occupy more space compared to the bonded orbitals and thus repulsions are greater.


Write a short note on 'Hydrogen bond'.

A hydrogen bond is a type of a dipole - dipole interaction
between $$\mathrm{H}$$ and more electro negative atom such as O, F
and N. A hydrogen bond is stronger than Van-der Waal's forces, but weaker
than Covalent bonds or ionic bands. There are two types of
hydrogen bond:
1. Intermolecular hydrogen bond : This type of bond is formed between two or more same or different molecules.
Example : Water, ammonia etc.

2. Intramolecular hydrogen bond: This type of bond is formed
between two different atoms of Same molecule
Example: 0 - nitrophenol


Class 11 Engineering Chemistry Extra Questions