Chemical Bonding and Molecular Structure Test -3

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Chemical Bonding and Molecular Structure Test -3

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Question 1
1 / -0

Bond angle helps us in

A
determining molecular size

B
determining the reactivity of the element

C
determining molecular shape

D
determining the stability of the element

Solution

Bond angles also contribute to the shape of a molecule. Bond angles are the angles between adjacent lines representing bonds. The bond angle can help differentiate between linear, trigonal planar, tetraheral, trigonal-bipyramidal, and octahedral. The ideal bond angles are the angles that demonstrate the maximum angle where it would minimize repulsion, thus verifying the VSEPR theory.
Question 2
1 / -0

The amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state is called

A
lattice enthalpy

B
Electron affinity

C
ionization enthalpy

D
bond enthalpy

Solution

Bond enthalpy, also known as bond energy, is the energy that is needed to break a particular bond in a gaseous compound. The unit that expresses bond enthalpy is kilojoules per mole, or kJ/mol.
Question 3
1 / -0

H.O.H bond angle in water is

A
110⁰

B
240⁰

C
416⁰

D
104.5⁰

Solution

Due to presence of two lone pairs on O in H₂O bond angle reduce to 104.5⁰ from 109⁰.
Question 4
1 / -0

For N₂, bond order is

A
1

B
2

C
3

D
4

Solution

Bond order is the number of chemical bonds between a pair of atoms. in diatomic nitrogen N≡N, Triple covalent bond is present in nitrogen so the bond order is 3;
Question 5
1 / -0

The product of the magnitude of the charge and the distance between the centres of positive and negative charge is called

A
covalent character

B
ionic character

C
Dipole moment

D
electronegativity

Solution

A dipole moment is a measurement of the separation of two opposite electrical charges. Dipole moments are a vector quantity. The magnitude is equal to the charge multiplied by the distance between the charges and the direction is from negative charge to positive charge:

µ=q x r
Question 6
1 / -0

the valence shell electron pair repulsion (vsepr) theory helps in the

A
ionic formation of molecules

B
prediction of lewis structure of molecules

C
electronic configurations of atoms

D
prediction of geometrical shapes of molecules

Solution

Valence shell electron pair repulsion (VSEPR) theory is a model used in chemistry to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms. It is also named the Gillespie-Nyholm theory after its two main developers.
Question 7
1 / -0

In the formation of hydrogen molecule, overlapping of atomic orbitals occur which results in the pairing of electrons. These are:

A
valence shell electrons irrespective of the spins

B
valence shell electrons with opposite spins.

C
lone pair of electons

D
valence shell electrons with same spins

Solution

overlapping of atomic orbitals having electrons of opposite spin take place in the formation of molecule to cancel the dipole moment.
Question 8
1 / -0

$$PCl_5$$ has a shape of trigonal bipyramid whereas $$IF_5$$ has the shape of a square pyramid. It is due to:

A
presence of unshared electron pair on $$I$$ which is oriented so as to minimize repulsion while $$P$$ in $$PCl_5$$ has no unshared pair

B
octet of $$P$$ is complete while that of $$I$$ is incomplete

C
$$P$$ and $$I$$ are of different groups

D
$$F$$ and $$Cl$$ have different extent of repulsion

Solution

$$ { PCl }_{ 5 } $$ = $$ \begin{matrix} & & Cl & & Cl \\ & & | & / & \\ Cl & - & P & - & Cl \\ & & | & & \\ & & Cl & & \end{matrix}$$
There are no lone pairs, hence no repulsion.
Whereas in $$ { IF }_{ 5 }$$; Iodine has lone pair electrons. the geometry is thus the geometry is arranged in such a way, so, as to minimize the repulsion between the lone pair and bounded pair of electrons.
It has a Square pyramid geometry.
Question 9
1 / -0

Which of the following molecule has a dipole moment ?

A
$$BF_3$$

B
$$SiH_4$$

C
$$PF_3$$

D
$$O_2$$
Question 10
1 / -0

Electron pairs occupy localised orbitals. Their orbitals are oriented in such a way that the repulsion between electron clouds becomes:

A
zero

B
minimum

C
maximum

D
infinite

Solution

For an atom to exist in equillibrium, there should be minimum repulsion between its electron pairs. Therefore, their orbitals containing electron pairs are oriented in such a way that the repulsion between electron clouds becomes minimum.
Question 11
1 / -0

Arrange the following compounds in increasing dipole moment.
toluene $$(I)$$, $$m$$-chlorobenzene $$(II)$$, $$o$$-dichlorobenzene $$(III)$$, $$p$$-dichlorobenzene $$(IV)$$

A
$$I < IV < II < III$$

B
$$IV < I < II < III$$

C
$$IV < I < III < II$$

D
$$IV < II < I < III$$

Solution

Due to symmetrical structure, p-dichlorobenzene have zero dipole moment.
Also, in toluene methyl group have $$+I$$ effect while $$Cl$$ have $$-I $$ effect so toluene have low dipole moment.
Question 12
1 / -0

The number of bonding pairs electron $$(X)$$ and lone pairs electron $$(Y)$$ around the central atom in the $${I}^-_3$$ ion are $$X +Y$$:

A
$$2 2$$

B
$$2 3$$

C
$$3 2$$

D
$$4 3$$

Solution

The number of bonding pairs electron $$(X)=2\times 2=4$$

Total no. of lone pairs electron $$Y=3\times 3 \times 2=18$$

$$X+Y= 4+18=22$$

Option A is the correct answer.
Question 13
1 / -0

Geometrical configuration of $$BF_3$$ and $$NF_3$$ molecules is:

A
the same because of same covalency of the central atom

B
different because $$BF_3$$ is polar and $$NF_3$$ is non-polar

C
different because $$BF_3$$ is non-polar and $$NF_3$$ is polar

D
none of the above is correct

Solution

$$B{ F }_{ 3 }$$ is a trigonal planar molecule
$$N{ F }_{ 3 }$$ is a pyramidal mol
Question 14
1 / -0

Bond length can be calculated by merely adding covalent bond radii which are $$H$$ = 0.28 $$\mathring {A}$$, $$N$$ = 0.70 $$\mathring {A}$$, $$O$$ = 0.66 $$\mathring {A}$$, $$Cl$$ = 0.99 $$\mathring {A}$$, $$(C=)$$=0.67$$\mathring {A}$$, (C$$\equiv $$)=0.61 $$\mathring {A}$$, (N$$\equiv $$)= 0.55 $$\mathring {A}$$ and $$(C-) = 0.77A^o$$. Calculate bond lengths in $$NH_{3}$$ :

A
$$N - H\, \, \, 0.98 \mathring {A}$$

B
$$N - H\, \, \, 0.90 \mathring {A}$$

C
$$N - H\, \, \, 0.58 \mathring {A}$$

D
None of these

Solution

Bond length can be calculated by merely adding covalent bond radii. Adding bond lenghts of $$N$$ and $$H,$$ the bond length in $$N-H$$ in ammonia is 0.98$$A^o$$.
Question 15
1 / -0

Which one would better represent bonding in the molecule $$HNO$$?
I : $$H - \ddot{N} = \ddot{O}:$$ and II : $$H - \ddot{O} = \underset{\cdot\cdot}{\ddot{N}}$$

A
$$ I > II $$

B
$$ II > I $$

C
$$ I = II $$

D
None of these

Solution

$$I$$ is better as octets of $$N$$ and $$O$$ are complete and duplet of $$H$$ is also complete.
Question 16
1 / -0

True structure is predicted by :

A
valence bond approach

B
Sidgwick approach

C
hybrid orbital formation

D
None of the above

Solution

Hybrid formation was introduced to explain molecular structure when the other theories failed to correctly predict them.

Hence the correct option is C.
Question 17
1 / -0

In which compound, are the bonds most polar?

A
$$H_2O$$

B
$$CO_2$$

C
$$CCl_4$$

D
$$ClF$$

Solution

Since, the electronegativity difference between $$H$$ and $$O$$ is maximum, hence, $$H_2O$$ bond is most polar.

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Chemical Bonding and Molecular Structure Test -3

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Chemical Bonding and Molecular Structure Test -3

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Question 1

determining molecular size

determining the reactivity of the element

determining molecular shape

determining the stability of the element

Solution

Question 2

lattice enthalpy

Electron affinity

ionization enthalpy

bond enthalpy

Solution

Question 3

1100

2400

4160

104.50

Solution

Question 4

1

2

3

4

Solution

Question 5

covalent character

ionic character

Dipole moment

electronegativity

Solution

Question 6

ionic formation of molecules

prediction of lewis structure of molecules

electronic configurations of atoms

prediction of geometrical shapes of molecules

Solution

Question 7

valence shell electrons irrespective of the spins

valence shell electrons with opposite spins.

lone pair of electons

valence shell electrons with same spins

Solution

Question 8

presence of unshared electron pair on $$I$$ which is oriented so as to minimize repulsion while $$P$$ in $$PCl_5$$ has no unshared pair

octet of $$P$$ is complete while that of $$I$$ is incomplete

$$P$$ and $$I$$ are of different groups

$$F$$ and $$Cl$$ have different extent of repulsion

Solution

Question 9

$$BF_3$$

$$SiH_4$$

$$PF_3$$

$$O_2$$

Question 10

zero

minimum

maximum

infinite

Solution

Question 11

$$I < IV < II < III$$

$$IV < I < II < III$$

$$IV < I < III < II$$

$$IV < II < I < III$$

Solution

Question 12

$$2 2$$

$$2 3$$

$$3 2$$

$$4 3$$

Solution

Question 13

110⁰

240⁰

416⁰

104.5⁰