Chemical Bonding and Molecular Structure Test 34

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Chemical Bonding and Molecular Structure Test 34

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of lone pairs in $$\displaystyle XeF_{2}$$ molecule and the lone pairs on its central atom is:

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Total no. of lone pairs molecule:$$9$$
Ratio of lone pairs in molecule to that of central atom is $$9:3$$
$$\Rightarrow 3:1$$
$$\therefore$$ correct answer is option A.
Question 2
1 / -0

In a molecule of phosphorus (V) oxide, there are:

A
$$4P-P, 10P-O$$ and $$4P = O$$ bonds

B
$$12P-O$$ and $$4P = O$$ bonds

C
$$2P-O$$ and $$4P = P$$ bonds

D
$$6P-P, 12P-O$$ and $$4P= P$$ bonds

Solution

As can be seen from the image, it contains $$12$$ $$P-O$$ bonds and $$4 \:P=O$$ bonds.
Question 3
1 / -0

An ionic compound is expected to have tetrahedral structure if r+ / r- lies in the range of:

A
0.414 to 0.732

B
0.225 to 0.414

C
0.115 to 0.225

D
0.732 to 1

Solution

For ionic solids, the ratio of the radius of cation to that of the anion is called radius ratio.
Radius ratio
Structural arrangement
Above 0.732
Cubic
0.414 – 0.732
Octahedral
0.225 – 0.414
Tetrahedral
0.155 – 0.225
Triangular
Question 4
1 / -0

The number of $$\displaystyle \angle FIF$$ angles which are less than $$\displaystyle 90^{\circ}$$ in $$\displaystyle IF_{5}$$ are :

A
8

B
7

C
6

D
5

Solution

Four angle in a plane and four angle due to repulsion of lone pair.
Question 5
1 / -0

Which of the following orbital combination does not form $$\pi$$-bond?

A
$$p_x+p_x$$ sideways overlapping

B
$$d_{\displaystyle x^2-y^2}+p_y$$ sideways overlapping

C
$$d_{xy}+d_{xy}$$ sideways overlapping

D
$$d_{yz}+p_y$$ sideways overlapping

Solution

$$d_{\displaystyle x^2-y^2}+p_y$$ sideways overlapping does not form $$\pi$$-bond as the two orbitals do not have proper orientation for the overlap.
Question 6
1 / -0

The correct statement regarding $$SO_2$$ molecule is:

A
two $$p\pi-d\pi$$ bonds

B
molecule has 2 lone pair, $$2\sigma$$ bonds and $$2\pi$$ bonds

C
two $$p\pi-p\pi$$ bonds

D
one $$p\pi-p\pi$$ bond and one $$p\pi-d\pi$$ bond

Solution

$$SO_2$$ molecule has one $$p\pi-p\pi$$ bond and one $$p\pi-d\pi$$ bond.
$${ SO }_{ 2 }=6+6(2)=18e-16e=2e\rightarrow 1$$ lone pair

$$\rightarrow $$ G.S.=[Ne]$$3{s}^{2}$$$$3{p}^{4}$$
2 Unpaired electron
$${1}^{st}$$ E.S.$$= [Ne] 3{s}^{2}3{p}^{3}3{d}^{1}$$
4 unpaired electron
$$O\rightarrow 1{ s }^{ 2 }2{ s }^{ 2 }2{ p }^{ 4 }$$

$$\rightarrow$$ Geometry of $$SO_2$$ is BENT/ANGULAR
$$\rightarrow$$ one $$p\pi-p\pi$$ and one $$p\pi-d\pi$$ Bond
$$\rightarrow$$ $$2\pi$$ Bonds
$$\rightarrow$$ $$1$$ lone pair on the central atom.
Question 7
1 / -0

Assuming the bond direction to the z-axis, which of the overlapping of atomic orbitals of two atoms ($$A$$) and ($$B$$) will result in bonding?
(I) $$s$$-orbital of $$A$$ and $$p_x$$ orbital of $$B$$
(II) $$s$$-orbital of $$A$$ and $$p_z$$ orbital of $$B$$
(III) $$p_y$$ orbital of $$A$$ and $$p_z$$ orbital of $$B$$
(IV) $$s$$-orbital of both ($$A$$) and ($$B$$)

A
I and IV

B
I and II

C
III and IV

D
II and IV

Solution

As s is nondirectional so it will form bond in all directions.
Also $$p_z$$ is directed on z-axis so it will form bond on z-axis.
So II and IV are correct.
Question 8
1 / -0

The distance between two neutral atoms in equilibrium is :

A
$$\displaystyle \sqrt\frac{2A}{B}$$

B
$$\displaystyle \left(\frac{2A}{B}\right)^\dfrac{1}{6}$$

C
$$\displaystyle \frac{2A}{B}$$

D
$$\displaystyle \left(\frac{2A}{B}\right)^\dfrac{1}{2}$$

Solution

The distance between two neutral atoms in equilibrium is $$\displaystyle \left(\frac{2A}{B}\right)^\dfrac{1}{6}$$.
This is obtained when the value of the force F is substituted as zero in the expression for the force which is $$F=\dfrac {12A}{r^{13}}-\dfrac {6B}{r^7}$$.
Question 9
1 / -0

Dipole moment of $$\displaystyle NH_{3}$$ is ________ than $$\displaystyle NF_{3}.$$

A
medium

B
less

C
more

D
none of the above

Solution

Dipole moment of $$NH_3$$ is more than $$NF_3$$. Because in $$NH_3$$, the Nitrogen is more electronegativity than the hydrogen, so electron density shift towards nitrogen. On the other hand, in $$NF_3$$, the electron equally distributed to the three fluorine atoms as it has more electronegative than nitrogen.
The correct option is C.
Question 10
1 / -0

In $$BrF_3$$ molecule, the lone pairs occupy equatorial position to minimize :

A
lone pair - bond pair repulsion only

B
bond pair - bond pair repulsion only

C
lone pair - lone pair repulsion and lone pair - bond pair repulsion

D
lone pair - lone pair repulsion only

Solution

Lone pair repulsion is greater than bond pair repulsion. Therefore, to attain stability, the lone pairs should be kept at a position farthest from each other, so they occupy the equatorial positions.