Chemical Bonding and Molecular Structure Test 13

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Chemical Bonding and Molecular Structure Test 13

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Weekly Quiz Competition

Question 1
1 / -0

The number of $$C - C$$ sigma bonds present in the given compound:

A
$$16$$

B
$$14$$

C
$$18$$

D
$$11$$
Question 2
1 / -0

The relation between $$x, y$$ and $$z$$ in bicarbonate ion with respect to bond length is:

A
$$x> y > z$$

B
$$x > z > y$$

C
$$z = y > x$$

D
$$x > y = z$$

Solution

$$x>y>z$$
Because double bond length is always less than single bond length . So, '$$y$$' value is small here.
Between '$$x$$' and '$$z$$', '$$x$$' has more value because of more electronegtivity of '$$O$$', it pulls electron of '$$H$$'. So its electronegativity decreases and bond length is more.
Question 3
1 / -0

$$S-S$$ bond is present in:

A
$$\alpha - (SO_3)_n$$

B
$$\gamma (S_3O_9)$$

C
$$H_2S_2O_3$$

D
$$H_2S_2O_8$$

Solution
Question 4
1 / -0

Which compound has non-zero dipole moment?

A

B

C

D

Solution

(Refer to Image) has non-zero dipole moment.
Question 5
1 / -0

The correct order of increasing bond length of $$C-H$$, $$C-O$$, $$C-C$$ and $$C=C$$ is :

A
$$C-H\ <\ C-O\ <\ C-C\ <\ C=C$$

B
$$C-H\ <\ C=C\ <\ C-O\ <\ C-C$$

C
$$C-C\ <\ C=C\ <\ C-O\ <\ C-H$$

D
$$C-O\ <\ C-H\ <\ C-C\ <\ C=C$$

Solution

The $$C-H$$ bond is the shortest due to the small size of both atoms & large electronegativity difference.

The double bonds are shorter than single bonds.

Hence $$C=C < C-O$$ and $$C-C$$

Due to electronegativity difference between $$C$$ and $$O$$

$$C-O < C-C$$

$$\therefore$$ Correct order of increasing bond length is :

$$C-H < C=C < C-O < C-C$$

Hence, the correct option is B.
Question 6
1 / -0

Which of the following molecule does not possess a permanent dipole moment?

A
$$CS_2$$

B
$$SO_{3}^{2-}$$

C
$$H_2S$$

D
$$SO_2$$

Solution

Structure of given molecules.
$${ CS }_{ 2 }$$ does not passes a permanent dipole moment.
Question 7
1 / -0

Which of the following is not correct?

A
$$\text{A sigma bond is weaker than } \pi \ bond$$

B
$$\text{A sigma bond is stronger than } \pi \ bond$$

C
$$\text{A double bond is stronger than a single bond}$$

D
$$\text{A double bond is shorter than a single bond}$$

Solution

The sigma bond is formed by the co-axial overlapping of the orbitals while pi bond is formed by the lateral overlapping of the orbitals.
Thus, the extent of overlapping is more in sigma bond than in pi bond.
Thus $$\sigma$$ bond is stronger than the $$\pi$$ bond.
Question 8
1 / -0

A: Tetracyanomethane
B: Carbondioxide
C: Benzene
C: $$1,3-buta-di-ene$$
Ratio of $$\sigma$$ and $$\pi$$:

A
in A and B are equal

B
in C and D are equal

C
is same in A, B, C, and D

D
in B and D are equal

Solution

$$a:$$ Tetracyanomethane, see the figure i. where $$\sigma$$ bond $$=8$$, ratio$$=1$$, $$\pi$$ bond $$=8.$$
$$b:$$ Carbondioxide, $$O=C=O\\ \sigma=2\\ \pi=2$$, ratio$$=1.$$
$$c:$$ Benzene, see the figure ii, where $$\sigma$$ bond$$=12,$$ $$\pi$$ bond $$=3$$, ratio $$=4.$$
$$d:$$ $$1,3-$$buta-di-ene, see the figure iii, where $$\sigma$$ bond $$=9,$$ $$\pi$$ bond $$=2,$$ ratio $$=\cfrac{9}{2}.$$
$$\therefore\text{ }A=B$$
Question 9
1 / -0

In which of the following molecules, bonding is not taking place in excited state:

A
$$CH_4$$

B
$$BF_3$$

C
$$IF_7$$

D
$$PCl_3$$
Solution
In the case of $$PCl_3$$, only outer electrons are going to be shown for simplicity.
Each atom in $$PCl_3$$ has $$2,8$$ electrons in inner layers of orbit.
Every atom has a noble gas structure.
So, in $$PCl_3$$ bonding is not taking place in excited state.
Question 10
1 / -0

From the given list of compounds whose proton bonding is highest?

A
$$H_2O$$

B
$$H_2S$$

C
$$NH_3$$

D
$$PH_3$$

Solution

1. $$H_2O$$ has the highest proton bonding.
2.(a) In a water molecule, the electrons of the hydrogen atoms are strongly attracted to the oxygen atom.
3. Therefore, the part of the water molecule containing the oxygen atom has a slightly negative charge, and the part containing hydrogen atoms has a slightly positive charge.
4.In $$NH_3, PH_3 , H_2S$$ hydrogen bonding is comparably low.