Chemical Bonding and Molecular Structure Test 15

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Chemical Bonding and Molecular Structure Test 15

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Weekly Quiz Competition

Question 1
1 / -0

The dipole moment of $$HBr$$ is $$0.78\ \times 10^{-18}$$ esu cm and interatomic spacing is $$1.41\ \mathring {A}$$ The $$\%$$ ionic character of $$HBr$$ is?

A
$$7.5$$

B
$$11.$$

C
$$15$$

D
$$27$$

Solution

$$D=$$ charge $$\times$$ distance
$$=1.6 \times 10^{-19}\times 1.41\times 10^{-8}$$
$$=\cfrac {2.256\times 10^{-27}}{3.33\times 10^{-30}}$$
$$=6.76 \times 10^2$$
% of ionic charge= $$0.78 \times 10^{-18} esu$$
$$=0.78$$
$$=\cfrac {0.78}{6.76}=11$$% .
Question 2
1 / -0

Which combination of atomic orbital is correct for $$\sigma$$-bond formation?

A
$$s - P_z$$

B
$$s - P_x$$

C
$$s - P_y$$

D
All are correct

Solution
Question 3
1 / -0

$$16$$g of sulphur completely reacts with $$38$$ g of $$F_2$$ to form $$SF_x$$. The value of $$x$$ is:
(Atomic mass of $$F = 19$$)

A
$$1$$

B
$$2$$

C
$$4$$

D
$$6$$

Solution

$$16g$$ of $$S$$ reacts with $$38g$$ of $$F_2$$
The ratio of moles reacted is $$\cfrac {16}{32}:\cfrac {38}{19}$$
$$\Rightarrow \cfrac {1}{2}:2$$
$$\Rightarrow 1:4$$
So one $$S$$ atom and $$4$$ atoms forms $$SF_4$$
So, $$x=4$$ .
Question 4
1 / -0

In the given structure of compound, the correct various bond moments direction involving atoms are shown as:

A
$$Br\overset{\longleftarrow}{-}N\overset{\longleftarrow}{-}CH_2\overset{\longrightarrow}{-}SiH_2\overset{\longleftarrow}{-}H_2\overset{\longrightarrow}{-}O\overset{\longleftarrow}{-}CH_3$$

B
$$Br\overset{\longleftarrow}{-}N\overset{\longleftarrow}{-}CH_2\overset{\longleftarrow}{-}SiH_2\overset{\longleftarrow}{-}CH_2\overset{\longrightarrow}{-}O\overset{\longleftarrow}{-}CH_3$$

C
$$Br\overset{\longrightarrow}{-}N\overset{\longleftarrow}{-}CH_2\overset{\longleftarrow}{-}SiH_2\overset{\longrightarrow}{-}CH_2\overset{\longrightarrow}{-}O\overset{\longleftarrow}{-}CH_3$$

D
$$Br\overset{\longleftarrow}{-}N\overset{\longrightarrow}{-}CH_2\overset{\longrightarrow}{-}SiH_2\overset{\longleftarrow}{-}CH_2\overset{\longrightarrow}{-}O\overset{\longrightarrow}{-}CH_3$$

Solution

Bond moments direction depends on the electronegativity of the elements.

The electronegativity of the given elements is-
$$O>N>Br>C>Si$$

option C is correct.
Question 5
1 / -0

On treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is:

A
$$C1-O2$$

B
$$O2-C3$$

C
$$C4-O5$$

D
$$O5-C6$$

Solution

On treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is $${ O }_{ 5 }-{ C }_{ 6 }$$.
Question 6
1 / -0

Among the following molecular orbitals, how many have at least one nodal plane ?
$$ \sigma ^*1s,\sigma 2s,\sigma ^{ * }2p_{ z },\pi 2p_{ x },\pi ^{ * }2p_{ y },\sigma 2p_{ z } $$

A
$$ 4 $$

B
$$ 5 $$

C
$$ 2 $$

D
$$ 6 $$

Solution
Question 7
1 / -0

Which of the following molecule is theoretically not possible?

A
$$SF_4$$

B
$$OF_2$$

C
$$OF_4$$

D
$$O_2F_2$$

Solution

$$O$$ belongs to period 2 having 2p valence shell and does not have vacant d-orbitals. Belonging to group 6, it already has 6 electrons in its valence shell and can accommodate only 2 more electrons to complete its octet. In the formation of $$OF_4$$ the number of electrons in its outermost shell is 10. Since, it does not have vacant d-orbitals to accommodate the extra electrons, formation of $$OF_4$$ is theoretically not possible.
Question 8
1 / -0

In the given, lewis structure of $${ S }_{ 2 }{ O }_{ 3 }^{ 2- }$$ formal charge present on sulphur atoms $$1$$ and $$2$$ respectively are:

A
$$zero, \ +1$$

B
$$+1,+1$$

C
$$+2,+2$$

D
$$zero, \ zero$$

Solution

Formal charge is expressed as (a lewis dot structure)
$$=$$ (total no. of valence electron in free atom) $$-$$ (total no. of non bonding electron) $$-\dfrac { 1 }{ 2 } $$ (total no. of bonding electron)
$$=V-L-\dfrac { 1 }{ 2 } \times S$$
$$L=$$ lone pair electrons, $$S=$$ bonding electrons
For Sulphur atom $$(1)$$
$$FC=6-4-\dfrac { 1 }{ 2 } \times 4\Rightarrow 6-4-2\Rightarrow 0$$
For Sulphur atom $$(2)$$
$$FC=6-0-\dfrac { 1 }{ 2 } \times 12\Rightarrow 6-6=0$$
Ans is $$(D)$$.
Question 9
1 / -0

1 mole of $$H_3PO_4$$ having

A
Two $$\pi $$- bond

B
six mole of $$\sigma $$-bond

C
Sixteen lone pair of electron

D
seven $$\sigma $$-bond

Solution
Question 10
1 / -0

Which bond angle $$\theta$$ would result the maximum dipole moment for the triatonuc molecules $$XY_2$$ as shown below?

A
$$90^o$$

B
$$120^o$$

C
$$150^o$$

D
$$180^o$$

Solution

The dipole moment of $$2$$ dipoles inclined at an angle $$\theta $$ is given by,
$$\mu =\sqrt { x^{ 2 }+{ y }^{ 2 }+2xycos\theta } ,\quad \quad cos90=0$$
$$\therefore$$ The dipole moment is maximum for $$\theta ={ 90 }^{ 0 }$$