Chemical Bonding and Molecular Structure Test 38

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Chemical Bonding and Molecular Structure Test 38

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Weekly Quiz Competition

Question 1
1 / -0

Hybridisation of carbon in $${ CH }_{ 2 }=CH-\overset { \oplus }{ { CH }_{ 2 } }$$ is:

A
$${ sp }^{ 2 }$$

B
$$sp$$

C
$${ sp }^{ 3 }$$

D
all of these

Solution

$$S{ N }_{ 1 }=\dfrac { 4+2 }{ 2 } =\dfrac { 6 }{ 2 } =3-s{ p }^{ 2 }\\ S{ N }_{ 2 }=\dfrac { 4+2 }{ 2 } =\dfrac { 6 }{ 2 } =3-s{ p }^{ 2 }\\ S{ N }_{ 3 }=\dfrac { 4+3-1 }{ 2 } =3-sp^{ 2 }$$
Question 2
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The correct order of stability of $$Al^+,Al^{+2},Al^{+3}$$ is:

A
$$Al^{+3}>Al^{+2}>Al^+$$

B
$$Al^{+2} > Al^{+3} > Al^+$$

C
$$Al^{+2} < Al^+ > Al^{+3}$$

D
$$Al^{+3} > Al^+ > Al^{+2}$$

Solution
Question 3
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Complete the balanced and following reaction:

$$Na_2HPO_4 + X \rightarrow Y + NaH_2PO_4$$

A
X = $$NaHCO_3$$

Y = NaCl

B
X = $$HCO_{3}^-$$

Y = $$NaHCO_3$$

C
X = $$NaHCO_3$$

Y = $$H_2CO_3$$

D
X = $$H_2CO_3$$

Y = $$NaHCO_3$$

Solution
Question 4
1 / -0

Bond-length of $$HCl$$ is $$1.275 \overset oA$$ $$(e=4.8 $$$$\times 10 ^{ -10 }$$ e.s.u.). If dipole moment $$=1.02 D$$ then $$HCl$$ is:

A
$$100\%$$ ionic

B
$$83\%$$ covalent

C
$$50\%$$ covalent

D
$$40\%$$ ionic

Solution
Question 5
1 / -0

The hybridisation of each carbon in the following compound is:
$$\underset {i}{CH_{3}} - \underset {ii}{\overset {\overset {O}{||}}{C}} - \underset {iii}{CH_{2}} - \underset {iv}{CN}$$

A
$$sp^{3}\ sp^{2}\ sp^{3}\ sp$$

B
$$sp^{3}\ sp^{3}\ sp^{2}\ sp$$

C
$$sp^{3}\ sp\ sp^{3}\ sp^{2}$$

D
$$sp^{3}\ sp^{2}\ sp\ sp^{3}$$

Solution

$$\underset {i}{CH_{3}} - \underset {ii}{\overset {\overset {O}{||}}{C}} - \underset {iii}{CH_{2}} - \underset {iv}{CN}$$
In (i), all single bonds occur therefore $$sp^{3}$$ hybridization.
In (ii), one double bond occurs therefore $$sp^{2}$$ hybridization.
In (iiI), all single bond occurs therefore $$sp^{3}$$ hybridization.
In (iv), one triple bond occurs, therefore $$sp$$ hybridization.
Question 6
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Out of given reaction which show change in hybridization of central atom__________.

A
$${ H }_{ 2 }{ BO }_{ 3 }$$ dissolve in water

B
$${ H }_{ 2 }{ SO }_{ 4 }$$ dissolve in water

C
$${ N }_{ 2 }{ O }_{ 5(g) }\longrightarrow { N }_{ 2 }{ O }_{ 5(s) }$$

D
$$P{ Br }_{ 5(g) }\longrightarrow P{ Br }_{ 5(s) }$$

E
$${ C }_{ 2 }{ H }_{ 6 }\xrightarrow [ bond\ cleavage\ of\ C-C\ bond ]{ Homolytic } $$

Solution

Correct Option $$- A$$
Reason:-
$$H_{3}BO_{3}\longrightarrow HBO_{2} + H_{2}O$$
$$H_{3}BO_{3}$$ has $$sp^{3}$$ hybridisation at the central atom and $$HBO_{2}$$ has $$sp^{2}$$ hybridisation at the central atom.
Question 7
1 / -0

Arrange the molecule in the increasing order of bond angle.

A
$${ H }_{ 2 }O< O{ F }_{ 2 }< O{ Cl }_{ 2 }$$

B
$$O{ Cl }_{ 2 }< { H }_{ 2 }O< O{ F }_{ 2 }$$

C
$$O{ F }_{ 2 }< O{ Cl }_{ 2 }< { H }_{ 2 }O\quad $$

D
$$O{ F }_{ 2 }<{ H }_{ 2 }O< O{ Cl }_{ 2 }$$

Solution

$$H_{2}O< OF_2<OCl_2$$
$$OCl_2$$ have a maximum bond angle as the size of $$Cl$$ atom is bigger due to which $$2\ Cl$$ atom repel each other, thus have bigger bond angle than $$H_{2}O$$ and $$OF_2$$.
Question 8
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Identify correct order of average molecular dipole moment.

A
$$P{Cl}_{3}{F}_{2}> {N}_{2}{O}_{4(g)}> {N}_{2}{O}_{4(s)}$$

B
$$P{Cl}_{3}{F}_{2}< {N}_{2}{O}_{4(g)}< {N}_{2}{O}_{4(s)}$$

C
$$P{Cl}_{3}{F}_{2}< {N}_{2}{O}_{4(g)}= {N}_{2}{O}_{4(s)}$$

D
All are incorrect.

Solution

Correct Option $$- D$$
Reason:- $$PCl_{3}F_{2}$$ has a symmetrical & thus has a zero dipole moment. Increasing dipole moments N from $$N_{2}O_{4}$$$$_{(s)}$$ to $$N_{2}O_{4}$$$$_{(g)}$$is due to the fact that dipole moment=charge*distance.
Question 9
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Which of the following is not the correct representation of sigma bond?

A

B

C

D

Solution

$$\sigma$$ bond is formed by the coaxial overlapping of the two orbitals.
The $$\pi$$ bond is formed by the lateral overlapping of the two orbital.
Thus, the overlapping in option D is not a sigma bond.
Question 10
1 / -0

The number of $$\sigma$$ and $$\pi$$ bonds in orange azo dye is:

A
$$27$$ and $$5$$

B
$$24$$ and $$7$$

C
$$26$$ and $$7$$

D
$$26$$ and $$6$$

Solution

1 single bond (-) contains 1 $$\sigma$$ bond and 1 double bond consists of 1 $$\sigma$$ and 1 $$\pi$$ bond.

The number of $$\sigma$$ and $$\pi$$ bonds in orange azo dye (p-hydroxyazobenzene) is $$26$$ and $$7$$.

Hence, option C is correct.