Chemical Bonding and Molecular Structure Test -7

Result

Chemical Bonding and Molecular Structure Test -7

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

When a gas phase atom in its ground state gains an electron. This is called

A
Electron affinity

B
lattice enthalpy

C
ionization enthalpyelectron gain enthalpy

D
ionization enthalpy

Solution

electron gain enthalpy is the energy released when an electron is gained by isolated gaseous atom to form isolated gaseous anion.
Question 2
1 / -0

Which of the following angle corresponds to sp² hybridisation?

A
120^∘

B
180^∘

C
90^∘

D
109^∘

Solution

sp² hybridisation have triangular planar structure so bond angle is 120^o
Question 3
1 / -0

The electronic configuration of the outer most shell of the most electronegative element is

A
2s²2p⁵

B
4s²4 p⁵

C
5s²5⁵

D
3s²3p⁵

Solution

Most electronegative element is F. Its atomic number is 9. So electronic configuration is 2s²2p⁵.
Question 4
1 / -0

CO is isoelectronic with

A
NO⁻₂ and SnCl₂

B
N₂and SnCl₂

C
NO⁺and N₂

D
SnCl₂and NO⁺

Solution

CO, N2 and NO⁺ have same number of electrons = 14.
Question 5
1 / -0

Which of the following species have the same shape?

A
O₃ and CO₂

B
NO⁻₂ and CCl₄

C
O₃and NO⁻₂

D
CCl₄ and NO⁻₂

Solution

O₃ and NO₂^- have same shape i.e planar.
Question 6
1 / -0

Hydrogen bonds are formed in many compounds e.g., H₂O, HF, NH₃.The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds. The correct decreasing order of the boiling points of above compounds is :

A
H₂O > HF > NH₃

B
NH₃> H₂O > HF

C
NH₃> HF > H₂O

D
HF > H₂O > NH₃

Solution

Boiling point order depend on hydrogen bonding strength which depends on the no. of hydrogen bonds formed and electronegative difference. H₂O and NH₃ forms 4 hydrogen bonds while HF forms only 2. But F is more electronegative than N.
Question 7
1 / -0

Hybridisation involves:

A
addition of an electron pair

B
combination and redistribution of atomic orbitals

C
removal of an electron pair

D
separation of orbitals

Solution

$$\bf{Hint:-}$$ Mixing of different energy orbitals to form new equal energy orbitals known as hybridization.

$$\bf{Explanation-}$$
$$\bullet$$ Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals.

$$\bullet$$ Atomic orbitals with similar energy combine to give degenerated hybrid orbitals.

$$\bullet$$ The number of combining atomic orbitals is equal to the hybridised orbitals.

$$\bullet$$ They involve redistribution of electrons of combined atomic orbital to hybrid orbitals.

$$\bf{Conclusion-}$$ Hence, option $$B$$ is correct.
Question 8
1 / -0

Length of the hydrogen bond range from $$1.75A^{o}$$ to _________.

A
$$2.5A^{o}$$

B
$$2.75A^{o}$$

C
$$2.6A^{o}$$

D
$$3.2A^{o}$$

Solution

As H-bond is a weak bond compared to other covalent bonds, its bond length is a little bit more than covalent bonds. The range of the length of a hydrogen atom is $$1\cdot 75\overset {o}A$$ to $$2\cdot 5\overset {o}A$$
Question 9
1 / -0

The molecular weight of an oxide of Hydrogen is $$34$$. Therefore the number of covalent bonds in its molecule are:

A
$$4$$

B
$$5$$

C
$$2$$

D
$$3$$

Solution

Oxide of hydrogen,
$$H_2O$$ $$H_2O_2$$
M.W-$$16$$ M.W-$$34$$
In $$H_2O_2$$ , three covalent bonds are present.
Question 10
1 / -0

Among the following ions, the $$p\pi - d\pi$$ overlap could be present in:

A
$$NO^{-}_{3}$$

B
$$PO^{3-}_{4}$$

C
$$CO^{2-}_{3}$$

D
$$NO^{-}_{2}$$

Solution

In $$NO_3^-, CO_3^{2-}, NO_2^-$$ only $$p-p$$ orbitals are involved in bonding.
In $$PO_4^{3-}$$, $$3d$$ orbitals of phosphorus are also involved in bonding.
Question 11
1 / -0

Diethyl ether is mostly used in solvent extraction due to all of the following reasons except:

A
Its solvation capacity is very high

B
being inert, it does not react with most organic compound

C
there are two lone pairs in it, therefore it acts as a strong nucleophile

D
its boiling point is low, therefore, it can be easily separated by distillation

Solution

$$C_2H_5OC_2H_5$$
There are two lone pairs in it but being inert, it does not react with most organic compound.
Question 12
1 / -0

The percentage of s- character of the hybrid orbitals in ethane, ethene and ethyne are respectively:

A
$$50, 75, 100$$

B
$$10, 20, 40$$

C
$$25, 33, 50$$

D
$$25, 50, 75$$

Solution

Ethane Ethene Ethyne

$$CH_3 - CH_3$$ $$CH_2 = CH_2$$ $$HC = CH$$
$$SP^3 \,\,\,\,\, SP^3$$ $$SP^2 \,\,\,\,\, SP^2$$ $$SP \,\,\,\,\, SP$$

$$\% S = 25\%$$ $$\% S = 33.33 \%$$ $$\% S = 50 \%$$
$$\% P = 75\%$$ $$\% P = 66.67 \%$$ $$\% P = 50 \%$$

Hence, option C is correct.
Question 13
1 / -0

The number of lone pairs on chlorine atom in $$\mathrm{Cl}\mathrm{O}^{-},$$ $$\mathrm{Cl}\mathrm{O}^{-}_{2},\ \mathrm{Cl}\mathrm{O}^{-}_{3},\ \mathrm{C}\mathrm{l}\mathrm{O}^{-}_{4}$$ ions are:

A
$$0, 1, 2, 3$$

B
$$1, 2, 3, 4$$

C
$$4, 3, 2, 1$$

D
$$3, 2, 1, 0$$

Solution

Lone pairs are in the order: 3, 2, 1 and 0 respectively as shown in the figure

Hence, option $$D$$ is correct..
Question 14
1 / -0

In which of the compound given below contains more than one kind of hybridization ($$sp,sp^{2}$$, $$sp^{3}$$ ) for carbon:

A
$$CH_{3}CH_{2}CH_{2}CH_{3}$$

B
$$CH_{3}-CH=CH-CH_{3}$$

C
$$CH_{2}=CH-CH=CH_{2}$$

D
$$H-C\equiv C-H$$

Solution

(i) $$\overset{sp^3}{CH_3}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_2}-\overset{sp^3}{CH_3}$$

Only $$sp^3$$ hybridized carbon.

(ii) $$\overset{sp^3}{CH_3}-\overset{sp^2}{CH}=\overset{sp^2}{CH}-\overset{sp^3}{CH_3}$$

Both $$sp^2$$ and $$sp^3$$ hybridized carbon.

(iii) $$\overset{sp^2}{CH_2}=\overset{sp^2}{CH}-\overset{sp^2}{CH}=\overset{sp^2}{CH_2}$$

Only $$sp^2$$ hybridized carbon.

(iv) $$H- \underset{sp}{C}\equiv \underset{sp}{C}-H$$

Only $$sp$$ hybridized carbon.

Option B is correct.
Question 15
1 / -0

The number of sigma bonds in $$1$$-butene is:

A
$$8$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution
Question 16
1 / -0

Which of the following does not exhibit hydrogen bonding in liquid phase?

A
$$Phenol$$

B
$$Liq.NH_{3}$$

C
$$Water$$

D
$$Liq.HCl$$

Solution

In liquid $$HCl$$, there are no water molecules. Hydrogen doesn't form H-bond with $$Cl$$ atom. So, liquid $$HCl$$ doesn't exhibit H-bond.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Chemical Bonding and Molecular Structure Test -7

Result

Chemical Bonding and Molecular Structure Test -7

Score

-

Rank

-

SHARING IS CARING

Question 1

Electron affinity

lattice enthalpy

ionization enthalpyelectron gain enthalpy

ionization enthalpy

Solution

Question 2

120∘

180∘

90∘

109∘

Solution

Question 3

2s22p5

4s24 p5

5s255

3s23p5

Solution

Question 4

NO−2 and SnCl2

N2 and SnCl2

NO+and N2

SnCl2 and NO+

Solution

Question 5

O3 and CO2

NO−2 and CCl4

O3 and NO−2

CCl4 and NO−2

Solution

Question 6

H2O > HF > NH3

NH3> H2O > HF

NH3> HF > H2O

HF > H2O > NH3

Solution

Question 7

addition of an electron pair

combination and redistribution of atomic orbitals

removal of an electron pair

separation of orbitals

Solution

Question 8

$$2.5A^{o}$$

$$2.75A^{o}$$

$$2.6A^{o}$$

$$3.2A^{o}$$

Solution

Question 9

$$4$$

$$5$$

$$2$$

$$3$$

Solution

Question 10

$$NO^{-}_{3}$$

$$PO^{3-}_{4}$$

$$CO^{2-}_{3}$$

$$NO^{-}_{2}$$

Solution

Question 11

Its solvation capacity is very high

being inert, it does not react with most organic compound

there are two lone pairs in it, therefore it acts as a strong nucleophile

its boiling point is low, therefore, it can be easily separated by distillation

Solution

Question 12

$$50, 75, 100$$

$$10, 20, 40$$

$$25, 33, 50$$

$$25, 50, 75$$

Solution

120^∘

180^∘

90^∘

109^∘

2s²2p⁵

4s²4 p⁵

5s²5⁵

3s²3p⁵

NO⁻₂ and SnCl₂

N₂and SnCl₂

NO⁺and N₂

SnCl₂and NO⁺

O₃ and CO₂

NO⁻₂ and CCl₄

O₃and NO⁻₂

CCl₄ and NO⁻₂

H₂O > HF > NH₃

NH₃> H₂O > HF

NH₃> HF > H₂O

HF > H₂O > NH₃