Chemical Bonding and Molecular Structure Test 33

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Chemical Bonding and Molecular Structure Test 33

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Weekly Quiz Competition

Question 1
1 / -0

The number of $$sp^{2}-sp^{2} \sigma$$ bond in given compound is:

A
$$4$$

B
$$5$$

C
$$3$$

D
$$6$$

Solution
Question 2
1 / -0

The ratio of number of lone pairs in the central atoms of $$NH_3$$, $$H_2O$$ and $$XeF_{2}$$ is:

A
$$3:2:1$$

B
$$2:1:3$$

C
$$1:2:3$$

D
$$2:3:1$$

Solution

$$NH_3$$: 3 bond pairs and one lone pair because, out of 5 valence electrons of nitrogen, only 2 electrons are left to form a lone pair.

$$H_2O$$: 2 bond pairs and two lone pairs because, out of 6 valence electrons of oxygen, 4 electrons are left to form 2 lone pairs.

$$XeF_2$$: 2 bond pairs and three lone pairs because out of 8 valence electrons of xenon, 6 electrons are left to form 3 lone pairs.

So, the ratio of lone pairs is $$1:2:3$$.
Question 3
1 / -0

Which one of the following pairs of molecules will have a permanent dipole moment for both members?

A
$$SiF_{4}$$ and $$NO_{2}$$

B
$$NO_{2}$$ and $$SF_6$$

C
$$NO_{2}$$ and $$O_{3}$$

D
$$SiF_{4}$$ and $$CO_{2}$$

Solution

$$SF_6$$, $$SiF_4$$, $$CO_2$$ are perfectly symmetrical molecules. Here the dipole in all the directions cancel out and the resutant dipole is zero. Resonance takes place in both $$NO_2$$ and $$O_3$$. The partial charges in both the molecules cause individual dipoles to exist which in turn creates a permanent dipole moment.
Question 4
1 / -0

In the analysis of an organic compound, it was found that it contains 7 carbon atoms, there are two C = C bonds and one $$C\equiv C$$ bond. This compound is a hydrocarbon. Hydrocarbons have only carbon and hydrogen elements. On structural analysis it was found that it is covalent in nature and expected structure is given above. The ratio between the pure and hybrid orbitals is?

A
$$7:10$$

B
$$8:7$$

C
$$5:9$$

D
$$1:2$$

Solution

In $$\displaystyle sp, sp^2 $$ and $$\displaystyle sp^3 $$ hybridization of $$C$$, the number of hybrid orbitals formed are $$2,3$$ and $$4 $$ respectively and the number of pure orbitals are $$2,1$$ and $$0$$ respectively.
In this compound, $$2$$ $$\displaystyle sp $$ hybridized, $$4$$ $$\displaystyle sp^2$$ hybridized and $$1$$ $$\displaystyle sp^3 $$ hybridized $$C$$ atom is present. The number of pure orbitals in these atoms are $$\displaystyle 4+4+0=8$$
Also there are $$6\: H$$ atoms which contribute $$6$$ pure orbitals. Hence,the total number of pure orbitals are $$\displaystyle 8+6=14 $$
The number of hybrid orbitals are $$\displaystyle 4+12+4+0=20 $$
Hence, the ratio of pure orbitals to hybrid orbitals is $$\displaystyle 14: 20 $$ or $$\displaystyle 7:10$$
Question 5
1 / -0

How many lone pairs of electrons are in resonance of the given compound?

A
$$4$$

B
$$2$$

C
$$3$$

D
$$6$$

Solution

Two (one of $$N$$-atom with double bond and one of $$O$$-atom with double bond) lone pairs of electrons are in resonance.
Question 6
1 / -0

Which of the following compounds has zero dipole moment?

A

B

C

D

Solution

Only linear groups and atoms, attached at para positions to each other will have zero dipole moment.
Question 7
1 / -0

Which of the following types of bonds are present in $$CuSO_4\cdot 5H_2O$$?
I: Electrovalent II: Covalent III: Coordinate

A
I, II

B
II, III

C
I, III

D
I, II, III

Solution

Here, $$CuSO_{4}$$ is ionic compound. When it is hydrated, it forms hydrogen bonding with one of the water molecule and coordinate bonds with the other 4 water molecules. It has covalent bonding in $$SO_4^{2-}$$.

∴CuSO4.5H2O contains all the three types of bond
i. Electrovalent (Ionic interaction between $$Cu^{2+}$$ & $$SO_4^{2-}$$)
ii.Covalent (with $$SO_4^{2-}$$)
iii.Co ordinate (H2O molecules to $$Cu^{2+}$$)
Question 8
1 / -0

In which of the following pairs, indicated bond is of greater strength?

A
I

B
II

C
Both are equal

D
None of the above

Solution

In compound (I), the double bond is tetra substituted. In compound (II), the double bond is trisubstituted.
In tetra substituted double bond, due to steric hindrance, the planar geometry is slightly twisted. Due to this the overlap of the p orbitals is not proper.
Hence, the double bond is slightly weaker in compound (I) than in compound (II).
Question 9
1 / -0

In $$XeF_2, XeF_4$$ and $$XeF_6$$, the number of lone pairs of $$Xe$$ is respectively:

A
$$2, 3, 1$$

B
$$1, 2, 3$$

C
$$4, 1, 2$$

D
$$3, 2, 1$$

Solution

number of lone pairs of $$Xe$$ are $$3,2$$ and $$1$$.
Question 10
1 / -0

The total number of valence electrons in $$4.2\ gm$$ of $${N}_{3}^{-}$$ ion are :

A
$$2.2\ N_A$$

B
$$4.2\ N_A$$

C
$$1.6\ N_A$$

D
$$3.2\ N_A$$

Solution

As each molecule of $${N}_3^-$$ has $$16$$ valence electrons.

Therefore, one mole will have $$16$$ mole of electrons.

Number of valence electrons = No. of moles $$\times N_A\times $$ valence electrons.
Here, $$N_A$$ is the Avagadro number.

$$= \dfrac{4.2}{42}\times N_A\times 16$$

$$= 1.6N_A$$

Hence, option C is correct.