Chemical Bonding and Molecular Structure Test 32

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Chemical Bonding and Molecular Structure Test 32

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Weekly Quiz Competition

Question 1
1 / -0

Number of S - S bond is $$H_{2}S_{n}O_{6}$$ :

A
n

B
(n - 1)

C
(n - 2)

D
(n + 1)

Solution
Question 2
1 / -0

Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false :
(i) The order of the repulsion between different pair of electrons is $$l_{p} - l_{p} > l_{p}- b_{p} > b_{p}- b_{p}$$
(ii) In general, as the number of lone pair of electrons on central atom increases, value of bond angle from normal bond angle also increases
(iii) The number of lone pair on O in $$H_{2}O$$ is 2 while on N in $$NH_{3}$$ is 1
(iv) The structures of xenon fluorides and xenon oxyfluorides could not be explained on the basis of VSEPR theory

A
TTTF

B
TFTF

C
TFTT

D
TFFF

Solution

correct statements are
(ii) (F) In general, as the number of lone pair of electrons on the central atom increases, the value of bond angle from normal bond angle decreases due to $$lp - lp>lp - bp$$

(iv) (F) Structures of xenon fluorides and xenon oxy fluoride are explained on the basis of VSEPR theory.
In $$SOBr_{2}$$ S-O bond has a maximum bond length in comparison to S-O bond lengths in $$SOF_{2}$$ and $$SOCl_{2}$$ because in $$SOBr_{2}$$ S-O bond has been formed by hybrid orbital containing less s-character.
Question 3
1 / -0

A : tetracyanomethane
B : carbondioxide
C : benzene
D : 1, 3-butadiene
Ratio of $$\sigma$$ and $$\pi$$ bonds is in order :

A
A = B < C < D

B
A = B < D < C

C
A = B = C= D

D
C < D < A < B

Solution
Question 4
1 / -0

In which of the following cases C-C bond length will be highest?

A
$$CH_3-CF_3$$

B
$$FCH_2-CH_2F$$

C
$$F_2CH-CHF_2$$

D
$$CF_3-CF_3$$

Solution

According to Bent's Rule more electronegative substituents attached to hybrid orbital that contains more p-character. Hence, more p-character, longer will be bond length.
Question 5
1 / -0

Directions For Questions

Polar covalent molecules exhibit dipole moment. Dipole moment is equal to the product of charge separation, $$q$$ and the bond length $$d$$ for the bond. Unit of dipole moment is debye. One debye is equal to $$10^{-18}\ esu\ cm$$.

Dipole moment is a vector quantity. It has both magnitude and direction. Hence, dipole moment of a molecule depends upon the relative orientation of bond dipoles, but not on the polarity of bonds alone. A symmetrical structure shows zero dipole moment. Thus, dipole moment helps to predict the geometry of molecules. Dipole moment values can be used to distinguish between cis-and trans-isomers; ortho-, meta- and para-forms of a substance, etc.

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A diatomic molecule has a dipole moment of 1.2 D. If the bond length is $$1.0 \times 10^{-8} \ cm$$, what fraction of charge does exist each atom?

A
0.1

B
0.2

C
0.25

D
0.3

Solution

Fraction of charge = $$\frac{1.2 \times 10^{-18}}{4.8 \times 10^{-10}\times 10^{-8}} = 0.25$$
Question 6
1 / -0

How many sigma bonds are present in a molecule of diethyl either $$C_{2}H_{5}OC_{2}H_{5}$$ ?

A
14

B
12

C
8

D
16

Solution
Question 7
1 / -0

Which of the following statements is incorrect for sigma and $$\pi$$ - bonds formed between two carbon atoms?

A
Sigma - bond is stronger than a $$\pi$$- bond

B
Bond energies of sigma and $$\pi$$- bonds are of the order of 264 kJ/mol and 347 kJ/mol

C
Free rotation of surrounding atoms about a sigma-bond is allowed but not in case of a $$\pi$$- bond

D
Sigma- bond determines the direction between carbon atoms but a $$\pi$$- bond has no primary effect in this regard

Solution
Question 8
1 / -0

Which of the following has no lone pair of electrons?

A
$$NO$$

B
$$CO$$

C
$$NH_3$$

D
$$O_2$$

Solution

Nitric oxide has no lone pair. It makes two covalent bonds and one coordination bond with oxygen so it uses it's four electron in bonding.
It has only one odd electron.
Hence option (A) is correct.
Question 9
1 / -0

How many sigma and pi bonds are present in toluene ?

A
$$10\sigma$$ and $$3\pi$$ bonds

B
$$12\sigma$$ and $$3\pi$$ bonds

C
$$15\sigma$$ and $$3\pi$$ bonds

D
$$6\sigma$$ and $$3\pi$$ bonds

Solution

Toluene is $$C_{6}H_{5}CH_{3}.$$

No. of $$\sigma$$ bonds is $$15.$$

No. of $$\pi$$ bonds is $$3.$$
Question 10
1 / -0

In $${NO_{3}}^{-}$$ ion, number of bond pair and lone pair of electrons on nitrogen atom are :

A
$$2,2$$

B
$$3,2$$

C
$$1,3$$

D
$$4,0$$

Solution

According to the VSEPR theory, $$NO_3^-$$ is sp² hybridized with 3 sigma bond pairs, 1 pi bond pair and no lone pair on the central atom as shown in the figure. Its shape is trigonal planar.
Hence, option D is correct.