Chemical Bonding and Molecular Structure Test 28

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Chemical Bonding and Molecular Structure Test 28

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Weekly Quiz Competition

Question 1
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If $$ \mu_1$$ and $$\mu_2 $$ are the two dipole moments in the molecule with an angle $$ \theta $$ between them, then the resultant dipole moment will be:

A
$$ \mu = \mu_1^2 + \mu_2^2 + 2\mu_1 \mu_2 cos \theta $$

B
$$ \mu = \mu_1^2 + \mu_2^2 -2\mu_1\mu_2 cos \theta $$

C
$$ \mu = (\mu_1^2 + \mu_2^2 + 2\mu_1\mu_1 cos \theta)^{ \frac{1}{2}} $$

D
$$ \mu = (\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 cos \theta)^{\frac{1}{3}} $$

Solution

Let AB & AC are two polar bonds inclined at angle $$\theta$$, their dipole moments are $$\mu_1$$ and $$\mu_2$$. The resultant dipole moment may be calculated using,

$$\mu=(\mu_1^2+\mu_2^2+2\mu_1\mu_2cos\theta)^{1/2}$$
Question 2
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The hybridization of I in ICl$$_2^+$$ is :

A
$$sp$$

B
$$ sp^2 $$

C
$$ sp^3 $$

D
$$ dsp^2 $$

Solution
Question 3
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Which of the following halides has different bond lengths?

A
$$ BCl_3 $$

B
$$ CCl_4 $$

C
$$ BeCl_2 $$

D
$$ PCl_5 $$

Solution
Question 4
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The hybridisation of oxygen atom in $$H_2O_2$$ is:

A
$$sp^3d$$

B
$$sp$$

C
$$sp^2$$

D
$$sp^3$$

E
$$sp^3d^2$$

Solution

each oxygen atom in peroxide is $$sp^3$$ hybridized. There is a covalent $$sp^3-sp^3$$ bond between the two oxygens. Each oxygen atom is covalently linked to its respective hydrogen via.

$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D)}$$
Question 5
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Which of the following does not have $$S-S$$ linkage?

A
$$S_{2}O_{8}^{2-}$$

B
$$S_{2}O_{6}^{2-}$$

C
$$S_{2}O_{5}^{2-}$$

D
$$S_{2}O_{3}^{2-}$$

Solution
Question 6
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The correct order of $$O-O$$ bond length in $$O_{2}, H_{2}O_{2}$$ and $$O_{3}$$ is:

A
$$O_{3}>H_{2}O_{2}>O_{2}$$

B
$$O_{2}>H_{2}O_{2}>O_{3}$$

C
$$O_{2}>O_{3}>H_{2}O_{2}$$

D
$$H_{2}O_{2}>O_{3}>O_{2}$$

Solution
Question 7
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Which of the following has the highest value of dipole moment?

A
HCl

B
HF

C
HI

D
HBr

Solution

We know dipole moment generally vary with separated charge and bond distance. The electronegativity order in halogens $$F>Cl>Br>I$$, but the bond distance order between $$HX\ (X=F,Cl,Br,I)$$ is $$HI>HBr>HCl>HF$$.

In this case, bond distance and electronegativity are contradictory factors. But we also know 'F' is highly electronegative in the periodic table for which' F' can attract bond pair electron in $$H-F$$ more than iodine in $$H-I$$.

So in $$H-F$$ bond 'H' becomes highly electropositive and 'F' becomes highly electronegative, in this case, the charge factor is more effective than bond distance.
Question 8
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Which of the following has $$\pi-d\pi$$ bonding?

A
$$NO_{3}^{-}$$

B
$$SO_{3}^{2-}$$

C
$$BO_{3}^{2-}$$

D
$$CO_{3}^{2-}$$

Solution
Question 9
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In $$OF_{2}$$ molecule, the total number of bond pairs and line pairs of electron present respectively are:

A
$$2,6$$

B
$$2,8$$

C
$$2,10$$

D
$$2,9$$

Solution
Question 10
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The geometry of $$H_{2}S$$ and its dipole moment are:

A
angular and non-zero

B
angular and zero

C
linear and zero

D
linear and non-zero

Solution