Chemical Bonding and Molecular Structure's Question 56 | Chemistry XI - Part 1 | NCERT Exemplar Questions

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Subjective

Match the shape of molecules in Column I with the type of hybridisation in Column II.

	Column I		Column II
A.	Tetrahedral	1.	sp$$^2$$
B.	Trigonal	2.	sp
D.	Linear	4.	sp$$^3$$

Explanation

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

A. Tetrahedral shape $-s p^3$ hybridisation

B. Trigonal shape $-s p^2$ hybridisation

C. Linear shape - sp hybridisation

MCQ (Single Correct Answer)

Assertion (A) Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.

Reason (R) This is because sodium and chloride ions acquire octet in sodium chloride formation.

A and R both are correct and R is the correct explanation of A

A and R both are correct, but R is not the correct explanation of A

$A$ is true, but $R$ is false

A and R both are false

MCQ (Single Correct Answer)

Assertion (A) Though the central atom of both $\mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ molecules are $s p^3$ hybridised, yet $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle is greater than that of $\mathrm{H}-\mathrm{O}-\mathrm{H}$.

Reason (R) This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.

$A$ and $R$ both are correct and $R$ is the correct explanation of $A$

$A$ and $R$ both are correct but $R$ is not the correct explanation of $A$

$A$ is true, but $R$ is false

A and R both are false

MCQ (Single Correct Answer)

Assertion (A) Among the two $0-\mathrm{H}$ bonds in $\mathrm{H}_2 \mathrm{O}$ molecule, the energy required to break the first $0-\mathrm{H}$ bond and the other $0-\mathrm{H}$ bond is the same.

Reason (R) This is because the electronic environment around oxygen is the same even after breakage of one $0-\mathrm{H}$ bond.

$A$ and $R$ both are correct and $R$ is the correct explanation of $A$

A and R both are correct, but R is not the correct explanation of A

$A$ is true, but $R$ is false

$A$ and $R$ both are false

Subjective

(a) Discuss the significance/applications of dipole moment.

(b) Represent diagrammatically the bond moments and the resultant dipole moment in $\mathrm{CO}_2, \mathrm{NF}_3$ and $\mathrm{CHCl}_3$.

Explanation

(a) The applications of dipole moment are

(i) The dipole moment helps to predict whether a molecule is polar or non-polar. As $\mu=q \times d$, greater is the magnitude of dipole moment, higher will be the polarity of the bond. For non-polar molecules, the dipole moment is zero.

(ii) The percentage of ionic character can be calculated as

$$\text { Percentage of ionic character }=\frac{\mu_{\text {observed }}}{\mu_{\text {ionic }}} \times 100$$

(iii) Symmetrical molecules have zero dipole moment although they have two or more polar bonds (in determination of symmetry).

(iv) It helps to distinguish between cis and trans isomers. Usually cis-isomer has higher dipole moment than trans isomer.

(v) It helps to distinguish between ortho, meta and para isomers. Dipole moment of para isomer is zero. Dipole moment of ortho isomer is greater than that of meta isomer.

(b)

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