Question

Which bond is more polar in the following pair of molecules?
a.\[{{H}_{3}}C-H\text{ },\text{ }{{H}_{3}}C-Br\]
b.\[{{H}_{3}}C-N{{H}_{2}}\text{ },\text{ }{{H}_{3}}C-OH\]
c.\[{{H}_{3}}C-OH\text{ },\text{ }{{H}_{3}}C-SH\]
A.
a.${{H}_{3}}C-H$
b.\[{{H}_{3}}C-OH\]
c.\[{{H}_{3}}C-OH\]
B.
a.\[{{H}_{3}}C-Br\]
b.\[{{H}_{3}}C-OH\]
c.\[{{H}_{3}}C-OH\]
C.
a.\[{{H}_{3}}C-H\]
b.\[{{H}_{3}}C-N{{H}_{2}}\]
c.\[{{H}_{3}}C-OH\]
D.
a.\[{{H}_{3}}C-H\]
b.\[{{H}_{3}}C-N{{H}_{2}}\]
c.\[{{H}_{3}}C-SH\]

Answer 1

Hint: For this we have to first know about polarity, polar molecule which attracts another molecule toward itself along which factor polarity of any molecule effect as well as we have to compare polarity of molecule.

Complete step-by-step answer:Let's start with the definition of polar molecule. Polar Molecules: A diatomic molecule which is bounded by a polar covalent bond is called a polar molecule. Polar means compound is having magnetic poles between elements, due electrical charges within molecules.

Polarity: The formation of poles between elements of a molecule is called Polarity. Factors which affect polarity:
Presence of lone pair: When there is presence of lone pairs in a molecule , the molecule generates a pole due to larger electron density on the lone pair. So polarity develops as well as electron density is higher on the lone pair side.
Presence of electron withdrawing group: When there’s presence of an electron withdrawing the group of molecule which generate pole, due to a larger ability to attract electrons of an electron withdrawing group. So, polarity develops along with electron density is higher on the electron withdrawing group side. Polarity is a vector quantity, if a molecule has geometrical symmetry in structure then they will cancel each other polarity with become non-polar and if the molecule is asymmetrical the bit will be polar.
From other, \[{{H}_{3}}C-H\] and \[{{H}_{3}}C-Br\] , \[{{H}_{3}}C-Br\] bond is more polar since $Br$ is more $EN$ than $H$ . Similarly, \[{{H}_{3}}C-N{{H}_{2}}\] and \[{{H}_{3}}C-OH\] , \[{{H}_{3}}C-OH\] bond is more polar since $O$ is more Electronegative than $N$ . Whereas, \[{{H}_{3}}C-OH\]and \[{{H}_{3}}C-SH\] , \[{{H}_{3}}C-OH\] bond is more polar since $O$ is more Electronegative than $S$
Hence, this question has multiple correct options, correct option is (b) i.e. \[{{H}_{3}}C-Br\] , \[{{H}_{3}}C-OH\] , \[{{H}_{3}}C-OH\]

Note: While finding, a molecule has polarity or not we have to think about geometry of molecule as well as if we have to compare polarity of two molecules, we have to keep in mind lone pair as well as electron withdrawing group both.