Polar molecule among the following is:

N_{2}

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H_{2} O

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C H_{4}

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C O_{2}

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K C l

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Solution

The correct option is B $H_{2} O$
Water is polar molecule
Electronegativity Difference Type of Bond Formed

0.0 to 0.2 nonpolar covalent
0.3 to 1.4 polar covalent
> 1.5 ionic
The electronegativity difference provides a way of predicting the kind of bond that will form between two elements. If the difference is between $0.5 - 1.4$ then the bond formed is polar covalent.
$N_{2}$ Electron negativity difference $= 3.0 - 3.0 = 0$ Bond $=$ Non-polar covalent
$H_{2} O$ Electron negativity difference $= 2.1 - 3.5 = 1.4$ Bond $=$ Polar covalent
$C H_{4}$ Electron negativity difference $= 2.1 - 2.5 = 0.4$ Bond $=$ Non-polar covalent
$C O_{2}$ Electron negativity difference $= 2.5 - 3.5 = 1.0$ Bond $=$ Non-polar covalent
$K C l$ Electron negativity difference $= 0.8 - 3.0 = 2.2$ Bond $=$ Ionic

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Similar questions

{B F}_{3}, {C O}_{2}, {S O}_{2}, {P C I}_{5}, {C I F}_{3}, {N H}_{3}, {C H}_{4}

How many covalent bonds are present in the following compounds?

(i) $C O_{2}$

(ii) $H_{2} O$

(iii) $C H_{4}$

(iv) $N_{2}$

(v) $N H_{3}$

Q. Complete the following reaction:

K_{2} C O_{3} + H C l \to

C O_{2} + H_{2} O + H_{2}

C H_{4} + H_{2} O

C H_{4} (g) + H_{2} O

3 F e + 4 H_{2} O (s t e a m)

C H_{4} + O_{2} \to C O_{2} + H_{2} O

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Electronegativity Difference	Type of Bond Formed
0.0 to 0.2	nonpolar covalent
0.3 to 1.4	polar covalent
> 1.5	ionic