To get this it helps to know the electronegativity numbers of the elements but it isn't required. You just need to know that Fluorine is the most electronegative element and that the farther away from Fluorine you are on the periodic table, the less electronegative you get. The one exception to this rule is hydrogen with actually has an electronegativity of 2.1 while lithium has one of 1.0. Also the higher difference in electronegativity between two atoms the more polar the bond is.
Now to start the question. H-Br could be a contender since H has an electronegativity number of 2.1 and Br is relatively close to Fluorine so we'll put that one aside for now. H-Cl knocks out A because both bonds have H but one bond has Br and the other has Cl. Cl is closer to Fluorine than Br so answer B is the contender now. For answer C, I and Br are too close to have a higher electronegativity difference than H-Cl so that one isn't it. Finally for answer D, I is much closer to Cl than H is so the electronegativity difference is much less, making your answer B.
Explanation:
( a )
<u>The four types of spread spectrum techniques are as follows -</u>
1. Direct sequence spread spectrum .
2. frequency hopping spread spectrum .
3. chirp spread spectrum .
4. time hopping spread spectrum .
( b )
<u>The Direct sequence spread spectrum was devised for eavesdropping in the military .</u>
In the field of telecommunications , the Direct sequence spread spectrum , it is the technique of spread spectrum modulation which is used to reduce the overall inference of the signal .
