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I find an answer by asking Google "Why does atomic radius decrease across a period?"...
I find an answer by asking Google "Why does atomic radius decrease across a period?"
And here's the answer :
Atomic radius decreases because you are adding protons to the nucleus as you move from the left to the right of the period. These additional protons pull with more force on the electrons outside the nucleus, drawing them all in closer, and making the radius of the atom smaller.
And my new question is, Why would these additional protons pull with more force?
2 Answers | add yours
Atomic radius decreases because number of protons are increasing in the the nucleus as we move from the left to the right in the same period. Since number of protons are equals to number of electrons in an atom.Thus number of electrons also inceases ,out side of nucleuos .There are force of attraction between protons and electrons , so they pull with more force towards each other .So radius of atom gets decreased. Thus size of atom get reduced or the atom becomes smaller in size. Electorn are negatively charged and protons are positively charge so ,they attract each other or there are force of attraction between them.
Posted by pramodpandey on May 13, 2013 at 8:35 AM (Answer #1)
All the positive charge (and mass) of an atom is seated in the nucleus, the negatively charged electrons revolve around it in definite shells. Attraction between the positively charged nucleus and negatively charged electrons is balanced by the centrifugal force of repulsion between the revolving electron/s at a particular distance (plus other repulsive interactions if present, for example, e- – e-repulsions, in multi-electron systems). Thus the two factors that govern the size of an atom are the nuclear charge and the size of the outermost shell.
In general, the atomic radii decrease with increasing atomic number in a period. This is because, while travelling from left to right along a period, the nuclear charge (as well as the total number of electrons) increases progressively by one unit, but the additional electron goes to the same principal shell. Thus the added electron cannot ‘screen’ the added nuclear charge appreciably. As a result, the attractive force (given by F = (ze × e)/(r˄2)) and consequently, the atomic radius goes on decreasing. This is evidenced in the observed atomic radii values of Li to F in the second period.
Posted by llltkl on May 13, 2013 at 4:53 PM (Answer #2)
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