When an atom forms a positive ion (by losing electrons). will the radius of an ion change from the radius of an atom? explain with the example of sodium atom (atomic no. 11) and sodium ion.

Radius of the positive ion will be smaller than the radius of its parent atom.

Sodium Na has atomic number 11. A neutral Na atom has 11 electrons. It has

Na tends to lose one electron. It would form an Na⁺ ion. The ion is positive and has ten electrons. It has

Three of the main shells of Na contain electrons. Only two main shells contain electrons in Na⁺. The energy of the second main shell is higher than the first. It is more distant from the nucleus than the first electron shell. Similarly, the third main shell is further away from the nucleus than the second.

The outermost shell in Na is the third main shell. The outermost shell in Na⁺ is the second main shell. As a result, the outermost electrons in Na are further away from the nucleus that electrons in Na⁺. As a result, the radius of Na appears to be larger than that of Na⁺.

When an atom loses electrons to form a positive cation, it tends to lose all electron in its outermost shell (the valence shell). That will give the ion an octet, which is more stable than having one or two electrons left in the valence shell. The rule on Na shall still apply. The positive ion will have one less main shell; its radius shall be smaller than that of the parent atom.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml