Based on the trends in first ionization energies which representative element (G

Question

Based on the trends in first ionization energies which representative element (Group IA and VIIA) would you expect to have the lowest ionization energy? The highest? Explain your choices. The bromine monofluoride molecule, BrF has a bond length of 1.78 A. Calculate the dipole moment, in Debyes, that results if the charges on Br and F were I^+ and I^+, respectively. (1 Debye (D) = 3.34 times 10^-30 C-m, I e = 1.6 times 10^-19 C). IA = 1 times 10^-10 m) (Include the equation, not just numbers) If the experimentally measured dipole moment is 1, 42 D, what is the magnitude (in units of e) of the charges on Br and F. (Include the equation, not just numbers) Complete and balance the following equations:

Explanation / Answer

5. Ionization energy decreases down the group as the number of shells are increasing, the distance between valence shell electron and nuclear reduces and attraction of +vely charged e- to +ve nucelus decreases. So lowest ionization energy would be for francium in Group IA.

Ionization energy increases from left to right in periodic table. So highest ionization energy would be for fluorine in Group VIIA.

6.

(a) dipole moment = qr

q = charge

r = distance

feeding values

dipole moment (u) = 1.6 x 10^-19 x 1.78 x 10^-10/3.34 x 10^-30

                              = 8.527 D

(b) charge q = u/r

                    = 1.42 x 3.34 x 10^-30/1.78 x x10^-10 x 1.602 x 10^-19

                    = 0.166 e

7. Balanced equations

(a) BaO(s) + 2HNO3(aq) --> Ba(NO3)2(aq) + H2O(l)

(b) Mg(s) + Cl2(g) --> MgCl2(aq)

(c) Cs(s) + 2H2O(l) --> Cs(OH)2(s) + H2(g)

(d) SrO(s) + H2O(l) --> Sr(OH)2(s)

(e) CO2(g) + H2O(l) --> H2CO3(aq)

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