Choose from the options given to correctly complete the following statements. Pr

Question

Choose from the options given to correctly complete the following statements. Provide a number in the boxes provided for the following questions How many neutral elements have a valence shell electron configuration ending in s^2p^3? How many electrons are in the ion As^3+? What is the oxidation number for sulfur in S_2O_6^2? Show the electron arrangement by filling in the orbital diagram below for phosphorous. Be sure to indicate the name of the orbitals (e.g., 1s, etc) on the lines below the boxes. Write the letter for the compound name in the box next to the appropriate formula. CuCI_2 dinitrogen oxide dinitrogen pentoxide Na_2SO_3 copper(II) chloride sodium sulfate N_2O_5 nitrogen pentoxide sodium sulfite copper(I) chloride sodium sulfide

Explanation / Answer

1. The enthalpy of fusion refers to converting a solid to liquid by application of heat energy while that of vaporization refer to converting a liquid to gas. Hence converting a liqud to gas demands more heat than converting a solid to liquid. Hence enthalpy of vaporiztino will be more than heat of fusion of substance.

2. nitrogen is smaller molecule than P. As a result, the hydrogen and nitrogen bind by strong dipole forces. in case of PH3 the bonds are less strong compared to NH3.

3. since 1/wave length = RH*(1/n12-1/n22), R is Rydberg constant

for transition from n=3 and n=2

1/wave length = RH*(1/22-1/32)= 0.14RH and wave length= 1/(0.14*RH)= 7.14/RH

for the transition from n=4 to n = 2, 1/wave length = RH*(1/4-1/16)= 0.1875RH

wave length = 1/(0.1875*RH)= 5.33RH

So wave length for the transition from 3 to 2 is more than 4 to 2.

4. pH= 10 meand pOH= 4 , [OH-] =10-4 = 0.0001, for pH= 7, pOH= 14-7= 7, [H+] =10-7

[OH[ concentration for pH 10 is greater than OH concentratino for pH 7.

5. S2P3 refer to group 5 elements. there are 5 elements in group 5. they are nitrogen, arsenic, antimony and bismuth

6. electronic configuration is [Ar] 3d10 4s2 4p3 when 3 electrons remove, it becomes [Ar] 3d10 4s2

7. for S2O6-2, overall is -2

for O, it is -2

hence 2* oxdiation state of S -12= -2

oxidation state of S= (-2+12)/2 =5

8, P electronic configuratino is 1s2 2s2 2p6 3s2 3p3

9. Cucl2 is copper (ii) chloride, Na2SO3 is sodium sulfite and N2O5 is dinitrogen pentaxide

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