Periodic Table Exam 1 and Problem Solutions

Periodic Table Exam 1 and Problem Solutions

1. X, Y and Z are in same period. Using given information below, find relation between metallic properties of them.

I. Atomic number of Y is 12

II. Formula of compound produced by X and Y is YX₂

III. Z^-2 and X^- have equal number of electrons

Solution:

I. ₁₂Y has electron configuration: 1s²2s²2p⁶3s²

Y is in 3. period and II A group. Thus, it has +2 value in compounds.

II. X in YX₂ compound is in VII A group and have -1 value in compound. Atomic number of X is 17.

III. Since their number of electrons are equal, Z is in 3. period and VI A group. Relation between metallic properties of elements;

Y>Z>X

Metallic property decreases when we go from left to right in same period.

2. Which ones of the following statements are true for element having atomic number 34?

I. It is in p block

II. It is nonmetal

III. Its valence electrons are all in p orbitals

Solution:

Electron configuration of element is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴

I. Element is in p block, since last electron is in p orbital. I is true.

II. In outer shell (4) it has 4+2=6 electrons. So it is nonmetal. II is true.

III. 4 valence electrons of element are in p orbital and 2 of them are in s orbital. So, III is false.

3. If an element has 15 filled and 1 half filled orbitals, which one of the following statements is false?

I. Atomic number of element is 31

II. It is in p block of periodic table

III. It is transition metal

IV. It is in 4. period

V. It is in III A group

Solution:

We write orbitals on electron configuration;

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p¹

1+1+3+1+3+1+5+1= Filled orbitals and 4p¹ is one half filled orbital

I. Number of electrons in neutral atom is equal to atomic number. Thus, if we sum number of electrons in given orbitals, we find atomic mass 31. I is true.

II. Since last electron of element is in p orbital, it is  p block element. II is true

III. Since its last electrons are in p orbitals, it is not transition metal. (If last electrons are in d orbitals, element becomes transition metal) III is false.

IV. Outer shell of element is 4, so it is i 4. period. IV is true.

V. It has 3 valence electrons; 2 in 4s and 1 in 4p, so it is in III A group. V is true.

4. Ionization energies of X and Y are given in the table below.

Using data given in the table, find which ones of the following statements are definitely true?

I. Oxidation number of X is 1

II. Compound formed by Y and ₇N is; Y₃N₂

III. X⁺ and Y⁺²are isoelectronic with same noble gas

Solution:

I. Since increase in the first IE₁ to second IE₂ is higher than others, oxidation number of X is 1. I is true.

II. Since increase in the second IE₂to third IE₃ is higher than others, oxidation number of Y is 2. In compounds Y takes +2 value. N has electron configuration;

1s²2s²2p³

N accepts 3 electrons and has value -3 in compounds. So, N and Y form following compound;

Y₃N₂ II is true

III. X lose 1 electron and Y 2 loses 2 electrons to have noble gas electron configuration. But, we can not definitely say they are isoelectronic.

5. Which one of the following elements has lowest first ionization energy?

I. 1s²2s²

II. 1s²2s²2p²

III. 1s²2s²2p⁴

IV. 1s²2s²2p⁵

V. 1s²2s²2p⁶3s²

Solution:

I. 1s²2s² is in 2. period and II A group

II. 1s²2s²2p² is in 2. period and IV A group

III. 1s²2s²2p⁴ is in 2. period and VI A group

IV. 1s²2s²2p⁵ is in 2. period and VII A group

V. 1s²2s²2p⁶3s² is in 3. period and II A group

In a periodic table, ionization energy decreases from top to bottom and right to left. Thus, element given in V has lowest first ionization energy.