# 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 YX_{2}

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

**Solution:**

**I.** _{12}Y has electron configuration: 1s^{2}2s^{2}2p^{6}3s^{2}

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

**II.** X in YX_{2} 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:

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}

**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;

1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{1}

1+1+3+1+3+1+5+1= Filled orbitals and 4p^{1} 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 _{7}N is; Y_{3}N_{2}

**III.** X^{+} and Y^{+2}are isoelectronic with same noble gas

**Solution:**

**I.** Since increase in the first IE_{1} to second IE_{2} is higher than others, oxidation number of X is 1. I is true.

**II.** Since increase in the second IE_{2}to third IE_{3} is higher than others, oxidation number of Y is 2. In compounds Y takes +2 value. N has electron configuration;

1s^{2}2s^{2}2p^{3}

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

Y_{3}N_{2} 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.** 1s^{2}2s^{2}

**II.** 1s^{2}2s^{2}2p^{2}

**III.** 1s^{2}2s^{2}2p^{4}

**IV.** 1s^{2}2s^{2}2p^{5}

**V.** 1s^{2}2s^{2}2p^{6}3s^{2}

**Solution:**

**I.** 1s^{2}2s^{2} is in 2. period and II A group

**II.** 1s^{2}2s^{2}2p^{2} is in 2. period and IV A group

**III.** 1s^{2}2s^{2}2p^{4} is in 2. period and VI A group

**IV.** 1s^{2}2s^{2}2p^{5} is in 2. period and VII A group

**V.** 1s^{2}2s^{2}2p^{6}3s^{2} 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.