Chemistry, 07.11.2019 01:31, SmokeyRN

Consider the following element combinations. classify the bonds formed between each pair as ionic, polar covalent, or nonpolar covalent qualitatively based solely on each element's position on the periodic table? do not conduct calculations. 1. rb and i 2. br and br 3. s and i 4. n and n 5. o and f 6. cs and cl 7. ca and o 8. p and i

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answered: Haven263

1. ${\text{Rb}} - {\text{I}}$ is an $\boxed{{\text{ionic bond}}}$ .

2. ${\text{Br}} - {\text{Br}}$ is a $\boxed{{\text{non - polar covalent bond}}}$ .

3. ${\text{N}} - {\text{N}}$ is a $\boxed{{\text{non - polar covalent bond}}}$ .

4. ${\text{O}} - {\text{F}}$ is a $\boxed{{\text{non - polar covalent}}}$ .

5. ${\text{S}} - {\text{I}}$ is a $\boxed{{\text{non - polar covalent}}}$ .

6. ${\text{Cs}} - {\text{Cl}}$ is an $\boxed{{\text{ionic bond}}}$ .

7. ${\text{Ca}} - {\text{O}}$ is a $\boxed{{\text{polar ionic compound}}}$ .

8. ${\text{P}} - {\text{I}}$ is a $\boxed{{\text{polar and covalent bond}}}$ .

Further explanation:

Ionic compound:

An ionic bond involves complete transference of electrons. The species that donates electron gets an overall positive charge and the species that gains electrons get an overall negative charge. The former is termed as cation and the latter is termed as anion. The cationic species often belong to either alkali metals or alkaline earth metals while anionic species belong to halogens or chalcogen groups.

A covalent bond is formed by sharing of electron via overlap of the orbitals of the combining atoms. For effective sharing of electron pairs each atom must be capable of sharing its valence electrons. For this to occur they must have similar electronegativities.If the electronegativity difference is more than the ionic bonding occurs that involves complete transference of electrons. For instance, ${\text{NaCl}}$ is an ionic compound.

An example of a covalent molecule is ${\text{HCl}}$ . Here the bond is polar and covalent. Another example of covalent molecule is methane where the difference in the C-H electronegativities is very small so it regarded as a pure covalent molecule.

Rubidium lies at the bottom of the alkali metal group while the iodine lies in group 17. Since the alkali metals posses a strong tendency to lose electrons while the lost electron can be readily accepted by halogen so the bond formed between Rb-I is an ionic bond.

The bond formed between two homonuclear is always a result of the sharing of electrons hence the bond in ${\mathbf{B}}{{\mathbf{r}}_{\mathbf{2}}}$ is a non-polar covalent bond. Similarly ${{\mathbf{N}}_{\mathbf{2}}}$ is a non-polar covalent molecule.

The sulfur belongs to the oxygen family so it not electropositive in character thus it is bonding with a halogen such as iodine and fluorine are polar owing to the electronegativity difference and covalent due to the sharing of electrons. So S-I bond as well as ${\mathbf{O - F}}$ bond are non-polar covalent bonds.

Calcium belongs to alkaline earth metals that possess two valence electrons and oxygen needs two electrons to complete its octet, hence the transfer of electron occurs from Ca to O to form calcium oxide. ${\mathbf{Ca - O}}$ is a polar covalent bond.

Cesium lies in the alkali metal group while the chlorine lies in group 17. Since the alkali metals have strong tendency to lose electrons while the lost electron can be readily accepted by halogen so the bond formed between Cs-Cl is an ionic bond.

The phosphorus belongs to the nitrogen family so it not much electropositive in character thus it is bonding with a halogen such as iodine is polar owing to the electronegativity difference and covalent due to the sharing of electrons. So bond in ${\mathbf{P}}{{\mathbf{I}}_{\mathbf{3}}}$ is regarded as polar and covalent bonds.

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical bonding

Keywords: Ionic bond octet rule, covalent bond, ionic bonding, overlap, valence electrons, Rb, I, Br, S, I , N, O, Cs, Cl, Ca, O, P, I, bonding electrons, non-bonding electrons alkali metals and alkaline earth metals.

answered: Guest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$