Question

classify the following as ionic or covalent compounds.
a \\(\ce{p2o5}\\)
ionic \\(\circ\\) covalent \\(\circ\\)
b \\(\ce{sno2}\\)
ionic \\(\circ\\) covalent \\(\circ\\)
c \\(\ce{fecl3}\\)
ionic \\(\circ\\) covalent \\(\circ\\)
d \\(\ce{na3p}\\)
ionic \\(\circ\\) covalent \\(\circ\\)
e \\(\ce{nh3}\\)
ionic \\(\circ\\) covalent \\(\circ\\)
f \\(\ce{ch3oh}\\)
ionic \\(\circ\\) covalent \\(\circ\\)

Explanation:

To classify compounds as ionic or covalent, we use the following rules:

• Ionic compounds: Form between a metal (or polyatomic cation) and a non - metal (or polyatomic anion), involving electron transfer.
• Covalent compounds: Form between non - metals, involving electron sharing.

Compound A: $\boldsymbol{P_2O_5}$

Phosphorus ($P$) and oxygen ($O$) are both non - metals. So, they share electrons to form a covalent compound.
Classification: Covalent

Compound B: $\boldsymbol{SnO_2}$

Tin ($Sn$) is a metal, and oxygen ($O$) is a non - metal. The metal (Sn) and non - metal (O) transfer electrons, so it is an ionic compound.
Classification: Ionic

Compound C: $\boldsymbol{FeCl_3}$

Iron ($Fe$) is a metal, and chlorine ($Cl$) is a non - metal. There is electron transfer between the metal (Fe) and non - metal (Cl), so it is an ionic compound.
Classification: Ionic

Compound D: $\boldsymbol{Na_3P}$

Sodium ($Na$) is a metal, and phosphorus ($P$) is a non - metal. Electron transfer occurs between the metal (Na) and non - metal (P), so it is an ionic compound.
Classification: Ionic

Compound E: $\boldsymbol{NH_3}$

Nitrogen ($N$) and hydrogen ($H$) are non - metals. They share electrons to form a covalent compound.
Classification: Covalent

Compound F: $\boldsymbol{CH_3OH}$ (Methanol)

Carbon ($C$), hydrogen ($H$), and oxygen ($O$) are non - metals. The atoms share electrons, so it is a covalent compound.
Classification: Covalent

Final Classifications:

Compound	Ionic	Covalent
B ($SnO_2$)	$\boldsymbol{\bigcirc}$	∘
C ($FeCl_3$)	$\boldsymbol{\bigcirc}$	∘
D ($Na_3P$)	$\boldsymbol{\bigcirc}$	∘
E ($NH_3$)	∘	$\boldsymbol{\bigcirc}$
F ($CH_3OH$)	∘	$\boldsymbol{\bigcirc}$

(Note: In the table, $\bigcirc$ represents the correct classification for each compound.)

Answer:

To classify compounds as ionic or covalent, we use the following rules:

• Ionic compounds: Form between a metal (or polyatomic cation) and a non - metal (or polyatomic anion), involving electron transfer.
• Covalent compounds: Form between non - metals, involving electron sharing.

Compound A: $\boldsymbol{P_2O_5}$

Phosphorus ($P$) and oxygen ($O$) are both non - metals. So, they share electrons to form a covalent compound.
Classification: Covalent

Compound B: $\boldsymbol{SnO_2}$

Tin ($Sn$) is a metal, and oxygen ($O$) is a non - metal. The metal (Sn) and non - metal (O) transfer electrons, so it is an ionic compound.
Classification: Ionic

Compound C: $\boldsymbol{FeCl_3}$

Iron ($Fe$) is a metal, and chlorine ($Cl$) is a non - metal. There is electron transfer between the metal (Fe) and non - metal (Cl), so it is an ionic compound.
Classification: Ionic

Compound D: $\boldsymbol{Na_3P}$

Sodium ($Na$) is a metal, and phosphorus ($P$) is a non - metal. Electron transfer occurs between the metal (Na) and non - metal (P), so it is an ionic compound.
Classification: Ionic

Compound E: $\boldsymbol{NH_3}$

Nitrogen ($N$) and hydrogen ($H$) are non - metals. They share electrons to form a covalent compound.
Classification: Covalent

Compound F: $\boldsymbol{CH_3OH}$ (Methanol)

Carbon ($C$), hydrogen ($H$), and oxygen ($O$) are non - metals. The atoms share electrons, so it is a covalent compound.
Classification: Covalent

Final Classifications:

Compound	Ionic	Covalent
B ($SnO_2$)	$\boldsymbol{\bigcirc}$	∘
C ($FeCl_3$)	$\boldsymbol{\bigcirc}$	∘
D ($Na_3P$)	$\boldsymbol{\bigcirc}$	∘
E ($NH_3$)	∘	$\boldsymbol{\bigcirc}$
F ($CH_3OH$)	∘	$\boldsymbol{\bigcirc}$

(Note: In the table, $\bigcirc$ represents the correct classification for each compound.)