Question

draw the structural formula for each of the following molecules
a) $\ce{ch4}$
b) $\ce{br2}$
c) $\ce{if}$
d) $\ce{sis2}$
e) $\ce{sbi3}$
f) $\ce{secl2}$

Explanation:

Step1: Analyze \( CH_4 \) (Methane)

Carbon (\( C \)) has 4 valence electrons. Hydrogen (\( H \)) has 1 valence electron. Carbon forms 4 single bonds with 4 hydrogen atoms. The structural formula is a tetrahedral arrangement with \( C \) at the center and \( H \) atoms at the vertices. Represented as: \( \text{H}-\text{C}-\text{H} \) (with two more \( H \) atoms attached to \( C \), one in front and one behind in 3D, but in 2D structural formula: \(

$$\begin{matrix} & \text{H} & \\ \text{H} & - & \text{C} - & \text{H} \\ & \text{H} & \end{matrix}$$

\))

Step2: Analyze \( Br_2 \) (Bromine Molecule)

Bromine (\( Br \)) has 7 valence electrons. Two \( Br \) atoms share a pair of electrons to form a single covalent bond. Structural formula: \( \text{Br} - \text{Br} \)

Step3: Analyze \( IF \) (Iodine Monofluoride)

Iodine (\( I \)) and Fluorine (\( F \)) each have 7 valence electrons. They share a pair of electrons to form a single covalent bond. Structural formula: \( \text{I} - \text{F} \)

Step4: Analyze \( SiS_2 \) (Silicon Disulfide)

Silicon (\( Si \)) has 4 valence electrons, Sulfur (\( S \)) has 6. \( Si \) forms double bonds with 2 \( S \) atoms (to satisfy octet: \( Si \) shares 4 electrons, each \( S \) shares 2, so \( O=\text{Si}=O \) analog, here \( S \) instead of \( O \)): \( \text{S}=\text{Si}=\text{S} \)

Step5: Analyze \( SbI_3 \) (Antimony Triiodide)

Antimony (\( Sb \)) has 5 valence electrons, Iodine (\( I \)) has 7. \( Sb \) forms 3 single bonds with \( I \) atoms (using 3 valence electrons, leaving one lone pair on \( Sb \)): \(

$$\begin{matrix} & \text{I} & \\ \text{I} & - & \text{Sb} - & \text{I} \\ & (\text{lone pair}) & \end{matrix}$$

\) (in 2D structural formula: \( \text{I} - \text{Sb} - \text{I} \) with one \( I \) above and one lone pair on \( Sb \))

Step6: Analyze \( SeCl_2 \) (Selenium Dichloride)

Selenium (\( Se \)) has 6 valence electrons, Chlorine (\( Cl \)) has 7. \( Se \) forms single bonds with 2 \( Cl \) atoms (using 2 valence electrons, leaving two lone pairs on \( Se \)): \(

$$\begin{matrix} & \text{Cl} & \\ \text{Cl} & - & \text{Se} - & \\ & (\text{two lone pairs}) & \end{matrix}$$

\) (structural formula: \( \text{Cl} - \text{Se} - \text{Cl} \) with two lone pairs on \( Se \))

Answer:

a) \( \boldsymbol{

$$\begin{matrix} & \text{H} & \\ \text{H} & - & \text{C} - & \text{H} \\ & \text{H} & \end{matrix}$$

} \) (or \( \text{H}_3\text{C}-\text{H} \) simplified as \( \text{CH}_4 \) structural formula with four \( \text{C-H} \) bonds)
b) \( \boldsymbol{\text{Br} - \text{Br}} \)
c) \( \boldsymbol{\text{I} - \text{F}} \)
d) \( \boldsymbol{\text{S}=\text{Si}=\text{S}} \)
e) \( \boldsymbol{\text{I} - \text{Sb} - \text{I}} \) (with one \( \text{I} \) above \( \text{Sb} \) and a lone pair on \( \text{Sb} \))
f) \( \boldsymbol{\text{Cl} - \text{Se} - \text{Cl}} \) (with two lone pairs on \( \text{Se} \))

(Note: For 3D structures, tetrahedral for \( CH_4 \), linear for \( Br_2 \) and \( IF \), linear (with \( sp \) hybridization) for \( SiS_2 \), trigonal pyramidal for \( SbI_3 \) (like \( NH_3 \) but \( Sb \) instead of \( N \)), bent (angular) for \( SeCl_2 \) (like \( H_2O \) but \( Se \) instead of \( O \) and \( Cl \) instead of \( H \)). The above are 2D structural formulas showing bonding.)