There are 10 hydrogen atoms that bind and there are 2 pairs of free electrons in the non-binding O atom
<h3>Further explanation</h3>
Aldehydes are alkane-derived compounds containing carbonyl groups (-CO-) where one bond binds to an alkyl group while another binds to a hydrogen atom.
The general structure is R-CHO with the molecular formula :

Naming is generally the same as the alkane by replacing the suffix with -al
Butanal or butyraldehyde is an aldehyde which has 4 C atoms
Inside the structure there are 3 atoms involved in bonding:
- 1. Atom C with 4 valence electrons, requires 4 electrons to reach the octet
- 2. Atom O with 6 valence electrons, requires 2 electrons to reach the octet
- 3. Atom H with 1 valence electron, requires 1 electron to reach a duplet
In describing Lewis's structure the steps that can be taken are:
- 1. Count the number of valence electrons from atoms in a molecule
- 2. Give each bond a pair of electrons
- 3. The remaining electrons are given to the atomic terminal so that an octet is reached
- 4. The remaining electrons that still exist in the central atom
- 5. If the central atom is not yet octet, free electrons are drawn to the central atom to form double bonds
In the Butanal structure (C₄H₈O) there is 1 double bond of the functional group (-CHO) between the C atom and the O atom
<h3>Learn more:
</h3>
Adding electron dots
brainly.com/question/6085185
Ionic bonding
brainly.com/question/1603987
Formal charge
brainly.com/question/7190235
Keywords: butanal, aldehyde, Lewis structure, a valence electron
The correct answer here is B - when chlorine accepts an electron to
complete its octet and becomes a chlorion ion it becomes an Anion. An
anion is a negatively charged ion. Chloride ions are an important
electrolyte within the body.
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.