Answer:
Explanation:
2-methyl-3-ethylpentane | C8H18 | ChemSpider.
Boiling Point: 115.3±7.0 °C at 760 mmHg
Molecular Formula: C8H18
Density: 0.7±0.1 g/cm 3
Enthalpy of Vaporization: 32.9±0.0 kJ/mol
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The formal charges of all nonhydrogen atoms are -1.
Solution:-
<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>
O-Cl-O 6-7 = -1x4 = -4 N 5-4=1 H-N-H 1-1=0
O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1
<u>6-6 = 0 1-2 = -1</u>
It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.
According to the lewis structure of SO2, The critical atom is sulfur and it is bonded with 2 oxygen atoms thru a double bond. each oxygen atom acquires 2 lone pairs of electrons and the primary sulfur atom has 1 lone pair of electrons.
Learn more about Nonhydrogen atoms here:-brainly.com/question/2822744
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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.