Add Formal Charges To Each Resonance Form Of HCNO Below., Please Add Formal Charges For Each Pictures.. For A , B And C.. Thank You..

Add Formal Charges To Each Resonance Form Of HCNO Below., Please Add Formal 
Charges For Each Pictures.. For A , B And C.. Thank You..

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Question: Add formal charges to each resonance form of HCNO be…
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Add formal charges to each resonance form of HCNO below., Please add formal charges for each pictures.. for A, B and C.. Thank you..
Add formal charges to each resonance form of HCNO below.
B.
HÖNSO:
HC
–ÖSNÖ
Based on the formal charges you added above, which structure is favored?
OA OB ос
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General guidance
Concepts and reason Each atom in a Lewis structure for a molecule or ion will have formal charge. Formal charge on each atom is due to difference in valence electrons in an isolated atom and number of electrons assigned to that atom in a Lewis structure. We use the following equation to calculate formal charge on an atom in Lewis structure. FC=V – N – Here, FC is formal charge on an atom, V are valence electrons of the atom, N are non-bonding electrons on the atom, and B are bonding electrons assigned to the atom. If a molecule has more Lewis structures, a Lewis structure with no formal charge on any of its atom is favorable than a Lewis structure with formal charge. A Lewis structure with formal charges +2,-2, or +3, -3 are not favorable. A favorable Lewis structure will have negative formal charge on more electronegative atom and positive formal charge on less electronegative atom.
Fundamentals Number of valence electrons in an atom is equal its group number in the periodic table. A single bond is formed with two electrons and they called bonding electrons. A double bond is formed using four electrons. A triple bond is formed using six electrons. Non-bonding electrons are placed on atoms as pairs. A molecule can have more than one Lewis structure. A Lewis structure with no formal charge on its atom is favorable structure. Each available Lewis structure for a molecule or ion is a resonance form. A new resonance form will be obtained by changing bonds and electron pairs around an atom in a resonance form.
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Step-by-step
FIRST STEP ALL STEPS | ANSWER ONLY
Step 1 of 4 A
The Lewis structure for the resonance form A is as follows:
А —o—N90: FC (onH) =1–0–
= 0
FC (onC) = 4 – 4 – 1
= -2
FC (onN) = 5 – 0 –
= +1
FC (onO) = 6 – 2 –
= +1 The resonance form A with formal charge on each atom is as follows:
А
0 -2 +1 +1 H-C-NEO:
Explanation Hint for next step Hydrogen (H) belongs to group and it has 1 valence electron. The H atom is attached to Catom with one bond so bonding electrons are 2. No non-bonding electrons on H. Carbon (C) belongs to group IV and it has 4 valence electrons. The Catom forms two bonds that account for 4 bonding electrons. Two lone pairs are on the Catom that account for 4 electrons. Nitrogen (N) belongs to group V and it has 5 valence electrons. The N atom has four bonds and bonding electrons are 8. Oxygen (O) belongs to group VI and it has 6 valence electrons. The O atom has three bonds and bonding electrons are 6 and non-bonding electrons are 2.
Step 2 of 4
The Lewis structure for the resonance form B is as follows:
B H—c51—:
FC (onH) = 1-0
= 0
FC (onC) = 4-0
= 0
FC (onN) = 5 – 0 –
=
+1
FC (ono) = 6 -6 –
= -1 The resonance form B with formal charge on each atom is as follows:
в
0 0 +1 H-CEN —
😀 0:
Explanation Hint for next step
The H atom has 1 valence electron, 2 bonding electrons, and O non-bonding electrons. The Catom has 4 valence electrons, 8 bonding electrons, and O non-bonding electrons. The N atom has 5 valence electrons, 8 bonding electrons, and O nonbonding electrons. The O atom has 6 valence electrons, 2 bonding electrons, and 6 non-bonding electrons.
Step 3 of 4 ^
The Lewis structure for the resonance form C is as follows:
HoczNzo FC (onH) =1-0 –
= 0
FC (onC) = 4 – 2 –
= -1
FC (onN) = 5-0
= +1
FC (ono) = 6 -4
=0 The resonance form C with formal charge on each atom is as follows:
0 H-
-1 +10 C=n= 0
Explanation | Hint for next step
The H atom has 1 valence electron, 2 bonding electrons, and O non-bonding electrons. The Catom has 4 valence electrons, 6 bonding electrons, and 2 non-bonding electrons. The N atom has 5 valence electrons, 8 bonding electrons, and O nonbonding electrons. The O atom has 6 valence electrons, 4 bonding electrons, and 4 non-bonding electrons.
Step 4 of 4 ^
The resonance form B is favored structure for HCNO. The resonance form B is as follows:
B 0 0 +1 H-CEN—0:
Explanation Common mistakes Oxygen is more electronegative than nitrogen. Resonance form B is favored as more electronegative atom bears negative formal charge and less electronegative atom bears positive formal charge.
Answer

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