Question

Part B: Calculate the number of valence electrons and then draw Lewis Structure by Formal Charge Rule and determine the Formal Charge on each atom. Lewis Structure Formula Valence Electrons (Formal Charge Rule) Formal Charge Include Resonance Structures (on Each Atom (If necessary) 5. SCI 6. CIO: 7. NCI: 8. SiH 9. NO2 10. PO

Answer 1

Part-8 6 Formula : sch (Atomic no: 16 Valence election in Sulfhuse (s)=6ė / is%25%25*3s 3p0 Valence elechons in Cl = 7e valence é Atomic no=17 15 25 266 35 365 Valence electrome in S¢ = 6+1x2= 20€ * All halogens F, Cl, Br, I Lewis structure: Contan 7 valenca é 7e - :d: S:ä: OR :Če-s-č: Formal charge : Foomula to find fromal Charge is : lone pair elections F.C = Cene Stoce Vy valence é → Lone paire s Bonding e J - Bonded *one Bond contain ze electons 0 . W same here V=7. v=6 し=6 L=4 S=2 s=4 is formed 2 bonds : 4é) F.C = 7-6-5X2 Fice 6-4-2xy =0 7-6-1 = 0 Formal charge once iso Formai Charge on sio | Co).. e) (6) Cl-sce v

Valence é in Oxygen = bé Atino: = 8 T25 269 Valence e in clos = 7+ 6x3+1 = 267 For negative charge Lewis structure : ••07_ O= ci – öz Resonance structures — Ср segir a sg 6 - - - toomal charge : = CI- 0 (0) . At nu=7 - T$ 25 26 s vale ② Na3 N contain 5 valance él Valenco è in Nelz = 5+7X3 = 26év Lewis Structure VEN: : OR ☺-Ü; : :

- Formal charge (O) (O) (O) ::(0) Sity Si contain 4 valé Latino = 14 1525 216 35 3p- Yuale J H contain I valé Valence e in Sitty = 4+4x1= 8e v Lewis Structure : :Si; } OR 4-şi-H Fomal charge u Hlo ② Nos Valence é in No = 5+6X2+] = 18€ For-ve charge Lewis structure : Resonance structures : o wło: ©= 0 –

Formel Charge | (-) (0) (0) 0– N= 0: ③ pays p contain 5 valence é At no=15 1525 2p% 38² 383 Svalce Valence e in foy = 5+6X4+3 = 32 év Lewis structure :0:- | 0 = > 6: 0:- Resonance structures <> 0== o Cô-e-o s LO-220 366) Formal charge : 0:40

Part A :D Co, co Valence election SX2 = 16e Lecois Structure Ö=(=Ö: Polarity : oxygen is more electronegative than carbon. so n partial negative charges on the both the oxygen partial positive charge on the carbon atom. there and Due to linear structure, dipole moment of one c=0 bond cancels the dipole moment of another c=0 bona. so, Net dipole moment in Zero. Therefore co molecule in non Polar (Bond polaity Non Polar ✓ н со CN do 1+4+5 Valence é = = 10e Lewis Starchuxe H-CED - Polarity: H-c = Ň Dipole moments directed on the same direction : dipole moment in not equals to zero. Hence of is a polar Polarity & Polar) -

② ho Valence e = 1X2 + 6 - 8 é H Lewis structure H o H Polarity porn Polar] » Hoot Valence é = 3x1 +6 -1 = . For the charge hX kan bare La = Polarity | Polar?