Question

Lewis Structures and Molecular Mo C. Build the models and give the shape for each. Shape Angular or bent Hydrogen sulfide trigonal pyramidal H-P-H H Phosphine Angular or bent -0-0- Ozone HC-c-H HaH Ethylene Shape around carbon bent Shape around carbon bent HO: H-C-C-H ai Acetaldehyde Shape around carbon @ tetrahedral Shape around carbon trigonal plane

Answer 1

H H here central atom s. and no. o electrons. around 5 in (6+1+1) no. of valence & othing in electon pair= 8 4 1 according to VSEPR tetrahedral geometry, and mo. of lone patr=2 bond pair= 2 Shape is bent a V-shapped

H-PH central atom Pin total es around 5+1 +1 + 2) Velence é s number of election pair= 8/2 = 4. of p 5 geometry in tetrahedral but bond pots=3 lone pair=1 Shape is pyramidal H

The total no.t en around OA (6+2) = 8 As, OA-Oc is coordinate covalent bend so, Oc doesn't Intribute amy e for this bend. Por ob is double bena no so from op de come e pais - 8/2=4 but among them - bend = 1 i-a-bond patr= (41) 23 ". Geometry is trigonal planar. band pair=2 lone parrat So, the shape is bent.

H 1 Valence & of C in 4 for central atom Ca total number of een around Co, (4+2 +1+1) = 8 (er due to double bond gesome ty te katedral 2 es come from Col but band pars - 4 but t-bend pelrat -a-bona pair =(4-1) = 3. - Ske.geometry i tregond planer. lone pelr=0 the shape around ca is also trigonal planar. for central Hom Cb, totalno, ofen (4+2 +1+1)-8 - pairs – 4 but I bona pels=1 ia-borid petr=3 no lone pain geometry and also shape around Co is trigonal plan planan

# :0 - Hắc – – H For control atom Ca, no. of total valence electrons. arround Ca in (4+1+1+1+1)=8 mo.of.e-peira 8/2-4 : geometry tetrahedral toe On no lone pats no, shape is tetrahedral For central atomap, no. of total valence e-s around Co (4+1 +2 +1) =8 Qen for o due to - B no. o e- palra 8%= 4 I double bena) e among them to bend pair=L : tota a-bend pets = (41) = 3 -- geometry is trigona planas. Due to presence of not lone pair the shape o rona Co also trigonal planar

Here for the determination of geometry pi bond doesnot contribute. So after the substruction of the number of pi bond calculation is done