First row: HCl, ZnCl2, FeCl3, AlCl3, BaCl2, PbCl4
Second row: H3P, Zn3P2, FeP, AlP, Ba3P2, Pb3P4
Third row: HNO3, Zn(NO3)2, Fe(NO3)3, Al(NO3)3, Ba(NO3)2, Pb(NO3)4
Fourth row: ZnO, Fe2O3, Al2O3, BaO, PbO2
Fifth row: HCaF2, Zn(CaF2)2, Fe(CaF2)3, Al(CaF2)3, Ba(CaF2)2, Pb(CaF2)4
Sixth row: H2SO4, ZnSO4, Fe2(SO4)3, Al2(SO4)3, BaSO4, Pb(SO4)2
Answer: molecular formula = C12H16O8
Explanation:
NB Mm CO2= 44g/mol
Mm H2O= 18g/mol
Moles of CO2 = 36.86/44=0.84mol
0.84mole of CO2 has 0.84 mol of C
Moles of H2O = 10.06/18= 0.56mol
1mol of H20 contains 1mol of O and 2 mol H,
Hence there are 0.56mol O and (0.56×2)mol H
Hence the compound contains
C= 0.84 mol H= 1.12mol O=0.56mol
Divide through by smallest number
C= 0.83/0.56= 1.5mol
H= 1.12/0.55= 2mol
O= 0.56/0.56= 1mol
Multiply all by 2 to have whole number of moles = 3:4:2
Hence empirical formula= C3H4O2
(C3H4O2)n = 288.38
[(12×3) + 4+(16×2)]n= 288.38
72n=288.38
n= 4
:. Molecular formula=(C3H4O2)4= C12H16O8
Hellium argon neon xenon krypton randon oxygen fluorine chlorine bromine
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
To calculate the hybridization of 
, we use the equation:
![\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom (S) = 6
N = number of monovalent atoms bonded to central atom = 0
C = charge of cation = 0
A = charge of anion = 0
Putting values in above equation, we get:
![\text{Number of electron pair}=\frac{1}{2}[6]=3](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5B6%5D%3D3)
The number of electron pair around the central metal atom are 3. This means that the hybridization will be 
and the electronic geometry of the molecule will be trigonal planar.
Hence, the correct answer is Option D.
