HOw many pi bonds does Sulfur Tetrafluoride have?
85238
Step 1: Draw the Lewis structure.\nStep 2: count the number of double bonds.\nStep 3: count the number of triple bonds.\nStep 4: total up number of pi bonds from steps 2 and 3.\n\nThere is 1 pi bond for every double bond and 2 pi bonds for every triple bond.
912679
PCl3 has the same shape as NH3 (tetrahedral with P in the center, the 3 Cl at the 3 corners and a lone pair at the fourth corner) So the polar covalent bonds don't cancel out each other\n\nSO3 is planar trigonal and thus the dipole moments cancel out each other and it is non-polar\n\nCF4 has a tetrahedral shape and since all bonds are the same they cancel each others dipole moment;the molecule is nonpolar\n\nCH2F2 has the approximately the same shape as CF4, but since not all bonds are the same their angle is a bit different and the sum of the dipole moments is not zero; thus the molecule is polar\n\nSO2 is a bent molecule (because of the sulphur's lone pair), think of it in a way as water with S in the place of O and O in the place of H so it is polar.\n\nCO2 is a linear molecule (O=C=O) so the dipole moments of the two identical polar bonds cancel out each other; the molecule is nonpolar\n\nO=C=S has different polar bonds, so although they have opposite directions the result will be a slight dipole moment, so the molecule is polar.\n\nIn all your questions you have to compare a polar and a nonpolar molecule. Of course the one with the strongest dipole-dipole interactions in each case is the polar molecule.
851509
There are two possibilities. Follow this link for a picture of the first possibility. http://img175.imageshack.us/img175/8223/ncssh2.jpg\n\nBasically, all 3 of these elements follow the octet rule (they want 8 electrons in their outermost shell), and as you can see, they all have 8. Now, S is carrying the - sign because it's formal charge is -1. Formal charge is calculated by Formal Charge=# of electrons the element has (S brings in 6) - the # of electrons it "owns" within the bond. Sulfur "owns" 7 in this case (3 lone pairs and one in a shared covalent bond--thus (3*2)+1=7.\n\nTherefore Formal Charge of Sulfur (in this case): is -1 because\n\n6-7=-1\n\nI probably made it sound more confusing than it is. Sorry about that.\n\nBut other than that, you know that Carbon should be the central atom here because it's coming in with only 4 electrons in it's valence shell, so it's more needy than Nitrogen with it's 5 and Sulfur with it's 6. So you stick it in the middle and then pretty much trial and error. \n\nThe other possibility is here.\nhttp://img175.imageshack.us/img175/6659/ncs2wf4.jpg\n\nThis time Nitrogen is carrying the -1 formal charge, because it comes into the bond with 5 electrons in it's valence shell, but it "owns" 6 in this case (2 lone pairs and 2 electrons out of the 4 in the N=C double bond..(2*2)+2=6. So...Formal charge of -1 comes from \n\n-1=5-6\n\nEven though these two molecules look different, they are in fact the same. This is a concept called resonance. \n\nI hope this wasn't too confusing.
525685
You can approach this is one of two ways:\n\n1)  By analogy with H2O (because O and S are both in the same group)\n\n2) From first principles ie octet rule, Lewis diagram, hybridisation, shape. \n\nNow S has 6 electrons in outer shell, while H has 1 each, therefore, S will share 1 e- from each H to complete its octet and each H will share an e- with S so H (which is too small to have 8e-s in outer shell) will have a full outer shell. So S is now covalently bonded to to H atoms, accounting for 4 e-s, with 2 non-bonding pairs. That makes 4 pairs of electrons around S all together, so hybridisation is sp3. Shape is tetrahedral minus two vertices ie bent (just like H2O).\n\nI hope this helps.
739949
For every molecule of H2SO4 there are 2 Hydrogen(H) atoms, 1 Sulfur(S) atom, and 4 Oxygen(O) atoms. So if there are 9.5 x 10^22 molecules of H2SO4 then are are also that number of sulfur molecules present.\n\nTo find how many grams 9.5 x 10^22 molecules weighs, use Avogadro's number. Dividing by Avogadro's number will tell you how many moles of Sulfur you have, and from that you can use the atomic mass of Sulfur to determine how many grams there are:\n\nAvogradro's number = 6.02 x 10^23\n\n(9.5 x 10^22) / (6.02 x 10^23) = 0.158 moles\n\nAtomic mass of Sulfur = 32.065 g/mol\n\nMultiply this by the number of moles of Sulfur you have to obtain the number of grams of Sulfur:\n\n32.065 * 0.158 = 5.067 g\n\nI hope this helps!
351716
In NH3, there are 3 bond pairs and 1 lone pair; there is repulsion between the lone pair and bond pair and between 2 bond pairs. Hence all the 4 pairs arrange themselves as far apart as possible to minimize the repulsion between them, while still maintaining the stability of the molecule. Hence, NH3 has a trigonal pyramidal shape which achieves the above purpose.\nSimilarly in H2S, there are 2 bond pairs and 2 lone pairs. Here there is repulsion between bond pair and bond pair; lone pair and lone pair and lone pair and bond pair, so the pairs spread out. In this case, a 'bent-V-shape' or a 'non-linear' shape for H2S is suitable as it minimizes repulsion and maximizes stability.
580069
Tetrahedral arrangement at the methyl carbon. (4 bonding pairs, no lone pairs)\nTrigonal planar arrangement at the carbonyl carbon. (3 bonding pairs, no lone pairs)\nBent arrangement at the single-bonded oxygen. (2 bonding pairs, 2 lone pairs)
327842
I am pretty sure that it is due to polarity.  Since sulfur forms nonpolar S8 molecules, they are nonpolar, where water has a + end and a -end, making it polar (has a dipole moment).  The general rule is like dissolves like.  Molecules that a alike can interact on  the molecular level and form solutions, unlike molecules can not.
279021
There's nothing surprising about the normal Group oxidation state of +4.\n\nAll of the elements in the group have the outer electronic structure ns2npx1npy1, where n varies from 2 (for carbon) to 6 (for lead). The oxidation state of +4 is where all these outer electrons are directly involved in the bonding.\n\nAs you get closer to the bottom of the Group, there is an increasing tendency for the s2 pair not to be used in the bonding. This is often known as the inert pair effect - and is dominant in lead chemistry.\n\nHowever, just giving it a name like "inert pair effect" explains nothing. You need to look at two different explanations depending on whether you are talking about the formation of ionic or covalent bonds.
806618
The only difference I can see would be the molecular geometry, due to the number of H's bonded onto the anion. Also, when drawing the Lewis dot structures, NH3 (N) would have one lone pair of electrons, HF (F) and H2O (O) would have two. This would dictate their structures according to the VSEPR rules.
98048
H:S:S with 2 dots on top and bottom of the S. S has 6 valence electrons and H has 1, a total of 8
362446
Sulfur with Six forms, Rohmbic, Orthorohmbic and 4 others I can not recall the names...But is it 6 for sure
825026
what you need to do, get a SI data book, and look up their natural state. not bonded with anything else, and possibly itself.\n\nPhophorus - white: molecular (P4). red: Chain. Black: Layer\nIodine - molecular (I2)\nMagnesium - hexagonal closest-packing\nNeon - cubic closest-packing\nArsenic -grey: 3d structure - Layer formation. yellow: molecular As4\nSulfur- molecular S8\nBoron - complicated structure.\nSelenium - grey: chain, Red: molecular: Se8\nGallium - Complicated\nCesium - body-centred cubic\nOxxygen - O2, O3 molecular
722082
s,p,d and f are designations for the four different types of orbitals.\n\ns is spherical in shape and has only one orientation in three dimensional space.\n\np has a dumbbell shape and has three orientations in 3-d space:\nPx,Py and Pz.\n\nd has a double dumbbell shape and has 5 orientations in 3-d space.\n\nf has a complex shape and has 7 orientations.\n\nEach orbital can hold 2 electrons. s having only 1 orientation can hold only 2 electrons. p having 3 orientations, has 3 orbitals, each having the capacity to hold 2 electrons, so it can totally hold (3*2=) 6 electrons.\nSimilarly d can hold (5*2=) 10 electrons and f can hold (7*2=) 14 electrons.
223542
thermodynamic factors as well as the orientation and relative energies of the pi molecular orbitals involved.
903420
In all cases the arragngement is tetrahedral and thus the ideal bond angle would be 109.5 degrees. In detail:\n\na) O has 2 unpaired e and 2 lone pairs. In H3O+, the 2 e participate in two single bonds with H and one lone pair participates in a dative bond with H+. Thus you have 3 single bonds + 1 lone pair so the arrangement is tetrahedral. lone pair-bond electron repulsion is stronger than bond electron-bond electron and thus the actual bond angles will be smaller.\nThe shape of the molecule will be trigonal pyramid (It is actually tetrahedral with O in the center and on one corner you have the lone pair but the lone pair is not "visible" in the molecular level)\n\nb) Si belongs to the same group as C. It forms 4 identical single  bonds in SiCl4 and has no lone pairs. Thus both the molecular shape and the electron arrangement are normal tetrahedral with bond angles 109.5\n\nc)N has 3 unpaired e and one lone pair. In NF3 it forms 3 single bonds through the unpaired e and so it has 3 bonds + 1 lone pair. Thus it is similar to H3O+ (tetrahedral electron arrangement, ideal bond angle 109.5, true bond angle smaller, molecular shape trigonal pyramid)
801097
Work through them one-by-one. You need p or d orbitals that are oriented properly. The s orbital is symmetric, so it won't work for a pi bond. \nA. s orbital is the wrong symmetry for a pi-bond.\nB. py and dxy do appear to be oriented properly\nC, D, E - all have as one choice the px. Since the bond is along this axis, the px is not oriented properly for a pi bond. (The pi-bond has to have lobes that stick out to the side perpendicular to the axis of the bond).