4. Which of the following functions are equal to their complex conjugate? In each case, real constant. k is a (a) eikx + e-ikx (b) eikx - e-ikx (c) eikx (d) ekx + e-kx

5. (a) Deduce the point group of the follo species : OH (i) HO HO, (ü) Co4(Cp)4 (iüi) PF3CI2 (one Cl axial and one Cl equatorial)

E10A.5(b) Assign the following molecules to point groups: (i) HF (ii) IF, (pentagonal bipyramid), (iii) CIF, (T-shaped), (iv) Fe,(CO), (3), (v) cubane, C,H, (vi) tetrafluorocubane, C,H̟F, (4). HO co Fe co F 3 4

For the aqueous [Ni(NH3 )6]2+ complex Kf=5.50 x 108 at 25C . Suppose equal volumes of 0.0024 M Ni(NO3 )2solution 0.94M NH3solution are mixed. Calculate the equilibrium molarity of aqueous Ni2+ ion. Round your answer to 2 significant digits.

For the aqueous [HgCl4]2- complex Kf=5.0 x1015 at 25°C.Suppose equal volumes of 0.0074M Hg(NO3)2  solution and 0.84M NaCl solution are mixed. Calculate the equilibrium molarity of aqueous Hg 2+ ion.Round your answer to 2 significant digits.

1. Calculate the stability field of gibbsite Al(OH)3 versus its aqueous species on the basis of the below listed four equilibrium equations. Al(OH)3(gibbsite) + 3H+ = A13+ + ЗН20() Ks0 = 109.36 Al" + H,O(1) = Al(OH)** + H* Al3* + 2H2O(1) = Al(OH)2* + 2H* Al³* + 4H2O(1) = Al(OH)4¯ + 4H* K. sl =104.99 K s,2 =10-10.13 K s,4 =10-22.05 2. Two solutions have the following activities: 0.02 mol L-' F and 0.01 mol L' Ca²+ 0.0001 molL-'F and 0.0002 mol L' Ca²+ i. ii. Calculate the saturation indices of the solutions versus fluorite (pKsp = 10.6) and how much fluorite would precipitate or dissolve in order to achieve equilibrium. -

(a) Identify the symmetry elements in ethene and in allene, and assign each molecule to a point group. (b) Consider the biphenyl molecule, Ph–Ph, in which different conformations are possible according to the value of the dihedral angle between the planes of the two benzene rings: if this angle is 0°, the molecule is planar, if it is 90°, the two rings are perpendicular to one another. For each of the following dihedral angles, identify the symmetry elements present and hence assign the point group: (i) 0°, (ii) 90°, (iii) 45°, (iv) 60°.

5. The most stable structure for N₂O4 is planar and belongs to the D2h point group, which has the character table shown below. D2h E C₂(X) C₂(V) C₂ (2) i o(xy) o(yz) o(x2) Ag 1 1 1 1 1 1 1 1 B1g 1 1 -1 -1 1 1 -1 -1 1 -1 -1 1 -1 1 1 -1 1 1 -1 -1 1 1 1 -1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 1 -1 -1 1 -1 1 1 B2g B3g Au B1u 1 B2u 1 B3u 1 1 -1 1 -1 1 -1 -1 1 -1 1 1 -1 5.1 How many translational degrees of freedom does N₂04 have? 5.2 How many rotational degrees of freedom does N₂04 have? 5.3 How many normal modes of vibration does N204 have? x², y², z² xy 5.5 How many of the vibrational modes of N₂04 are IR-active? XZ yz 5.6 How many of the vibrational modes of N₂O4 are Raman-active? Z y X Rz Ry Rx 3. 5.4 The displacement vectors for the atoms of N₂O4 span 3Ag + 3B1g + 2B2g + B3g + Au + 2B1u + 3B2u + 3B3u. What are the symmetry species of the normal modes of vibration? ☐☐☐

Please help me Point group of square Antiprismatic is D4d

Consider the three following molecules and answer the questions below.HC≡CHH2C=CH2H3C-CH3 (3 conformers with different symmetry)(a) For each molecule, including all conformers as indicated, draw the molecule, assign it toits point group, and draw on the diagram the symmetry elements that were used to makethe assignment. State in each case whether or not the species is dipolar and/or chiral.(b) If one hydrogen atom in each molecule is substituted by a chlorine atom, sketch the newmolecules and assign them to their point group. For this set of mono-chloro compounds: state the most common point group state how many are chiral state how many are dipolar state which molecule/conformer has the highest symmetry point group state which molecule/conformer has the lowest symmetry point group identify which of these molecules/conformers have three different H atomenvironments(c) For the dichlorinated versions of the molecules/conformers considered in part (a): state the most common point…