"1A. Radius of the bohr orbit = n^2 x k,where k is first bohr radiusRadius of the bohr orbit = (3)^2 x 0.0529 = 0.4761nm1/wavelength = R(1/nf^2- ni^2),where R is a constant ie 1.097 x 10^71/wavelength = 1.097 x 10^7(1/1^2-1/3^2)= 0.97511 x 10^71/wavelength =0.97511 x 10^7wavelength =1/0.97511 x 10^7= 1.025 nmb,Cr-{Ar}3d54s1Cr+-{Ar}3d5Cr-{Ar}3d4c, see in the attachmentd.Electronic coniguration of p is {Ne}3s2 3p3For 3s2 – n =3,l=0,m1=0,ms =+1/2For 3p2 – n=3,l=1,ml=+1,ms=+1/2e,Free energy charge Delta G = -nFEcellwhere n = products- reactants = 2-3 = -1delta G = (-1) x 96487 x 0.76 = 7330.12kj/mol f,Ecell = E0 (cu2+/cu)-E(2Br-/Br2)copper occurs at cathode cu2+(aq) +2e ------- cu(s)Br2 occurs at anode 2Br-(aq) ----------- Br2(liq) +2eg,1. Dichloro tetraammine Iron(iii)nitrate2. Tris ammine diaquo sulphate cobalt(ii)bromide3. Potassium manganic cyanide chloride4.Tris(tri ethylene diammine)diquao chromium(iii) Bromideh, The smallest part of a crystal is called the unit. Is formed by a combination of atoms andmolecules. The structure of crystal may be formed for repetition of these unit.I, If the atoms of the alloying elements are significantly smaller than atoms the metal matrixinterstitial solid solution forms, in the case of the solute atoms array are located in large gapbetween solvent atoms . Diagram in the attachmentj, Primitive cubic system (hp) consists of a lattice point on each corner of the cube. Each atom toa lattice point is then divided equally between eight adjacent cubes, and the unit of the cellstherefore contains in total an atom (1 /8 x 8).k,1.{Fe(NH3)6}cl 2.{Mn(c2o4)2(H2O)2}3.{Co(en)3}(NO3)34.{Cr(NH3)4Cl2}BrI, Oxidation number of {CO(NH3)5NO2} = X + (0) X 3 +(-1) =0x= +1k4{Fe(CN)6} = 4 + X+(-6) = 0X = 6-4 = 22Aa,In the attachmenta,There is one double bond in resonance over the four P-O bonds in the tetrahedral [PO4]^3- ionso the bond order is 1.25.In case of the ammonia, three 2p orbitals atom of the nitrogen are combined with 2s orbitally toform hybrid four sp3 orbitals. The pair of electrons not endorsed will occupy an orbital hybrid.Again a hybrid orbital sander for every atom and a pair of not joining electrons.. b,Theground stateelectronic configuration of the atom of nitrogen is: 1s2 2s2 2px1 2py12pz1. Since there are three electrons not paired in the sublevel of 2p, the atom of nitrogen canform three bonds with three atoms of hydrogen. This one will give the molecule of ammoniawith 90 degree bond angles . Nevertheless, but the bond angles observed 107o48 '.* Therefore, it was proposed that, the atom of Nitrogen surrenders to the hybridization sp3 of2s and three 2p orbitals to give four sp3 orbitals, that are arranged in the symmetry tetrahedral. this arrangement will give more stability to the molecule due to the minimalrepulsions.* The atom of nitrogen forms 3 bonds of ssp3-s with three atoms of hydrogen using three halfhybrid sp3 full orbitals. There is also a lone pair in the nitrogen that belongs to the fully filled the sp3 orbital. . * The atom of nitrogen forms 3 bonds of ssp3-s with three atoms of hydrogen using three halfhybrid sp3 full orbitals. There is also a lone pair in the atom of nitrogen that consists o filled the sp3 orbital.*the molecule of ammonia is trigonal pyramidalwith a lone pair in the atom of nitrogenc,Oxygen is paramagnetic due to presence of unpaired electron.Bond order of O2 is 2 3A.Given c = 3 x 10^8ms-1,h= 6.624 x 10^-34js,E = 2.18 x 10^-18 j/(5-3)^2 = 0.545 x 10^-18Wave length = c/frequencyFrequency = E/h =0.545 x 10^-18 /6.624 x 10-34 =0.0822 x 10^16Wave length = 3 x10^8/0.822 x 10^16 = 3.649 xx 10^-8nmb, Reduction half reactions2H++ 2e ------- H2Oxidation half reactionsZn------------- Zn2++2eGalvanic cell consists of1, Anode -Zn(s) ------------- Zn2+(aq)+2e2. cathode2H++ 2e ------- H23. Salt bridge ie used to complete a electrical circuit. "