Question

Determine the number of distinct quantum states for each of the following configurations. In each case, list the spectroscopic symbols of the allowed states. (a) two equivalent d electrons [e.g. (3d)?] (b) two non-equivalent d electrons [e.g. (3d)-(4d'] (c) the configuration (4p)(48) (d) the configuration (2p)? (3p)1

Answer 1

I 2D "Faus a) Two-equivalent d electrons (e.g (3012)] Because of the equivalent electrons we use the Brzeit scheme are follows notation 8,5%, sea yg OS mon = 1/2 ms. = I12 ТР &=2, 7 mire = 2,1,0,-1,-2. 1 3 F 22 =2 m = 2, 1, 0,-1,-2 How we constreact. The following table to obtain Ms and ML values met 2 1 0 -1 -21 mg, Ye Ye ! M4 4 3 2 1 02 . Malo 12 3 2 1 0 1 1 & 1 0 1 -2 0 men 1 O -1 -2 -3 - Table-2 0-1 -2 -3 -4 -2 amin Table-1 According to Breit's prescription if we permit unlike Ms vales es fie #1) then the varues of Me = 4, 3, 2, 1, 0,-1, -2,-3,-4 arce forbidden , since the uppere hart and lower half of the diagonal line of table - 1 are identical considering any one we have the following combination of Me and Ms. Ms = t1 M2 = 3, 2, 1, 0; -1, -2, -3. $=1 and L=3 I=1L-sl to 1L431 = 2, 3, 4 . The states are 352, 314

If we permit like values of Ms (1,eo) we have the 5 sets of Me values. So we have the following combination of Me and My vino broo Mg = I1 Me = 1,0,-| S=1 t=1 32 IL-S1 to 1L731 = 0, 1, 2 The states are spor 0, 1, 2 & we premit like values of Ms (eo) we have the s sets of Me values. So we have-lae following combination of M. and Ms . Mg = 0 out My = 4, 3, 2, 1, 0,-1,-2, -3, -4 wah 520 L=4 3= 1L sl to 11451 - 4 7 word The state is G taca dan mement) dritten Mg 20 M, = 3, 2, 1, 0,-1,-2,-3 this one is rejected as the Me values are appeared previousing Mg 20 M2=2,1,0,-1,-2 520 2 ' 2 = I= 2 the state is Do My=0 M₂ = 1,0, 1 this is also rejected as Me values are appeared previouste

Mg=0 Meso 302=0 il limone units Jo The state is to in the total no. of distinct quantum states are o and are given by 362 363, 3F4 , 3P, 3p, 3p Gy Dog So 12 Hand's Rull Rule-1 - Fore many electron atom out of all possono spectroscopic states, the states having out nighest multiplicity will be the lowest one Toli in energy order. Koerin Rule-2:- fore, a given multiplicity the states having to highest orbital quantum no. (L) will be the lowest one. ele-3: fore a given and the states having lowest I will be the rowest one Bore normale multiplate (less than half billed) for inverted (Ahalt filled multiplate the state with highest I will be the m lowest one Applying Hund's rule we can arrange the above allowed states according to their energy morce than

b) Two non-equivalent d electrons [eng (3d) (40)] l=2 lg=2 2= 120-lg 1 to 1 lethel = 0, 1, 2, 3, 4 sool so 18-21 to 18, +22)] S2=Y Stare [2st ] 130 > > 164 The no. of distinct quantum states are 18. 3Por 20% 3fu BDI 383' Seas 0, 1,2 1,2,3 2,3,4 3,4,5 3 4 3p 30, 3 P2 30, 31, 3D %F2, 363, PFA ²G 3G 3G1s c) The configuration (40)' (46) 20=1 2 = 16 la 1 to 1 Relal = 2, 3, 4 $, - % s=0, 1 Is= 18,-821 to 13, +4217 Sq=%

J = [11-100 124317 state [25+L, 2 no, of IGU in then distinct quantum states are 12 1,2,3 3D, 302, 303 2,3,4 3,4,5 d) the configuration (2P)? (BPJ' Fore two equivalent creation (2072 X21 m2 =-1,0,1 me=-1,01 8,2% mo, = 772 ms₂ = 1/2 Now construct the following table to obtain Mg and Ľ varues. m2, 1 0 -IL mse l Yq Ya M2 2 1 0 1 Mg 1 o 1 0 -1 0 - 0 -1 -2-1 fore unlike varries of My li, e II) Mg = 11 521 Me = 1,0, 1 Ľal

fore like value of My of (ie M820) M820 ME = 2,1,0,1,-2 s'o Mezo ME = 1,0,- this is rejected because it is previously obtain. M Ľ=0 M520 520 Ľ=0 fore (28) ² (3P)' were bore (3P)' Rz = 1 and 8z = Ya L=12 - 13 ) to / Litle I 0,1,2 1,2,3 Sz|6-831 to 154831 So And the possible values of L= 0, 1, 2, 3 the possible values of s = % 13%

J = | L - | to /PS از state [ 23.112] م او او وید اور و حا و ۹۰ 9( مع دمع امه بره ۱۶۱ (417 د سرود) درنده و 2 / 5 دا . وا 85( حما ود در - 3 % " " % % % ۱۹۷ 4931, 40512) ا 94 / 1 . The number of distinct quantum states are is.
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