Question

Identify the missing particle in the nuclear reaction below: 2U +n +_+13Cs +3fn o Br ORD Sr OKT Question 13 If CN is a strong held ligand, rank the following coordination compounds in order of decreasing number of unpaired d-electrons: [Fe(CN) [Pd(CN) [Mo(CNJ) O [Fe(CNJP >[PICNJ - (MoCNJ [ PCNJ) [Fe(CNJ (MOCNJ O [Mo(CNJ) [Fe(CNJP >IPICNJA [Fe(CNJP > M(CNJ [PdCNJF O [Mo(CNJI* > [PdNJ] > [Fe(CNJP MacBook Air 80 73 F4

Answer 1

Let the missing particle is YA X Therefore, the given equation becomes у 235 U 92 + in A + 137 Cs + 55 3 'n 0 0 Now, total number of protons should be same in both side of the above equation. Hence, 92 + 0 = x + 55+ 0 Or, x = (92 - 55) = 37 Similarly, total mass number of the nuclide should be same in both side of the above equation. Hence, 235 + 1 = y + 137 + (3 x 1) Or, y = (236 - 140) = 96 Hence the missing particle is 96 Rb 37 Question 13 In [Fe(CN)613- complex, CN is a strong field ligand and therefore will form low spin complex by using inner 3d-orbitals. The central metal has oxidation number +3 calculated as: X+6 (- 1) = -3, or, x = +3 Fe has 4s23d electronic configuration and therefore Fe3+ will be 3d”. Upon pairing it has one unpaired electron that makes it paramagnetic complex. 4p 3d 4s Fe in ground state 111111 Fe in +3 state Fe in [Fe(CN).13- 1111 A A A CN CN CN CN-CN CN Hybridization = d²sp3 In [Pd(CN)al?- complex, CN is a strong field ligand and therefore will form low spin complex by using inner 4d-orbitals. The central metal has oxidation number +4 calculated as: x + 6 (-1) = -2, or, x = +4 Pd has [Kr]4d10 electronic configuration and therefore Pd4+ will be 4d. Upon pairing it has zero unpaired electron. 5s 5p 4d Pd in ground state 11111 Pd in +4 state Pd in [Pd(CN).]2- 711 CN CN CNCNCN-CN- Hybridization = d²sp3 In [Mo(CN)614- complex, CN is a strong field ligand and therefore will form low spin complex by using inner 4d-orbitals. The central metal has oxidation number +2 calculated as: X+6 * (-1) = - 4, or, x = + 2 Mo has [Kr]4d5s' electronic configuration and therefore Mo2+ will be 4d4. Upon pairing it has two unpaired electron 4d 5s 5p Mo in ground state Mo in +2 state Mo in [Mo(CN)614 11 CNCN CN CN CN CN- Hybridization = d²sp3 Therefore, the order of decreasing number of unpaired d-electrons is [Mo(CN).14-> [Fe(CN).]3- [Pd(CN).]?- Number of unpaired d-electrons is 2 1 0