Question

8. (a) Identify the atomic orbitals having the following quantum numbers: (a) n 3,1 0, (b) n-4,1-2 ()n 6,1-1 (b) Give all possible combination of quantum numbers for the following atomic orbitals: (i) 4s (ii) 6d 9..Write down the electronic configuration for each of the following atoms: (b) Cr (a) Ti (d) Cu 10.Sketch the following atomic orbitals: a) an s orbital b) a px orbital c))a py orbital orbital a p, orbital e) ) a dsy Bonus Q. Up to 10 marks may be added to your score if you correctly answer Q. 11 11. The iodine liberated by the addition of an excess of KI and H2SOs to 25.00 mL of 0.0256 M KIOs required 22.55 mL of a solution for Na2S2O, for titration using starch as the indicator. What is the molarity of the Na2S20, solution? 2 KIO,(ag)+ 5 KI(aą)+ 3 H2SOs(ag)-3 1(ag)+3 K So(ag)+ 6 H,00

Answer 1

n = Principle quantum number which tells about the energy shell, represented by n and has value n = 1,2,3...

l = azimuthal quantum number, which tells us about the subshell or orbital and has value n-1

If l = 0, orbital is s

For, l = 1, orbital is p

For l = 2, d and so on

m_l = magnetic quantum number which tells us about the orientation of orbital and has value -l to 0 to l+1

m_s = spin quantum number and has value +1/2 or -1/2

8.(a) (a) 3s

(b) 4d

(c) 6p

8.(b)(i) n= 4, l = 0, m_l =0 and m_s = +1/2

or n= 4, l = 0, m_l =0 and m_s = -1/2

(ii) n= 6, l = 2, m_l =-2 and m_s = +1/2

n= 6, l = 2, m_l =-2 and m_s = -1/2

n= 6, l = 2, m_l =-1 and m_s = +1/2

n= 6, l = 2, m_l =-1 and m_s = -1/2

n= 6, l = 2, m_l =0 and m_s = +1/2

n= 6, l = 2, m_l =0 and m_s = -1/2

n= 6, l = 2, m_l =1 and m_s = +1/2

n= 6, l = 2, m_l =1 and m_s = -1/2

n= 6, l = 2, m_l =2 and m_s = +1/2

n= 6, l = 2, m_l = 2 and m_s = -1/2

9.(a) Electronic configuration of Ti ( Z = 22) =[Ar] 3d²4s² or 1s²2s²2p⁶3s²3p⁶3d²4s²

(b) Cr (Z = 24) = [Ar] 3d⁵4s¹ or 1s²2s²2p⁶3s²3p⁶3d⁵4s¹

(c) Ca( Z = 20) = [Ar] 4s² or 1s²2s²2p⁶3s²3p⁶4s²

10.

lo (a) Cc) S ơrbita P Cdumble) C domble CSphercal) Ce) day dobie dumble Cs Scamnatbe) CamSc ape

11. Millimoles of KIO₃ = Molarity * volume = 0.0256 M * 25 mL = 0.64 mmol

From reactions

0.64 mmol KIO₃ * ( 3 mol I₂ / 2 mol KIO₃) * ( 2 mol Na₂S₂O₃ / 1 mol I₂) = 1.92 mmol  Na₂S₂O₃

volume of solution = 22.55 mL

Molarity of Na₂S₂O₃ = moles/ Volume = 1.92 mmol / 22.55 mL = 0.0851 M