Answer to Question #279243 in General Chemistry for Emre

Question

Answer the following questions by showing electron configuration (spdf notation)

(a) Which of the following is (are) paramagnetic? Zn, I, K+, O2-, Al

(b) Which of the following has the largest number of unpaired electrons: S, Cr2+, P, Rb,

Kr?

(c) Choose the paramagnetic atom or ion: Zn, Zn2+, Mn2+, Ar, Al3+.

(d) Which of the following has the smallest value of the first ionization energy: Ar, Cs, N,

Cu, Rb?

(e) List in order of increasing size: Cl- , Ar, K+, S2- , Ca2+. (Show the parameter that effects the size as orbital-electron skeckout)

Answer 1

a)

I and Al are paramagnetic because they have unpaired electrons.

$Al=[Ne]3s^23p^1$

$I=[Kr]4d^{10}5s^{2}5p^{5}$

b)

Cr²⁺ is the highest with 4 unpaired electrons.

S=[Ne]3s²3p⁴ 2 unpaired electrons

P=[Ne]3s²3p⁴ 3 unpaired electrons

Rb=[Kr]5s¹ 1 unpaired electron

Cr²⁺=[Ar]3d⁶ 4 unpaired electrons.

c)

Mn²⁺ is the only paramagnetic one in the list. The others have paired electrons. [Ar]3d⁵. With 5 unpaired electrons

d)

Cs has the least ionization energy because of least value of Zeff due to highest atomic radius resulting from many energy levels [Xe]6s¹

e)

Ca²⁺,K⁺,Ar,Cl^-,S^2- is the correct order of arrangements. All the atoms have 18 electrons but difference in number of protons. Clearly, Ca²⁺ has more protons and experience highest Zeff than S^2-