Question

21. a. Following is a list of electron configurations. Group the configurations in pairs that would represent similar chemical properties of the atoms. Give the chemical symbols for the elements represented by each electron configuration.

1s2 2s2 2p6 3s2 1s2 2s2 2p3 1s2 2s2 2p6 3s2 3p6 4s2 3d104p6

1s2 2s2 3s2 3p3 1s2 2s2 1s2 2s2 2p6

b. A 2+ ion derived from a transition metal has 3 electrons in the 3d subshell. What element made this ion?

c. In each pair, select the atom or ion with the larger atomic (ionic) radius: Mg and Mg2+, C and Sn, Rb and Sn, Be2+ and O2-

d. Which of the following elements would have the largest 1st ionization energy? Write the chemical equation representing the 1st ionization of this element. Na, Cl, Al, S or Cs

e. Which of the following elements would have the most negative electron affinity? Write the chemical equation representing the electron affinity of this element. Na, Cl, Al, S or Cs

Answer 1

Ans 21 :

a)

The atoms that have the same number of valence electrons in their last orbit will have the similar chemical properties.

So these along with their chemical symbols are :

(Mg) = 1s2 2s2 2p6 3s2

(N) =1s2 2s2 2p3 and (P) =1s2 2s2 2p6 3s2 3p3

(Kr) =1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 and (Ne) =1s2 2s2 2p6

b) Since the ions has 3 electrons in 3d , so the element must have 5 electrons in 3d. So the electroonic configuration of the element will be : 1s2 2s2 2p6 3s2 3p6 4s2 3d5.

The element will be : Manganese (Mn)

c) The element loses electrons to become smaller and gains electrons to become larger.

So the elements with larger radius will be :

Mg

Sn

Rb

O^2-

d) The inization energy increases from left to right and bottom to top in a peridoic table.

So the element with the largest 1st ionization energy is Cl.

Cl = Cl⁺ + 1e^-

e) Electron affinity also increases from left to right and bottom to top in a periodic table.

So the element with most negative electron affinity will be :

Cl + 1e^- = Cl^-